this video is for those of you who may have a test on acids and bases or if you just want to review this topic so it's a list of multiple choice questions and i recommend that you try these problems before viewing the solution so let's start with number one which of the following is strong acid is it ammonia acidic acid sodium hydroxide perchloric acid or hydrofluoric acid now you need to know the six common strong acids these include hi hbr hcl h2so4 hclo4 and also hno3 if it's not one of these acids chances are it's a weak acid or something else now we could eliminate antichoice a nh3 ammonia is a weak base and b acetic acid that's a weak acid sodium hydroxide is a strong base the right answer is d perchloric acid that's one of the strong acids and e hydrofluoric acid is a weak acid now let's move on to number two the h3o plus hydronium concentration is four point seven times ten to minus three moles per liter calculate the ph of the solution now there's a simple formula that will help us to get the answer the ph is negative log of the hydronium ion concentration so it's going to be negative log of 4.7 times 10 to the minus 3. now if you don't have access to a calculator what do you think the ph is going to be what's the range of where it might be so focus on this number it's either going to be between two to three or three to four now if this number is greater than one pick the smaller range so the answer is going to be somewhere between two and three so it's going to be b if you type in negative log of 4.7 times 10 to the minus 3 this will give you 2.3279 and so b is the right answer for this problem number three the ph of the solution is 5.4 what is the poh of the solution now you need to know that the ph plus the poh must add to 14 at 25 degrees celsius so to calculate the poh of the solution it's simply 14 minus the ph of the solution so in this example that's 14 minus five point four fourteen minus five is nine and nine minus point four is eight point six and so this is the answer so c is the right answer in this problem number four which of the following is the bronze stellar base in the reaction shown below the broadcast base is the proton acceptor and the broad salary acid is the proton donor so notice that hf donates a proton as it becomes f minus so hf is the acid because it gives away a hydrogen water is the bracillary base because it accepts a hydrogen as it turns to hydronium now whenever you add a hydrogen to something you create the conjugate acid and if you take away hydrogen as in the case of hf the result is the conjugate base so the answer for this problem is water water is the bronze salary base now here's a question for you what is the conjugate acid of water and what's the conjugate base of it so anytime you want to write the conjugate acid of something add a hydrogen so h3o plus is the conjugate acid of water and hydroxide is the conjugate base to write the conjugate base take away hydrogen so let's use ammoni for example the conjugate acid is nh4 plus and the conjugate base take away a hydrogen is nh2 minus so with that in mind let's say if we have h2 po4 minus what do you think the conjugate acid and the conjugate base for this substance would be so for h2po4 minus if you add a hydrogen it becomes h3po4 that's the conjugate acid if you take away hydrogen it becomes hpo4 to the negative two so now you know how to identify the conjugate acid and the conjugate base of a substance number five the hydroxide ion concentration is 3.7 times 10 to the negative four moles per liter or molarity calculate the hydronium ion concentration now you need to know that water reacts with itself and it ionizes it turns into two ions it becomes atrial plus and hydroxide so this water molecule is the acid well it could be either or but actually this one's acting as the bronchiology base because as it turns into h3o plus it accepts a proton this one is acting as the acid because it loses the hydrogen turning into hydroxide so that's the bronze lary acid this is the conjugate acid and this is the conjugate base now if we write the equilibrium expression for this reaction now these two are in a liquid phase and this is an aqueous phase we know that k is going to be the products divided by the reactants the products are on the right side so that's going to be h3o plus times hydroxide divided by the reactants which would normally be divided by water but we cannot include any liquids or solids in the equilibrium expression so this is known as the autoionization constant of water sometimes referred to as the ion product constant for water and at 25 degrees celsius this value is temperature dependent it's 1 times 10 to the minus 14. so this is the equation that we need to calculate the h2o plus concentration so it's going to be kw which is 1 times 10 to the minus 14 divided by the hydroxide ion concentration so we're going to divide it by 3.7 times 10 to the minus 4. and so this is going to be 2.7 times 10 to the minus 11 which means that answer choice d is the answer number six the ph of the solution is 4.2 calculate the h3o plus concentration so here's the formula that you need the h2o plus concentration is equal to 10 raised to the negative ph so it's going to be 10 raised to the negative 4.2 and that comes out to be 6.31 times 10 to the negative 5. and so that's it for this problem so answer choice d is the answer number 7 the ph of the solution is 9.5 at 25 degrees celsius is the solution acidic basic or neutral now you need to be familiar with the ph scale at a ph of 7 the solution is set to be neutral when a ph is less than seven it's acidic if it's greater than seven it's basic so the ph of this particular solution is 9.5 so that's greater than seven so what we have is a basic solution number eight the hydroxide ion concentration is 2.6 times 10 to the negative nine moles per liter calculate the ph of the solution so there's different ways in which we could do this now what i'm going to do is calculate the poh first which is negative log of the hydroxide ion concentration so that's going to be negative log of 2.6 times 10 to the minus 9. and so that's 8.5 now the ph of the solution is going to be 14 minus the poh so that's 14 minus 8.585 and so that's 5.415 which we can round that to 5.4 so that's the ph of the solution therefore b is the correct answer number nine the concentration of h3o plus is 2.7 times 10 to the minus four and the solution is the solution acidic basic or neutral now here's what you need to know if the concentration is greater than 1 times 10 to the minus 6 at 25 degrees celsius and we're going to assume the temperature is 25 unless specified otherwise if it's greater than this number the solution is acidic if it's equal to this value then the solution is neutral and if it's less than 1 times 10 to the minus seven we have a basic solution so ten to negative four is that greater than or less than ten to negative seven two point seven times ten to minus four is greater than one times ten to minus seven and you could use the number line so let's say this is zero this is three here's negative four and here's negative seven as you go to the right the numbers become greater in value so negative four is higher than negative seven because it's to the right so therefore what we have is an acidic solution which means answer choice a is the right answer number 10 which acid is stronger is it hf hydrofluoric acid or hcn hydrocyanic acid and we're given the ka values for these two acids the ka value for hf is 7.2 times 10 to the negative four and the ka value for hcn is 6.2 times 10 to the minus 10. so given this information how can we determine which acid is stronger now you need to know that acid strength increases with higher ka values so the acid with the higher ka value is a stronger acid so it's not going to be c or d those reasons are the wrong reasons 10 to the negative 4 is higher than 10 to negative 10. so therefore hf is the stronger acid since it has the higher ka value number 11 what is the ph of a .025 molar hydrochloric acid solution now you need to realize that hcl is a strong acid so when it reacts with water it goes to completion so you should only have one arrow as opposed to two arrows this is not a reversible reaction and strong acids ionizes almost 100 so let's say if we make an ice table initially we started with 0.025 now the reaction is going to go to completion at the end this is going to be zero so the change is negative 0.025 and on the right side the change is positive 0.025 so what we need to realize is that the concentration of h3o plus is the same as the concentration of hcl which is 0.025 so therefore we can calculate the ph of the solution it's simply going to be negative log of the hydronium ion concentration or negative log of 0.025 and so that is 1.6 which means that answer choice a is the correct answer 12 calculate the ph of a 0.75 molar hypochlorous acid solution and we're given the ka the acid dissociation constant of hocl it's 3.5 times 10 to the negative eight so what's the ph of the solution now hypochlorous acid is a weak acid and so it reacts with water but reversibly so for weak acids there should be two arrows as opposed to one and so it ionizes according to this equation now we need to make an ice table so initially we have 0.75 moles per liter of hocl and of the products initially we have zero so the reaction has no choice but to initially shift to the right so the products will increase by x and the reactant will decrease by x now let's add the first two rows now the acid dissociation constant which is an equilibrium constant is equal to the products divided by the reactants now everything is in the aqueous phase except water water is in a liquid phase so it should not be included in the equilibrium expression now the ka value is 3.5 times 10 to the negative 8. atrial plus and the hypochlorite ion they're both equal to x so x times x is x squared and hocl the equilibrium value is going to be 0.75 minus x now because the ka value is very small we could ignore this x it's not going to be significant so now let's cross multiply so we're going to have 1 times x squared which is x squared and then 3.5 times 10 to the negative eight multiplied by point seven five so that's two point six two five times ten to the negative eight so now we can take the square root of both sides so x is equal to 1.62 times 10 to the negative 4. now notice that x is equal to the hydronium ion concentration so therefore we can now calculate the ph of the solution so it's going to be negative log of 1.62 times 10 to the minus 4. and so that works out to be 3.79 and so that's the ph of the solution which we can round it to 3.8 so d is the right answer number 13 which acid is stronger is it hydrochloric acid or hydrobromic acid now you need to know the trend for acids that is associated with the periodic table so let's say this is carbon nitrogen oxygen fluorine chlorine bromine and iodine acid strength increases as you go to the right and as you go down if if these elements are directly attached to the hydrogen atom so it deals with binary acids so what you need to know is that hi is a stronger acid than hbr which is stronger than hcl which is stronger than hf and hf is a stronger acid in water water is more acidic than ammonia and ammonia is more acidic than methane so clearly we can see that hbr is more acidic than hcl so hcl is not the stronger acid so the answer has to be between b and d now electronegativity increases as you go up so clearly chlorine has a higher electronegativity value than bromine so the reason is not due to electronegativity the reason is due to size a strong acid means that the conjugate base is very weak so the stronger the acid the more stable the conjugate base compare bromide to chloride bromide is a lot bigger than chloride and because that ion is bigger it has more volume to stabilize the negative charge so because this ion is more stable the acid is stronger it's very easy for hbr to give off the hydrogen because it leads to a very stable product and that's why hbr is more acidic than hcl it's due to the stability of the conjugate base so d is the right answer hbr is the stronger acid because the bromide ion is larger than the chloride ion and as a result the bromide ion is more stable than the chloride ion so the weaker the conjugate base the stronger the acid number fourteen what is the ph of a three molar ammonia solution and we're given the kb of ammonia so first let's write a reaction ammonia is a weak base and so it's going to react with water reversibly to produce the ammonium ion that's the conjugate acid plus hydroxide so we need to make an ice table so the initial concentration of ammonia is 3.0 and the products initially are zero so the reaction is going to shift to the right increasing the products by x and decreasing the reactant by x so let's add up the first two roles and the base dissociation constant kb is going to be the products which is nh4 plus and then times hydroxide divided by the reactant which is ammonia nh3 now we're given kb it's 1.8 times 10 to the minus 5. and so that's going to equal these two which are both x so x times x is x squared and the equilibrium value for nh3 is 3 minus x so if this value is large and if this value is small you could ignore x and also kb is relatively small so x can be ignored for this problem so let's cross multiply this is going to be 1 times x squared and then 1.8 times 10 to the minus 5 multiplied by 3 which is 5.4 times 10 to the minus 5. now let's take the square root of both sides so the square root of 5.4 times 10 to the negative 5 is equal to 7.348 times 10 to the minus 3. so that's the value of x and we can see that x is equal to the hydroxide ion concentration so once we have the hydroxide ion concentration we can easily calculate the ph of the solution but first we need to calculate the poh so that's going to be negative log of the hydroxide ion concentration so that's negative log of 7.348 times 10 to the minus 3. and so the poh of the solution it's 2.134 now to calculate the ph it's 14 minus the poh of the solution so that's 14 minus 2.134 and so this will give you 11.866 and that's the answer so we could round that to 11.9 and that's it for this problem number 15 what is the ph of a one molar sodium fluoride solution and we're given the acid dissociation constant of hf so fluoride is the base in the salt solution and it's going to react with water it's going to produce the conjugate acid hf and the conjugate base hydroxide so once again we need to make an ice table so the initial amount of fluoride is 1.0 this is going to be zero the products will increase by x and the reactants will decrease by x now because we have hydroxide in this expression we need to use kb for this reaction however we're given ka so we need to calculate kb kb times ka is equal to kw the autoionization constant of water and kw is 1 times 10 to the minus 14. so kb is this number divided by the ka value so that's 1 times 10 to the negative 14 divided by 7.2 times 10 to the minus four and so kb is one point three eight nine times ten to negative eleven so now we can write the kb expression for this reaction so it's going to be hf times the hydroxide concentration divided by fluoride and so this is going to equal x squared divided by the fluoride concentration which is one minus x now this number is very small so we could definitely ignore this x value so if we cross multiply this is going to be x squared which is equal to kb times one now let's take the square root of both sides so x is going to be 3.727 times 10 to the negative six and so we can see that x is the hydroxide ion concentration when dealing with kb so now we can calculate the poh of the solution so the poh is the negative log of the hydroxide ion concentration which is 3.727 times 10 to the negative 6. and so the poh is 5.429 now the ph of the solution is going to be 14 minus the poh so that's 14 minus 5.429 and so the ph is 8.571 and so this is the answer which correlates to answer choice b so b is the right answer for this problem number 16 which of the following substances is not amphoteric so what does it mean for a substance to be amphiteric an amphoteric substance is a substance that can act as an acid and as a base water fits this description when water acts as an acid it releases the hydrogen ion and turns into the conjugate base hydroxide when water acts as a base it acquires a proton turning into the conjugate acid h2o plus so that's how you can tell if a substance is amphoteric you have to ask yourself can a substance give away a hydrogen and can it receive one now looking at answer choice b this substance is also amphoteric it can receive a hydrogen turning into the conjugate acid atrial excuse me h3po4 and it can give away a hydrogen becoming the conjugate base hpo4 two minus so the same is true for bicarbonate it can act as an acid turning into carbonate and it can act as a base becoming carbonic acid now sulfate is not amphoteric it can behave as a base turning into the conjugate acid by sulfate however it doesn't have any hydrogens to give away so it cannot there's no conjugate base for sulfate so sulfate can only act as a base and it cannot act as an acid it can receive a proton but it cannot give one away so d is the right answer this substance is not amphoteric number 17 which acid is stronger is it chloric acid or chloric acid so this is a free response problem now you need to know that hclo3 is a stronger acid than hclo2 both of these acids are known as oxyacids they're acids that contain oxygen and for oxyacids as the number of oxygen atoms increases the acidity goes up so what this means is that hclo4 is more acidic than hclo3 which is more acidic than hclo2 and that's more acidic than hclo now hcl is not part of this trend in fact hdl is stronger than the first three acids hclo4 and hcl are considered strong acids but hlo3 not exactly it's like on a borderline now some other examples are nitric acid which is a known strong acid and nitrous acid which is a weak acid sulfuric acid is a strong acid but sulfurous acid is relatively weak so as you can see as the number of oxygen atoms increase the acidity increases so for this problem hclo3 chloric acid is a stronger acid than hglo2 chloros acid number 18 the ph of a 0.4 molar hx solution is 3.5 what is the ka value of hx so go ahead and try this problem so hx is a weak acid and like all weak acids it's going to react with water reversibly forming the conjugate acid h3o plus and the conjugate base x minus so as always we're going to make an ice table and that's a terrible looking e and so this is going to be 0.400 and then the reaction is going to shift to the right increasing the products decreasing the reactants so ka is going to be h3o plus times x minus divided by hx so h3o plus and x minus are both equal to x so we can say that ka is x squared divided by hx which is 0.4 minus x so if we could find the value of x then we can calculate the acid dissociation constant ka now we could find x by knowing the fact that we have the ph of a solution and notice that x is equal to the hydronium ion concentration to calculate the hydronium ion concentration it's simply 10 raised to the negative ph so that's going to be 10 raised to the negative 3.5 and so that's going to be 3.16 2 times 10 to the minus 4 which is equal to x so we're going to take this value and plug it into this equation and that's how we can calculate the acid association constant so it's going to be 3.162 times 10 to the negative 4 squared divided by the concentration minus that same number and make sure to put these two numbers in parenthesis so the answer is 2.5 times 10 to the negative 7. so that's the acid association constant for this particular problem which means that d is the right answer number 19 the ph of a .25 molar weak based solution is 9.75 what is the kb value of the weak base so let's call the weak base b it's going to react with water and it's going to turn into the conjugate acid which we'll call bh plus or hb plus and then the conjugate base hydroxide so if we make the ice table this is going to be point 25 0 0 plus x minus x and so forth after a while you just know what to do the kb value for this reaction is going to be the concentration of the conjugate acid times the hydroxide ion concentration divided by the concentration of the base so kb is going to be x squared divided by 0.25 minus x so the key is to calculate the value of x if we could find it we can easily calculate the value of kb now we're given the ph of the solution however notice that x represents the hydroxide ion concentration so we need to find the poh first the poh of the solution is going to be 14 minus the ph which is 9.75 so fourteen minus nine is five and five minus point seven five is four point twenty five now that we have the poh we can calculate the hydroxide ion concentration and we know it's 10 raised to the negative poh so that's going to be 10 to the negative 4.25 and so that's going to be 5.623 times 10 to the negative 5. so that's the value of x which is what we need to plug it into that equation so this is going to be 5.623 times 10 to the negative 5 squared divided by 0.25 minus x so the kb value for this particular base is 1.265 times 10 to the negative 8. and so that's the answer which you could round that to 1.27 times 10 to the negative 8. so c is the right answer in this problem number 20 which base is stronger ammonia or methylamine and we're given the kb values for nh3 and ch3 nh2 the kb value for ammonia is 1.8 times 10 to the negative five and the kb value from methylamine is 4.4 times 10 to the negative 4. so what you need to realize is that as the kb value increases the base strength increases as well so the stronger base is associated with the higher kb value 10 to negative 4 is higher than 10 to negative 5 so therefore methylamine is the stronger base number 21 the pka values of four acids ha hb hc and hd are 4.6 2.5 7.3 and 6.4 which of these acids is the strongest acid now we know the relationship between ka and acid strength as the ka value increases the acid strength increases so the stronger acid is associated with the high ka value now what about pka and acid strength these two possess an inverse relationship that is as the pka decreases the acid strength increases so the stronger acid is associated with the lower pka value so since we want to identify the strongest acid we need to know which one has the lowest pka value and that's the second one hb so therefore hb is the strongest acid the weakest acid will be the one with the highest pka value and so that would be hc hd is the weakest acid hb is the strongest acid so if you had to rank it it would be hb that's the strongest acid the second strongest acid would be h a it goes in order 2.5 that's less than 4.6 which is less than 6.4 and that's less than 7.3 now 6.4 corresponds to hd and 7.3 corresponds to hc so we can see that hb is clearly the strongest acid since it has the lowest pka value and hc is the weakest acid because it has the highest pka value number 22 which of the following salts will produce a basic solution is it sodium chloride ammonium chloride aluminum chloride sodium iodide or sodium nitrite sodium chloride will produce a neutral solution the ph will be around seven ammonium chloride will produce an acidic solution the ph will be less than seven when you place ammonium in water it will re-aquifer reversibly it's a weak acid and so it's going to create h3o plus and the weak base nh3 so because it generates the hydronium ion it's going to produce an acidic solution now aluminum chloride also works the same way the aluminum plus three cation acts as the lewis acid in fact in water it ionizes according to this equation there's six water molecules attached to it and it causes one of the water molecules to be ionized so notice that this five water molecules left but one of the water molecules will lose an h plus and as a result the charge on aluminum will change from plus three to plus two the total charge of the equation is plus three and so that's how the aluminum plus three cation creates an acidic solution so metal cations with a very high positive charge are acidic in water so for example fe plus three is another lewis acid it will create an acidic solution now what about sodium iodide this will be neutral and put a for acidic and for neutral but the answer is e sodium nitrite it's a basic solution when nitrite is put into water it's going to grab a proton and form nitrous acid and hydroxide now here's a question for you can chloride react the same way of water and produce hcl and hydroxide because it has a negative charge just like nitrite can that happen and the answer is no now let's talk about why the conjugate base of a strong acid will not form a basic solution however the conjugate base of a weak acid will form a basic solution you can write that down if you want to now hno2 is a weak acid we know that and the ka value for that if you want to look it up is four times ten to negative four and keep in mind it's not one of the six strong acids so it's safe to say it's a weak acid we know that weak acids react with water reversibly so this will produce h3o plus and nitrite so notice that nitrite is in this equilibrium expression and the both and the double arrow symbol means that the reaction can go to the right and to the left so because that reaction is reversible that means that if we place nitrite in water it will react to form the conjugate acid hno2 and so that's why we can say that the conjugate base of a weak acid will produce a basic solution because the reaction is reversible it can go both ways now the conjugate base of a strong acid will not produce a basic solution here's why we know that if we put hcl with water hcl is a strong acid it ionizes completely strong acids should not have a double arrow it's not reversible so it only goes in one direction which means that if we put chloride in water it's not going to form the strong acid it's just it's not going to happen so to write this that's supposed to be hcl and oh minus this reaction will not work so this reaction only goes towards the chloride ion chloride will not grab a hydrogen from water and turn into hdl so make sure you understand that the conjugate base of a strong acid will not form a basic solution but the conjugate base of a weak acid will and so that's why we can eliminate nacl because hcl is a strong acid and nai also because hi is a strong acid however this was the answer because hno2 is a weak acid so e is the right answer for this problem number 23 calculate the ph of a 0.5 molar ammonium chloride solution and we're given the kb for nh3 it's 1.8 times 10 to the negative 5. so go ahead and try this problem so ammonium chloride is an acidic salt solution but we don't have to worry about the chloride part ammonium will react with water reversibly to produce ammonia and hydroxide actually h3o plus in the hydroxide i take that back this is an acid so it's going to generate the hydronium ions nh4 plus is 0.5 and this is going to be zero and we know the reaction is going to shift to the right and so this is how we can complete the ice table now this reaction is associated with ka because it generates h3o plus so we can say that ka is going to be the products which is x times x divided by ammonium which is going to be 0.5 minus x now we don't have the ka value for this reaction we do have kb for nh3 so we need to calculate ka ka is going to be kw which is 1 times 10 to the negative 14 divided by kb which is one point eight times ten to the negative five and so that's equal to 5.556 times 10 to the negative 10. now let's cross multiply so we're going to have 1 times x squared which is x squared and because kb is very small we could ignore this x value so it's going to be 0.5 times this number as we cross multiply so 0.5 times 5.556 times 10 to the negative 10 that's 2.778 times 10 to the negative 10. now let's take the square root of both sides so x is going to be 1.67 times 10 to the negative 5. now keep in mind that x in this example is the hydronium ion concentration and so the ph which is the negative log of h3o plus that's going to be the negative log of 1.67 times 10 to the negative 5. and so the ph of the solution is 4.78 and so this is the answer which means that b is the right answer choice number 24 calculate the percent dissociation of a two molar acetic acid solution and we're given the ka value for acetic acid it's 1.8 times 10 to the negative five so i'm going to write acetic acid as h a so like all weak acids it reacts with water to create the hydronium ion and the conjugate base so the initial concentration is two this is going to be zero the products will increase by x and the acid concentration will decrease by x so we could say that ka is going to equal x squared divided by 2 minus x now we need to calculate the value of x so this is equal to 1.8 times 10 to the minus 5. and ka is small so we can ignore that x now if we cross multiply this is going to be x squared which is equal to 2 times 1.8 times 10 to the negative 5. so that's going to be 3.6 times 10 to the minus 5. and then let's take the square root so x is equal to 6 times 10 to the negative three now the percent dissociation is equal to the h plus concentration which is the same as the h3o plus concentration divided by the acid concentration times a hundred percent which i'm running out of space so the h2o plus concentration is the same as x so it's 6 times 10 to the minus 3 divided by the acid concentration which is 2 and then multiplied by 100 percent so this is equal to 0.3 percent and that's how you can calculate the percent dissociation of a weak acid so b is the right answer twenty-five the percent dissociation of a point two five molar a check solution is point thirteen percent calculate the ka value of hx so how can we calculate ka given percent dissociation well let's write the percent dissociation formula it's equal to the h2o plus concentration which is basically x divided by the acid concentration which is hx in this problem times 100 now the percent dissociation is point 13 percent and the acid concentration is 0.25 so our goal is to calculate x to do that we need to multiply both sides by .25 divided by 100. so on the left side it's going to be .25 over 100 times point 13. on the right side these two will cancel and these two will cancel giving us x so x is going to be 0.25 times point 13 divided by 100 and so that works out to be 3.25 times 10 to the negative 4. so now we could calculate ka so let's write the reaction associated with this problem so this is point 25 this is going to be zero and then you know how to finish it at this point so clearly we can see that ka is going to be x squared divided by the reactant 0.25 minus x so let's take this answer and plug it into this equation so ka is going to be 3.25 times 10 to the negative 4 squared divided by the acid concentration minus x and so ka is going to be 4.23 times 10 to the negative 7 which means that e is the right answer you