Chemical Reactions and Environmental Engineering

Okay, so it's time and we should get started. Okay, so in this class we're going to continue our discussion on the, regarding the basic concepts of the chemical systems. All right, so I'm listing the updated reading requirement or reading assignment, so it's going to be also in chapter two but a different range of pages. And you can also skip some contents because we're not going to use them in our homework or in our exam. I think I probably lost my screen, right? So let me try this. OK, let me try again. OK, now it should be good. All right, yeah. So you can skip a few pages regarding the onXtrain because that's a.. basically over the scope of this class. So if we do a quick recap of our last class, we mainly reviewed some basic properties of the elements or of molecules. So for example, we talked about what is the atomic number, what is the mass number, or what is the atomic weight. So we mentioned the atomic weight is the situation It's a situation where we consider the isotopes because there are different distributions of the isotopes so the averaged weight is going to have some digits after or some have some digits after the integer number there. So we also talk about what are the ionic and covalent bonds, right? It's mainly because of the different electron negativity of the elements. So we show the simple method to calculate or to determine what bonds they should belong to. We can just calculate the difference of the electron negativity of the elements. And then in that way we can calculate. So basically if the difference is larger than 1.5, or larger or equal to 1.5, then the bond is ionic. And if the difference is below 1.4 or equal to 1.4, the bond is covalent bond. So we also generally introduced the typical types of the chemical reactions. So for environmental engineering, we're mainly going to learn about these reactions, right? So there are precipitation dissolution reaction, the acid-base reaction, or the oxidation reduction reaction. So for the precipitation dissolution reaction, we mentioned that a major application is the softening of the water, because the water has a lot of calcium ions and magnesium ions. So before we send these water into the households, we first basically form solids or form the sediments so that they can get removed in the water, right? And then we're basically purifying the drinking water and send it to different house, right? And there are also acid-base reactions in water treatment because some of the water streams may be too acidic, right? Then because of that, we need to add in some base to the water to basically adjust the pH. So we're going to talk about the pH or how to calculate that in this class as well. So there are also reactions that's categorized as oxidation reduction reaction. We mentioned that a major application is to remove the organics in the water because there are quite many organics being dumped into the wastewater. For example, the leaves or the biomass, right? And then from the households industry, there are also a lot of organics being introduced or discharged into the water. So we have to basically oxidize them before we send them into the water bodies. And also because these organics are generally quite toxic to humans. So in the last class we also mentioned a little bit about balancing the equations. We said that it has a lot of meanings for engineering systems. Right. So if we go back to the example of the water softening. right? So here I'm showing the reaction, basically the precipitation dissolution reaction here, right? So basically we're saying that the major reason that's causing the hardness or the scale on the faucet is because of these calcium bicarbonates, right? It's mainly because of the calcium ions. So as I said, it's an interesting approach. that people use to solve this problem. So actually people introduce more calcium ions into the water to reduce the calcium ion concentration. But actually what's taking place is the reaction between the bicarbonate ion and the hydroxide ion. So what happens is this bicarbonate ion can react with the hydroxide ion to form water. and carbonate ion. So these two ions are going to react at a one-to-one ratio. So whenever you have one number of the bicarbonate ion, it's going to react with one number of the OH, or the hydroxide ion. And you're going to form water, one number of water, and one number of the carbonate ion. So because of that, we know that these two should always have the molar ratio. or the number ratio of one to one, right? So if you look back on this equation here, since this compound have two number of the carbonate ion and this compound have two number of the OH ion, so these two should react at a one to one ratio. So here it should be one and here it should be one, okay? And further, if you count on the right hand side, so you see that since we have two of these ions here, is going to form two of the water and two number of the carbonate ion. So because of that, for this reaction, it's going to be forming two numbers of water and two numbers of the calcium carbonate. So this is how we balance this equation. And this also has some meaning for the water treatment, mainly because we know that if we introduce this amount of the calcium hydroxide, we're going to form two numbers of the calcium carbonate as the sediments, right? Because of that, we need to adjust the engineering system so that it can accommodate this number of sediments so it doesn't clog the entire system, right? So this is an application of basically how, why we need to do the equation balancing or why we need to care about the stoichiometry, okay? So similarly you can look at this equation here. So actually we can also use this reaction to remove the hardness or remove the calcium ion. So what happens is the calcium bicarbonate can also react with the sulfuric acid. Okay so you see it's also going to form a solid. So basically we can also use this reaction to reduce or to remove the calcium ion from the water. and the result is that it's going to form carbonic acid. So to balance this equation, we also need to write out the ionic equation. So again, this ionic reaction involves the bicarbonate ion again. So the bicarbonate ion is actually a quite interesting species of ion. So it can react both with the base and with the acid. So you know that the acid, the major The major component of the acid is these proton ions, right, or its hydrogen ions. So what happens is that they too are going to react as a one-to-one ratio and it's going to form carbonic acid. And further the carbonic acid can get dissociated into water and carbon dioxide, okay, so if the water is becoming more and more acidic, okay. So basically from this ionic reaction we know that the bicarbonate ion is going to react with the hydrogen ions or the proton at a one-to-one ratio, right? So if you look back on this equation we have two numbers of the bicarbonate ions, two numbers of the hydrogen ions, right? So they too should react at a one-to-one ratio. So because of that we're going to form one number of the calcium sulfate. right? So this is going to form a sediment and then at the same time we're going to form two number of the carbonic acid. Okay, so actually as you can see for this reaction we can also reduce the hardness. We can also basically remove these carbonic ions. We can also remove these calcium ions, right? So why don't we use that in water treatment? So the major reason is because the sulfuric acid is too expensive. Okay, so for this reaction we're adding in calcium hydroxide. So as we have introduced, The calcium hydroxide, H2, they can be formed by calcium oxide dissolving into water, right? And further, the calcium oxide is just coming from the baking of the limestone. So we can go to mines, right? It's going to dig a lot of rocks. So most of them are limestone. And then by just heating up these limestone to, let's say, 1000 Celsius, we're going to form this calcium oxide, which means that it's going to be very cheap in producing these raw materials, right? Because it has a low price, we can directly use them in water treatment, but for the sulfuric acid, you need probably some complex method to form these compounds, right? So that's why we use this method for basically softening the water, right? So we also showed how to balance these equations. So basically for these two equations, we need to look at the ionic reactions. So the stoichiometry is determined by the stoichiometry for these ionic reactions. So here comes a simple one. So what about the iron getting oxidized into iron oxide? So we know that there are three number of oxygen elements on the right hand side, right? So the number in here should be 3 over 2, over oxygen, right? And then we have two number of iron, so the number in front of the iron should be 2. Okay, so if you want to rewrite this, then it's going to be 4 iron reacting with 3 oxygen to form 2 iron oxide. So this is quite straightforward. We can directly count the number of atoms on the left-hand side and right-hand side to solve this balancing. So what about this reaction here? Barium chloride reacting with sodium sulfate. So this is quite similar to the ionic reactions that we talked about in the beginning because we know that the barium ions can react with the sulfate ions. and to form barium sulfate, which is a solid. So because these two ions are reacting at a one-to-one ratio, so these two compounds are going to react as a one-to-one ratio as well. We're going to form one number of the barium sulfate, and then by counting the sodium and chloride, we know that it's going to form two number of sodium chloride. So this is how we can balance this reaction. So you can see that Actually, I would say most of the equation balancing or the reaction balancing is quite straightforward, as long as you know how the ions are reacting with each other. So what about this one? I guess that this reaction is a little bit challenging for us. So what happens is nitrogen dioxide is reacting with water to form nitric acid and also nitrogen monoxide. Okay, so why is it important? We can use this to control the nitrogen dioxide, which is a polluted gas species. So this reaction takes in the gas phase. Okay, so this is actually the same reaction that's happening in the atmosphere. So I always say something about the acid rain, right? So if we emit the nitrogen dioxide into the atmosphere, they can react with water vapor and further form these nitric acid. So it's forming the acid ring. Okay, so how to balance this reaction here? Okay, so you see that it is a little bit complex because for a lot of the elements, let's say there are just three elements here, hydrogen, oxygen, and nitrogen. Okay, so for Oxygen and nitrogen, they're showing up in almost all of the compounds, right? So for example, nitrogen is showing up here, here, and here. Oxygen is showing up in all of these compounds, okay? So we cannot assume that these two are reacting at a one-to-one ratio because they can actually react in any ratio, right? It can be... Let's say if we assume this to be one, this can be one, two, three, four, anything. So now under this situation, we need to look for the unique elements. So the unique element is the element that just shows up once on the left-hand side and right-hand side. So you can see that we mentioned oxygen and nitrogen show up many times. So the unique element in this reaction is hydrogen. Okay, so you see that the hydrogen just show up in the water on the left hand side and then in the nitric acid on the right hand side. So now let's assume that the number in front of the water is one and then because it just shows up once, so the number in front of the Nitric acid is 2. So now we don't know these two numbers. We can assume that to be then to be x and y here and we can basically set up a equation. So we have two unknowns and two equations each for the oxygen and nitrogen. So for oxygen to set up this balance it's going to be 2 multiplied by x plus 1. That's for the the water and on the right hand side that's going to be 2 multiplied by 3 plus y. So for nitrogen that's going to be 2 multiplied by x equal to 2 plus y. So if we just simplify these equations a little bit. So that's 2x equal to 5 plus y, and 2x equal to, let's see, okay, so I think I made a mistake here. So for nitrogen, okay, so for nitrogen here, it should just be x. Okay so on the left hand side for nitrogen, that's a single single nitrogen atom right, so that should be x equal to 2 plus y. So if we just use this equation to minus this second equation we have x equal to 3 right. and then if x equal to 3 then y is equal to 1 if we plug in this number here. So basically by setting up these equations we can further solve what is the number in front of these chemical compounds. So now we know the x and y are going to be 3 in here and 1 in here. So this is how we approach a little bit more complex chemical reactions. So now if we summarize how to balance the equation, okay, so for the simple ones we should rely on the ionic reactions. So based on the stoichiometry of these ions, right, we can calculate what is the stoichiometry for the compounds, or we can directly count. kinds of elements, so similar to this one. All right, so for a little bit complex ones, then we need to first find the unique element, and then we should set up equations, right, or to solve equations. Okay, so this is how we can balance the equation. As we said, they have very important engineering applications. So later in this class, you can find out basically how we use the balanced equations to calculate or to design these water treatment or air pollution control devices. So that is something quite simple. So I'm pretty sure that all of you guys know what a MO means. So a MO, means the number of stuff. Okay, so let's say if we mention a mole of carbon, it means 6.02 multiplied by 10 to the 23rd of the carbon atoms. So that turns out to have a mass of 12.011 grams. Okay, so we mentioned why this number has quite a lot of digits after it, after the integer, right? It's because of the isotopes, right? So this is associated with the atomic weight. So last class we mentioned, we also mentioned that there's a similar definition for the molecules, right? So the molecular weight. Molecular weight means that we just need to sum up all the atomic weight. So we showed that for the calcium carbonate, so it's a value of around 1.100. 0.08, right? So based on the molecular weight, people also define the molar weight. So basically, the molar weight means the weight per mole of stuff, okay? So it has the unit of gram per mole, right? So for example, for carbon dioxide, it has the molar weight of 44.01 gram per mole. And similarly for the calcium carbonate, it has the molar weight of 100.08 gram per mole. Okay, so this is a molar weight. So in water treatment and wastewater treatment, there are also a term called the molarity. Okay, so the molarity basically means the concentration of some species in water. So it means the number of moles per liter. So generally it has the unit of mole per liter. And then more specifically, when people use the term molarity, then the unit is going to be 1m. So let's say, for example, we have a bottle of water. So the bottle of water has some sodium chloride being dissolved. If we say the molarity of the sodium chloride, let's say. molarity of sodium chloride is 1m. It simply means that the concentration of the sodium chloride in the water is one mole per liter. So this is what 1m means. But for a lot of the water systems, we're not dealing with a very high concentration of these compounds. So that's why people also design the unit of 1 milliamp. Okay, so that's just one thousandth of this concentration here. So what that means is one millimole per liter, or you can think of it as 0.001 mole per liter. Okay, so this is what the molarity means. So in the future if you see the concentration of certain species, let's say it's 0.2 m. So that simply means the concentration is 0.2 mol per liter. Okay so as you can see from this slide, I use the mol in the full term or in the full name. And sometimes I use the mol in let's say just the three letters. So in this class I'll be using this full name and this type of mol interchangeably. So basically they mean the same thing. I know that in the textbook probably when they describe the mole they're going to use the entire name but when they're shown in the units we're just going to show the first three letters here. So this is about the the moles, the concentration, and the molarity. So now here is something a little bit complex. So this is the chemical equilibrium. I hope that most of you know what this means but if you forgot So basically chemical equilibrium means the final state or determines the final state of a reaction. So, for example, if we go back to these reactions that we talked about earlier. So although we write out these reactions with the symbol of the equate. or the equation symbol. The equation symbol doesn't mean that this reaction will happen completely. It doesn't mean that, let's say, if we have one mole of calcium, let's say one mole of the calcium bicarbonate and one mole of the sulfuric acid, it doesn't mean that these two will react completely and get into zero. Okay, so for all the reactions, there's basically there's none of the reaction that can happen completely. So there is always a final state that both the reactants and the products coexist. So we're talking about the final state, basically taking a very long time. It can be months or days, months, years or thousands of years. Okay, so we're talking about that final state. All right. So the chemical equilibrium determines what the final state looks like. Okay, so for example if you have this reaction here, so A number of A reacting with B number of B to form C number of C and D number of D. Okay, so what people, the way people define this final state is by creating a number that's called the equilibrium constant. So the equilibrium constant is calculated by the concentration of C to the power of C, and then multiply the concentration of D to the power of D, divide by the concentration of A to the power of A, multiply the concentration of B to the power of B. So basically, it's a product to certain power for the product of the reaction. divide by the product regarding the reactants of this reaction. So the equilibrium constant, as it says, is always a constant. It may be dependent on the temperature, actually it is dependent on the temperature, but at certain temperature this is always a constant. So one thing we need to notice is that all of these must be shown in the unit of M. So recall that for molarity we have this unit here. m, right? It basically means mole per liter. Okay. So just to do a quick practice. So let's say that if we have a reaction that has all of these coefficients to be one. So basically it's a plus b going to c plus d. Okay. So for this reaction, if we know that, let's say under 25 Celsius, the k is equal to one. So further we have the concentration of A and B and the concentration of C. So what is the concentration of D? So let's do a quick pooling on this problem here. I think it should be quite straightforward. So let's give it 10 more seconds. Okay. All right, we'll stop here. So I will share the results. So most of you guys are correct. So it should be 2M. Okay. So if we set up... this equilibrium constant, we know that k is equal to c to the power of one multiplied by d to the power of one divided by a to the power of one multiplied by b to the power of one, mainly because the lower case a, b, and c, and d are all equal to one, right? So because of that, this term is equal to one, okay? So further we know the concentration of a and b. these are 1, 1, so c is equal to 0.5, so because of that d should be equal to 2m. Okay, so basically what this means is that for this reaction, although we're using this equate sign here, it doesn't mean that this reaction will happen completely. Actually none of the reactions are going to happen. happen completely, right? So there's always some reactants and some products coexisting in the system, okay? So I also want to put a note here. So for this equilibrium constant here, that's defined here, for solids or pure liquid or gas, so if they happen to show up in this reaction here, then their term basically it's the the concentration to the power of x should be always equal to one. So we're referring to basically the formation of solids, right? So if you recall, for the water softening, right, so there are some solids being formed in that reaction there. So if that's the case, then we don't need to consider the concentration of the solid, or we don't need to consider the concentration of the pure liquid. Okay, so we're going to give some examples in the next few slides. So what is the application of the chemical equilibrium? So there are three major applications in environmental engineering. So one is a solubility product, one is the acid-base equilibrium or equilibria, and one is the gas-liquid equilibrium. So first about the solubility product. So the solubility product means basically any solid are going to dissolve some part of the solid into the water. So you may wonder, well, the rocks are rocks, right? There's no way that they can dissolve in water. So why do we care about this dissolution here? So actually, when we talk about the water hardness, right, the white scale that's forming on the faucet, as we mentioned, it's because of the calcium ions. Actually, these calcium ions are coming from these rocks. So there are always some part of the solid that can dissolve into the water. Although this dissolution is just happening for a little bit, it's not completely dissolved. So there's a small fraction of the solid that can get dissolved into the water. So, for example, if we're talking about the dissolution of the calcium carbonate, so we know that this is... basically the limestone, if you dissolve or if you just directly emerge this limestone into water, there are some dissolution. So how to determine the concentration of the ions? So we're going to use this chemical equilibrium, or more specifically, we're going to use the concept of the solubility product. So if you recall, for any reaction, For any reaction, the equilibrium constant is written in this way. So similarly, we can write out the equilibrium constant for this reaction. So that's going to be the concentration of the calcium ions multiplied by the concentration of the carbonate ions divided by the concentration of the calcium carbonate. Okay, so all of them are to the power of one. So further, If you follow the note that we mentioned here, so for solids, pure liquid and gas, their term should be equal to one. So basically the concentration of this solid to the power of the coefficient should be equal to one. So if we apply that into this dissolution here, then this term in the denominator should be one, right? So basically for this reaction here or the dissolution of the solid, then the equilibrium constant is just a product of the products that's coming out of this reaction here, right? So more specifically, we're going to add a subscript sp here. to show that it means solubility product. So basically for this reaction here, people have measured or determined what is the equilibrium constant or the solubility product or the Ksp. So it turns out that for this reaction, the Ksp has a value of around 10 to the negative 8.34. Okay so this is a very small value, right? So what that means is that So the product of these two concentrations should be equal to this value. right? And then you can find out how small this is. So let's assume further assume that these two concentrations are equal to each other because they're dissolving at a one-to-one ratio, right? So what you have is the calcium ion to the power of two is equal to 10 to the negative 8.34. So further the calcium ion's concentration is equal to the square root of 10 to negative 8.34. Okay, so this turns out to be around, let's say, let's say around 10 to the negative four. Okay, so this is a very small, very low concentration. So that's why these solids, they're not dissolving a large fraction of them. It's not like restoring a rock into the water and suddenly it's like salt that dissolved completely. This is also the reason why we have these trace amounts of the calcium ions in our water. Similarly, we can have the dissolution reaction of the aluminum hydroxide. We know that aluminum hydroxide has the form of Al and this is aluminum and has three hydroxide ions. Then they dissolve following this reaction. Similarly, we can write out the equilibrium constant, or more specifically the Ksp, the solubility product. Again, we apply the concept that since this is a pure solid, so the term in the denominator should be equal to one. So that's why the Ksp is equal to aluminum concentration multiplied by the OH concentration to the power of three. So it turns out that this solubility product is further very low value. This is extremely low. 10 to the negative 32.9. So basically by using this concept here we can also calculate what is the concentration of the aluminum and what is the concentration of the hydroxide. So here I'm just going to give a very simple application of this concept. So let's say we're adding barium sulfate, which is a solid, into one liter of water. until it saturates. What this means is that until we cannot dissolve any more solid. So what is the concentration of the barium ion? So basically it's asking for this concentration here. We know that the barium sulfate dissolves into barium ion and the sulfate ion at a one-to-one ratio and we know the Ksp is equal to 1.08. multiplied by 10 to the negative 10. Okay, so the way we solve this is similar to these ones. We can just write out the Ksp, right, so that's just barium ion concentration multiplied by not carbonate ion, it's sulfate ions, right, and this value should be 1.08 multiplied by 10 to the negative 10. So further, since we're directly dissolving this salt in a pure water, right? So we know that according to the stoichiometry, every one mole or every one number of the barium ion is going to correspond with one number of the sulfate ion because there are no sulfates pre-existing in the water, right? So that's why the barium ion concentration should be equal to the sulfate ion concentration. So because of that, we have the barium ion concentration to the power of 2 equal to 1.08 multiplied by 10 to the negative 10. So further we can calculate, we can take the square root and then calculate the barium ion concentrations. So the things will be a little bit different. Let's say if we add barium sulfate into one liter of the sodium sulfate solution, let's say if we assume that the sodium sulfate concentration is 1m, okay, so what's going to happen? So under this situation, we know that there are pre-existing sulfate ions in the water, right, so we cannot apply this relationship anymore. So what we should do is basically, we know that since there is already one mole per liter of the sulfate, in the water right so we know that the basically the barium are in concentration should be equal to the sulfate ion concentration minus 1m, right? So this 1m is coming from the sodium sulfate that's in the solution. So under this situation, if it's not pure water, if it's 1m of the sodium sulfate, then this thing should become barium sulfate ions multiplied by the barium ion sulfate, barium ion. bearing ion plus one equal to 1.08 multiplied by 10 to the negative 10. Okay, so instead of using this relationship, we're going to use this one because there is pre-existing one m of sulfate in the solution, right? And then you can further solve this. If you have this relationship, then you can find that the barium ion concentration will be further lower, actually significantly reduced. So if we just look at the water system, then we can just use this method. So it turns out that the barium ion concentration is 1.04 multiplied by 10 to the negative 5, which is a very low concentration. So this is an example for the... for the solubility product. Okay. So, but then you may wonder, where do we get all of these solubility product values? So in the textbook, there is a table that's showing all of these common solids, for example, aluminum hydroxide, calcium carbonate, ferric oxide, right? So all of these came with this value of PKS. Okay. So the PKS is a way to show the solubility product. So you can think of this P here as an operator, as a type of calculation. So we define this calculation symbol P as pKs equal to the negative log 10 of Ks. So basically the Ks or the Ksp is a solubility product and the pKsp or the pKs it's just negative log 10 of this value, right? So for example, if we have a KSP that's equal to 10 to the negative 10, then the pKSP is just negative log 10 of this thing here, okay? So you're going to get a value of 10, right? So this is how we define this symbol P here. So what this means is that the higher or the larger the pKs, the more difficult to dissolve this solid. So you can see the aluminum hydroxide has a very large value, the ferric hydroxide has a very large value, but compared to that, the calcium carbonate has a relatively lower value, which means that it's a little bit easier to dissolve this solid. So we're going to come back to this operator P here again later on. Okay, so as I said, so one of the equilibrium is the precipitation dissolution equilibrium, where we introduce the solubility product. So another equilibrium is the acid-base equilibrium. So what happens is that for any acid, they can dissolve into a hydrogen ion and some anion. So hydrogen ion is the cation and anion is the corresponding ion. For example, the hydrochloric acid is going to get dissolved into the hydrogen ion or the proton ion and the chloride ion. So this A here is just chloride ion. So for any of these dissolution of the acid, then we can also define an equilibrium constant Ka. So following the definition of the equilibrium constant, then it should be the concentration of the proton ion to the power one, the concentration of the anion to the power one, and divide by the concentration of the acid. So because of that you can have very strong acids or very weak acids. So that's shown in this equilibrium constant here. So if you recall, in here, we mentioned that the larger the PKS, then the more difficult it is. to dissolve these solids, right? So the same thing happens to these acids or these base. So we can also define the calculation pKa. So what this means is the higher the value, the more difficult to dissolve. So the more the difficult to dissolve, then the weaker the acid. So here we have a table that's showing the pKa values of common acids. You see that the strong acids, they have such a low value of the pKa that some of them even go to negative three. Okay, so a value of negative three means that the basically the Ka value is equal to 10 to the third. Right, so if you recall the pKa is equal to negative log 10. of Ka, right? So if you have a Ka value of 10 to the third, then this is going to be negative 3, okay? So if you have a very large Ka value, it just means that the concentration of the proton ions multiplied by the concentration of the anion is much much higher compared to the remaining acid concentration. So that's why the strong acids will have a very small value of the pKa. while the weak acids are going to have larger values of the pKa. So here a very important system is the carbonic acid because in the atmosphere there are carbon dioxide existing in the system. So the carbon dioxide, once they dissolve into the water, they can form carbonic acid and further get dissolved into the proton and the bicarbonate ion. And further the bicarbonate ion can dissolve into proton ion and the carbonate ions. So for these two reactions, as you can see, they have a pretty large pKa value, which means that they are weak acids, so the concentration of these proton ions are going to be lower in the system. So then a very easy application is the water. So what about water system? We know that water can be written as H connected to an O, oxygen, and further hydrogen, right? We know it's composed of this structure, okay? So water, you can treat it as a very weak acid, okay? So basically the water or the hydrogen here, they can get dissolved into the water, right, to form the protons. So for this reaction here, we can also So basically water is getting dissolved into proton and the hydroxide ion. So for this reaction, we can also write out its equilibrium constant. So that's going to be hydrogen multiplied by OH divided by the concentration of water. So if you recall the note that's earlier, so for pure liquid, this term should be regarded as one. So that's why. we can define a equilibrium constant for water, Kw. So W stands for water here. So that turns out to be 10 to the negative 14. So people have done these measurements. So now if you imagine we have a cup of water, let's say it's just water inside, it doesn't have any ions inside. Then let's see what's the concentration of the hydrogen and OH. So what happens is they are going to dissolve at a molar ratio one to one, right? So basically the concentration of hydrogen ions is going to be 10 to negative seven because basically this is going to be squared because these two are the same, right? And then you can solve that. So further let's again use the definition of the p. So let's say we have the pKa pKs, right? We can define the pH which is negative log 10 of the hydrogen ion concentration. So if you plug it in, you can get pH is equal to seven. So this is how the pH of water come out, right? So if you have a neutral water, then they should have the pH of seven. If you have acid, basically a higher concentration of the hydrogen ions, then you're going to have a lower pH. Reversely, you're going to have a higher pH, okay? So in this lab session, you're going to learn about the pH. And there's also a term called the alkalinity. So the alkalinity means how much hydrogen ions, it can react within the water system. So later on in this lab session, you're going to learn more about the alkalinity. And we're also going to cover that in our next class. Okay. So let's pick up the topic next Monday. All right. I'll see you on next Monday then. Professor, for the lab, is it in person or online? It is in person. Okay, awesome. Yeah, it is in-person lab. I'll put a notice on Canvas again. Okay, just double checking. All right, thank you.

And you can also skip some contents because we're not going to use them in our homework or in our exam. I think I probably lost my screen, right? So let me try this.

OK, let me try again. OK, now it should be good. All right, yeah. So you can skip a few pages regarding the onXtrain because that's a.. basically over the scope of this class.

So if we do a quick recap of our last class, we mainly reviewed some basic properties of the elements or of molecules. So for example, we talked about what is the atomic number, what is the mass number, or what is the atomic weight. So we mentioned the atomic weight is the situation It's a situation where we consider the isotopes because there are different distributions of the isotopes so the averaged weight is going to have some digits after or some have some digits after the integer number there. So we also talk about what are the ionic and covalent bonds, right? It's mainly because of the different electron negativity of the elements.

So we show the simple method to calculate or to determine what bonds they should belong to. We can just calculate the difference of the electron negativity of the elements. And then in that way we can calculate. So basically if the difference is larger than 1.5, or larger or equal to 1.5, then the bond is ionic.

And if the difference is below 1.4 or equal to 1.4, the bond is covalent bond. So we also generally introduced the typical types of the chemical reactions. So for environmental engineering, we're mainly going to learn about these reactions, right?

So there are precipitation dissolution reaction, the acid-base reaction, or the oxidation reduction reaction. So for the precipitation dissolution reaction, we mentioned that a major application is the softening of the water, because the water has a lot of calcium ions and magnesium ions. So before we send these water into the households, we first basically form solids or form the sediments so that they can get removed in the water, right?

And then we're basically purifying the drinking water and send it to different house, right? And there are also acid-base reactions in water treatment because some of the water streams may be too acidic, right? Then because of that, we need to add in some base to the water to basically adjust the pH. So we're going to talk about the pH or how to calculate that in this class as well. So there are also reactions that's categorized as oxidation reduction reaction.

We mentioned that a major application is to remove the organics in the water because there are quite many organics being dumped into the wastewater. For example, the leaves or the biomass, right? And then from the households industry, there are also a lot of organics being introduced or discharged into the water. So we have to basically oxidize them before we send them into the water bodies.

And also because these organics are generally quite toxic to humans. So in the last class we also mentioned a little bit about balancing the equations. We said that it has a lot of meanings for engineering systems.

Right. So if we go back to the example of the water softening. right?

So here I'm showing the reaction, basically the precipitation dissolution reaction here, right? So basically we're saying that the major reason that's causing the hardness or the scale on the faucet is because of these calcium bicarbonates, right? It's mainly because of the calcium ions.

So as I said, it's an interesting approach. that people use to solve this problem. So actually people introduce more calcium ions into the water to reduce the calcium ion concentration. But actually what's taking place is the reaction between the bicarbonate ion and the hydroxide ion. So what happens is this bicarbonate ion can react with the hydroxide ion to form water.

and carbonate ion. So these two ions are going to react at a one-to-one ratio. So whenever you have one number of the bicarbonate ion, it's going to react with one number of the OH, or the hydroxide ion. And you're going to form water, one number of water, and one number of the carbonate ion.

So because of that, we know that these two should always have the molar ratio. or the number ratio of one to one, right? So if you look back on this equation here, since this compound have two number of the carbonate ion and this compound have two number of the OH ion, so these two should react at a one to one ratio. So here it should be one and here it should be one, okay? And further, if you count on the right hand side, so you see that since we have two of these ions here, is going to form two of the water and two number of the carbonate ion.

So because of that, for this reaction, it's going to be forming two numbers of water and two numbers of the calcium carbonate. So this is how we balance this equation. And this also has some meaning for the water treatment, mainly because we know that if we introduce this amount of the calcium hydroxide, we're going to form two numbers of the calcium carbonate as the sediments, right?

Because of that, we need to adjust the engineering system so that it can accommodate this number of sediments so it doesn't clog the entire system, right? So this is an application of basically how, why we need to do the equation balancing or why we need to care about the stoichiometry, okay? So similarly you can look at this equation here. So actually we can also use this reaction to remove the hardness or remove the calcium ion. So what happens is the calcium bicarbonate can also react with the sulfuric acid.

Okay so you see it's also going to form a solid. So basically we can also use this reaction to reduce or to remove the calcium ion from the water. and the result is that it's going to form carbonic acid. So to balance this equation, we also need to write out the ionic equation. So again, this ionic reaction involves the bicarbonate ion again.

So the bicarbonate ion is actually a quite interesting species of ion. So it can react both with the base and with the acid. So you know that the acid, the major The major component of the acid is these proton ions, right, or its hydrogen ions. So what happens is that they too are going to react as a one-to-one ratio and it's going to form carbonic acid.

And further the carbonic acid can get dissociated into water and carbon dioxide, okay, so if the water is becoming more and more acidic, okay. So basically from this ionic reaction we know that the bicarbonate ion is going to react with the hydrogen ions or the proton at a one-to-one ratio, right? So if you look back on this equation we have two numbers of the bicarbonate ions, two numbers of the hydrogen ions, right? So they too should react at a one-to-one ratio.

So because of that we're going to form one number of the calcium sulfate. right? So this is going to form a sediment and then at the same time we're going to form two number of the carbonic acid. Okay, so actually as you can see for this reaction we can also reduce the hardness. We can also basically remove these carbonic ions.

We can also remove these calcium ions, right? So why don't we use that in water treatment? So the major reason is because the sulfuric acid is too expensive. Okay, so for this reaction we're adding in calcium hydroxide. So as we have introduced, The calcium hydroxide, H2, they can be formed by calcium oxide dissolving into water, right?

And further, the calcium oxide is just coming from the baking of the limestone. So we can go to mines, right? It's going to dig a lot of rocks. So most of them are limestone. And then by just heating up these limestone to, let's say, 1000 Celsius, we're going to form this calcium oxide, which means that it's going to be very cheap in producing these raw materials, right?

Because it has a low price, we can directly use them in water treatment, but for the sulfuric acid, you need probably some complex method to form these compounds, right? So that's why we use this method for basically softening the water, right? So we also showed how to balance these equations. So basically for these two equations, we need to look at the ionic reactions.

So the stoichiometry is determined by the stoichiometry for these ionic reactions. So here comes a simple one. So what about the iron getting oxidized into iron oxide? So we know that there are three number of oxygen elements on the right hand side, right?

So the number in here should be 3 over 2, over oxygen, right? And then we have two number of iron, so the number in front of the iron should be 2. Okay, so if you want to rewrite this, then it's going to be 4 iron reacting with 3 oxygen to form 2 iron oxide. So this is quite straightforward.

We can directly count the number of atoms on the left-hand side and right-hand side to solve this balancing. So what about this reaction here? Barium chloride reacting with sodium sulfate. So this is quite similar to the ionic reactions that we talked about in the beginning because we know that the barium ions can react with the sulfate ions.

and to form barium sulfate, which is a solid. So because these two ions are reacting at a one-to-one ratio, so these two compounds are going to react as a one-to-one ratio as well. We're going to form one number of the barium sulfate, and then by counting the sodium and chloride, we know that it's going to form two number of sodium chloride.

So this is how we can balance this reaction. So you can see that Actually, I would say most of the equation balancing or the reaction balancing is quite straightforward, as long as you know how the ions are reacting with each other. So what about this one?

I guess that this reaction is a little bit challenging for us. So what happens is nitrogen dioxide is reacting with water to form nitric acid and also nitrogen monoxide. Okay, so why is it important?

We can use this to control the nitrogen dioxide, which is a polluted gas species. So this reaction takes in the gas phase. Okay, so this is actually the same reaction that's happening in the atmosphere.

So I always say something about the acid rain, right? So if we emit the nitrogen dioxide into the atmosphere, they can react with water vapor and further form these nitric acid. So it's forming the acid ring. Okay, so how to balance this reaction here? Okay, so you see that it is a little bit complex because for a lot of the elements, let's say there are just three elements here, hydrogen, oxygen, and nitrogen.

Okay, so for Oxygen and nitrogen, they're showing up in almost all of the compounds, right? So for example, nitrogen is showing up here, here, and here. Oxygen is showing up in all of these compounds, okay? So we cannot assume that these two are reacting at a one-to-one ratio because they can actually react in any ratio, right?

It can be... Let's say if we assume this to be one, this can be one, two, three, four, anything. So now under this situation, we need to look for the unique elements.

So the unique element is the element that just shows up once on the left-hand side and right-hand side. So you can see that we mentioned oxygen and nitrogen show up many times. So the unique element in this reaction is hydrogen.

Okay, so you see that the hydrogen just show up in the water on the left hand side and then in the nitric acid on the right hand side. So now let's assume that the number in front of the water is one and then because it just shows up once, so the number in front of the Nitric acid is 2. So now we don't know these two numbers. We can assume that to be then to be x and y here and we can basically set up a equation.

So we have two unknowns and two equations each for the oxygen and nitrogen. So for oxygen to set up this balance it's going to be 2 multiplied by x plus 1. That's for the the water and on the right hand side that's going to be 2 multiplied by 3 plus y. So for nitrogen that's going to be 2 multiplied by x equal to 2 plus y.

So if we just simplify these equations a little bit. So that's 2x equal to 5 plus y, and 2x equal to, let's see, okay, so I think I made a mistake here. So for nitrogen, okay, so for nitrogen here, it should just be x. Okay so on the left hand side for nitrogen, that's a single single nitrogen atom right, so that should be x equal to 2 plus y. So if we just use this equation to minus this second equation we have x equal to 3 right.

and then if x equal to 3 then y is equal to 1 if we plug in this number here. So basically by setting up these equations we can further solve what is the number in front of these chemical compounds. So now we know the x and y are going to be 3 in here and 1 in here. So this is how we approach a little bit more complex chemical reactions.

So now if we summarize how to balance the equation, okay, so for the simple ones we should rely on the ionic reactions. So based on the stoichiometry of these ions, right, we can calculate what is the stoichiometry for the compounds, or we can directly count. kinds of elements, so similar to this one.

All right, so for a little bit complex ones, then we need to first find the unique element, and then we should set up equations, right, or to solve equations. Okay, so this is how we can balance the equation. As we said, they have very important engineering applications.

So later in this class, you can find out basically how we use the balanced equations to calculate or to design these water treatment or air pollution control devices. So that is something quite simple. So I'm pretty sure that all of you guys know what a MO means. So a MO, means the number of stuff.

Okay, so let's say if we mention a mole of carbon, it means 6.02 multiplied by 10 to the 23rd of the carbon atoms. So that turns out to have a mass of 12.011 grams. Okay, so we mentioned why this number has quite a lot of digits after it, after the integer, right? It's because of the isotopes, right? So this is associated with the atomic weight.

So last class we mentioned, we also mentioned that there's a similar definition for the molecules, right? So the molecular weight. Molecular weight means that we just need to sum up all the atomic weight.

So we showed that for the calcium carbonate, so it's a value of around 1.100. 0.08, right? So based on the molecular weight, people also define the molar weight. So basically, the molar weight means the weight per mole of stuff, okay? So it has the unit of gram per mole, right?

So for example, for carbon dioxide, it has the molar weight of 44.01 gram per mole. And similarly for the calcium carbonate, it has the molar weight of 100.08 gram per mole. Okay, so this is a molar weight.

So in water treatment and wastewater treatment, there are also a term called the molarity. Okay, so the molarity basically means the concentration of some species in water. So it means the number of moles per liter. So generally it has the unit of mole per liter. And then more specifically, when people use the term molarity, then the unit is going to be 1m.

So let's say, for example, we have a bottle of water. So the bottle of water has some sodium chloride being dissolved. If we say the molarity of the sodium chloride, let's say. molarity of sodium chloride is 1m. It simply means that the concentration of the sodium chloride in the water is one mole per liter.

So this is what 1m means. But for a lot of the water systems, we're not dealing with a very high concentration of these compounds. So that's why people also design the unit of 1 milliamp. Okay, so that's just one thousandth of this concentration here. So what that means is one millimole per liter, or you can think of it as 0.001 mole per liter.

Okay, so this is what the molarity means. So in the future if you see the concentration of certain species, let's say it's 0.2 m. So that simply means the concentration is 0.2 mol per liter.

Okay so as you can see from this slide, I use the mol in the full term or in the full name. And sometimes I use the mol in let's say just the three letters. So in this class I'll be using this full name and this type of mol interchangeably.

So basically they mean the same thing. I know that in the textbook probably when they describe the mole they're going to use the entire name but when they're shown in the units we're just going to show the first three letters here. So this is about the the moles, the concentration, and the molarity. So now here is something a little bit complex. So this is the chemical equilibrium.

I hope that most of you know what this means but if you forgot So basically chemical equilibrium means the final state or determines the final state of a reaction. So, for example, if we go back to these reactions that we talked about earlier. So although we write out these reactions with the symbol of the equate.

or the equation symbol. The equation symbol doesn't mean that this reaction will happen completely. It doesn't mean that, let's say, if we have one mole of calcium, let's say one mole of the calcium bicarbonate and one mole of the sulfuric acid, it doesn't mean that these two will react completely and get into zero. Okay, so for all the reactions, there's basically there's none of the reaction that can happen completely.

So there is always a final state that both the reactants and the products coexist. So we're talking about the final state, basically taking a very long time. It can be months or days, months, years or thousands of years. Okay, so we're talking about that final state. All right.

So the chemical equilibrium determines what the final state looks like. Okay, so for example if you have this reaction here, so A number of A reacting with B number of B to form C number of C and D number of D. Okay, so what people, the way people define this final state is by creating a number that's called the equilibrium constant. So the equilibrium constant is calculated by the concentration of C to the power of C, and then multiply the concentration of D to the power of D, divide by the concentration of A to the power of A, multiply the concentration of B to the power of B.

So basically, it's a product to certain power for the product of the reaction. divide by the product regarding the reactants of this reaction. So the equilibrium constant, as it says, is always a constant. It may be dependent on the temperature, actually it is dependent on the temperature, but at certain temperature this is always a constant.

So one thing we need to notice is that all of these must be shown in the unit of M. So recall that for molarity we have this unit here. m, right? It basically means mole per liter. Okay.

So just to do a quick practice. So let's say that if we have a reaction that has all of these coefficients to be one. So basically it's a plus b going to c plus d. Okay.

So for this reaction, if we know that, let's say under 25 Celsius, the k is equal to one. So further we have the concentration of A and B and the concentration of C. So what is the concentration of D? So let's do a quick pooling on this problem here.

I think it should be quite straightforward. So let's give it 10 more seconds. Okay. All right, we'll stop here.

So I will share the results. So most of you guys are correct. So it should be 2M. Okay.

So if we set up... this equilibrium constant, we know that k is equal to c to the power of one multiplied by d to the power of one divided by a to the power of one multiplied by b to the power of one, mainly because the lower case a, b, and c, and d are all equal to one, right? So because of that, this term is equal to one, okay?

So further we know the concentration of a and b. these are 1, 1, so c is equal to 0.5, so because of that d should be equal to 2m. Okay, so basically what this means is that for this reaction, although we're using this equate sign here, it doesn't mean that this reaction will happen completely.

Actually none of the reactions are going to happen. happen completely, right? So there's always some reactants and some products coexisting in the system, okay?

So I also want to put a note here. So for this equilibrium constant here, that's defined here, for solids or pure liquid or gas, so if they happen to show up in this reaction here, then their term basically it's the the concentration to the power of x should be always equal to one. So we're referring to basically the formation of solids, right?

So if you recall, for the water softening, right, so there are some solids being formed in that reaction there. So if that's the case, then we don't need to consider the concentration of the solid, or we don't need to consider the concentration of the pure liquid. Okay, so we're going to give some examples in the next few slides.

So what is the application of the chemical equilibrium? So there are three major applications in environmental engineering. So one is a solubility product, one is the acid-base equilibrium or equilibria, and one is the gas-liquid equilibrium.

So first about the solubility product. So the solubility product means basically any solid are going to dissolve some part of the solid into the water. So you may wonder, well, the rocks are rocks, right?

There's no way that they can dissolve in water. So why do we care about this dissolution here? So actually, when we talk about the water hardness, right, the white scale that's forming on the faucet, as we mentioned, it's because of the calcium ions. Actually, these calcium ions are coming from these rocks.

So there are always some part of the solid that can dissolve into the water. Although this dissolution is just happening for a little bit, it's not completely dissolved. So there's a small fraction of the solid that can get dissolved into the water.

So, for example, if we're talking about the dissolution of the calcium carbonate, so we know that this is... basically the limestone, if you dissolve or if you just directly emerge this limestone into water, there are some dissolution. So how to determine the concentration of the ions? So we're going to use this chemical equilibrium, or more specifically, we're going to use the concept of the solubility product.

So if you recall, for any reaction, For any reaction, the equilibrium constant is written in this way. So similarly, we can write out the equilibrium constant for this reaction. So that's going to be the concentration of the calcium ions multiplied by the concentration of the carbonate ions divided by the concentration of the calcium carbonate.

Okay, so all of them are to the power of one. So further, If you follow the note that we mentioned here, so for solids, pure liquid and gas, their term should be equal to one. So basically the concentration of this solid to the power of the coefficient should be equal to one. So if we apply that into this dissolution here, then this term in the denominator should be one, right?

So basically for this reaction here or the dissolution of the solid, then the equilibrium constant is just a product of the products that's coming out of this reaction here, right? So more specifically, we're going to add a subscript sp here. to show that it means solubility product.

So basically for this reaction here, people have measured or determined what is the equilibrium constant or the solubility product or the Ksp. So it turns out that for this reaction, the Ksp has a value of around 10 to the negative 8.34. Okay so this is a very small value, right? So what that means is that So the product of these two concentrations should be equal to this value. right?

And then you can find out how small this is. So let's assume further assume that these two concentrations are equal to each other because they're dissolving at a one-to-one ratio, right? So what you have is the calcium ion to the power of two is equal to 10 to the negative 8.34.

So further the calcium ion's concentration is equal to the square root of 10 to negative 8.34. Okay, so this turns out to be around, let's say, let's say around 10 to the negative four. Okay, so this is a very small, very low concentration. So that's why these solids, they're not dissolving a large fraction of them. It's not like restoring a rock into the water and suddenly it's like salt that dissolved completely.

This is also the reason why we have these trace amounts of the calcium ions in our water. Similarly, we can have the dissolution reaction of the aluminum hydroxide. We know that aluminum hydroxide has the form of Al and this is aluminum and has three hydroxide ions.

Then they dissolve following this reaction. Similarly, we can write out the equilibrium constant, or more specifically the Ksp, the solubility product. Again, we apply the concept that since this is a pure solid, so the term in the denominator should be equal to one. So that's why the Ksp is equal to aluminum concentration multiplied by the OH concentration to the power of three. So it turns out that this solubility product is further very low value.

This is extremely low. 10 to the negative 32.9. So basically by using this concept here we can also calculate what is the concentration of the aluminum and what is the concentration of the hydroxide. So here I'm just going to give a very simple application of this concept.

So let's say we're adding barium sulfate, which is a solid, into one liter of water. until it saturates. What this means is that until we cannot dissolve any more solid.

So what is the concentration of the barium ion? So basically it's asking for this concentration here. We know that the barium sulfate dissolves into barium ion and the sulfate ion at a one-to-one ratio and we know the Ksp is equal to 1.08.

multiplied by 10 to the negative 10. Okay, so the way we solve this is similar to these ones. We can just write out the Ksp, right, so that's just barium ion concentration multiplied by not carbonate ion, it's sulfate ions, right, and this value should be 1.08 multiplied by 10 to the negative 10. So further, since we're directly dissolving this salt in a pure water, right? So we know that according to the stoichiometry, every one mole or every one number of the barium ion is going to correspond with one number of the sulfate ion because there are no sulfates pre-existing in the water, right?

So that's why the barium ion concentration should be equal to the sulfate ion concentration. So because of that, we have the barium ion concentration to the power of 2 equal to 1.08 multiplied by 10 to the negative 10. So further we can calculate, we can take the square root and then calculate the barium ion concentrations. So the things will be a little bit different. Let's say if we add barium sulfate into one liter of the sodium sulfate solution, let's say if we assume that the sodium sulfate concentration is 1m, okay, so what's going to happen?

So under this situation, we know that there are pre-existing sulfate ions in the water, right, so we cannot apply this relationship anymore. So what we should do is basically, we know that since there is already one mole per liter of the sulfate, in the water right so we know that the basically the barium are in concentration should be equal to the sulfate ion concentration minus 1m, right? So this 1m is coming from the sodium sulfate that's in the solution. So under this situation, if it's not pure water, if it's 1m of the sodium sulfate, then this thing should become barium sulfate ions multiplied by the barium ion sulfate, barium ion. bearing ion plus one equal to 1.08 multiplied by 10 to the negative 10. Okay, so instead of using this relationship, we're going to use this one because there is pre-existing one m of sulfate in the solution, right?

And then you can further solve this. If you have this relationship, then you can find that the barium ion concentration will be further lower, actually significantly reduced. So if we just look at the water system, then we can just use this method. So it turns out that the barium ion concentration is 1.04 multiplied by 10 to the negative 5, which is a very low concentration. So this is an example for the...

for the solubility product. Okay. So, but then you may wonder, where do we get all of these solubility product values? So in the textbook, there is a table that's showing all of these common solids, for example, aluminum hydroxide, calcium carbonate, ferric oxide, right?

So all of these came with this value of PKS. Okay. So the PKS is a way to show the solubility product.

So you can think of this P here as an operator, as a type of calculation. So we define this calculation symbol P as pKs equal to the negative log 10 of Ks. So basically the Ks or the Ksp is a solubility product and the pKsp or the pKs it's just negative log 10 of this value, right? So for example, if we have a KSP that's equal to 10 to the negative 10, then the pKSP is just negative log 10 of this thing here, okay?

So you're going to get a value of 10, right? So this is how we define this symbol P here. So what this means is that the higher or the larger the pKs, the more difficult to dissolve this solid. So you can see the aluminum hydroxide has a very large value, the ferric hydroxide has a very large value, but compared to that, the calcium carbonate has a relatively lower value, which means that it's a little bit easier to dissolve this solid. So we're going to come back to this operator P here again later on.

Okay, so as I said, so one of the equilibrium is the precipitation dissolution equilibrium, where we introduce the solubility product. So another equilibrium is the acid-base equilibrium. So what happens is that for any acid, they can dissolve into a hydrogen ion and some anion.

So hydrogen ion is the cation and anion is the corresponding ion. For example, the hydrochloric acid is going to get dissolved into the hydrogen ion or the proton ion and the chloride ion. So this A here is just chloride ion. So for any of these dissolution of the acid, then we can also define an equilibrium constant Ka.

So following the definition of the equilibrium constant, then it should be the concentration of the proton ion to the power one, the concentration of the anion to the power one, and divide by the concentration of the acid. So because of that you can have very strong acids or very weak acids. So that's shown in this equilibrium constant here. So if you recall, in here, we mentioned that the larger the PKS, then the more difficult it is. to dissolve these solids, right? So the same thing happens to these acids or these base.

So we can also define the calculation pKa. So what this means is the higher the value, the more difficult to dissolve. So the more the difficult to dissolve, then the weaker the acid. So here we have a table that's showing the pKa values of common acids. You see that the strong acids, they have such a low value of the pKa that some of them even go to negative three.

Okay, so a value of negative three means that the basically the Ka value is equal to 10 to the third. Right, so if you recall the pKa is equal to negative log 10. of Ka, right? So if you have a Ka value of 10 to the third, then this is going to be negative 3, okay? So if you have a very large Ka value, it just means that the concentration of the proton ions multiplied by the concentration of the anion is much much higher compared to the remaining acid concentration.

So that's why the strong acids will have a very small value of the pKa. while the weak acids are going to have larger values of the pKa. So here a very important system is the carbonic acid because in the atmosphere there are carbon dioxide existing in the system. So the carbon dioxide, once they dissolve into the water, they can form carbonic acid and further get dissolved into the proton and the bicarbonate ion.

And further the bicarbonate ion can dissolve into proton ion and the carbonate ions. So for these two reactions, as you can see, they have a pretty large pKa value, which means that they are weak acids, so the concentration of these proton ions are going to be lower in the system. So then a very easy application is the water. So what about water system?

We know that water can be written as H connected to an O, oxygen, and further hydrogen, right? We know it's composed of this structure, okay? So water, you can treat it as a very weak acid, okay?

So basically the water or the hydrogen here, they can get dissolved into the water, right, to form the protons. So for this reaction here, we can also So basically water is getting dissolved into proton and the hydroxide ion. So for this reaction, we can also write out its equilibrium constant. So that's going to be hydrogen multiplied by OH divided by the concentration of water. So if you recall the note that's earlier, so for pure liquid, this term should be regarded as one.

So that's why. we can define a equilibrium constant for water, Kw. So W stands for water here.

So that turns out to be 10 to the negative 14. So people have done these measurements. So now if you imagine we have a cup of water, let's say it's just water inside, it doesn't have any ions inside. Then let's see what's the concentration of the hydrogen and OH. So what happens is they are going to dissolve at a molar ratio one to one, right?

So basically the concentration of hydrogen ions is going to be 10 to negative seven because basically this is going to be squared because these two are the same, right? And then you can solve that. So further let's again use the definition of the p.

So let's say we have the pKa pKs, right? We can define the pH which is negative log 10 of the hydrogen ion concentration. So if you plug it in, you can get pH is equal to seven.

So this is how the pH of water come out, right? So if you have a neutral water, then they should have the pH of seven. If you have acid, basically a higher concentration of the hydrogen ions, then you're going to have a lower pH. Reversely, you're going to have a higher pH, okay? So in this lab session, you're going to learn about the pH. And there's also a term called the alkalinity.

So the alkalinity means how much hydrogen ions, it can react within the water system. So later on in this lab session, you're going to learn more about the alkalinity. And we're also going to cover that in our next class. Okay.

So let's pick up the topic next Monday. All right. I'll see you on next Monday then. Professor, for the lab, is it in person or online? It is in person.

Okay, awesome. Yeah, it is in-person lab. I'll put a notice on Canvas again.

Okay, just double checking. All right, thank you.

Transcript for:Chemical Reactions and Environmental Engineering

Transcript for:
Chemical Reactions and Environmental Engineering