in this video we're going to talk about how to calculate the ph of a weak acid solution as well as the ph of a weak based solution we're also going to talk about percent ionization and some other problems associated with weak acid bases but let's review some basics now if you recall when a strong acid like hydrochloric acid reacts with water because it's a strong acid it will dissociate completely so we could use a single arrow to represent this reaction so if the initial concentration of hcl let's say it's point zero five as it dissociates completely the hydronium ion concentration will be the same so you can just calculate the ph of the solution using this formula the ph is equal to negative log of h3o plus and if you need to calculate the poh let's save a strong base solution you could use this negative log o h minus and remember the ph plus the poh adds up to 14 at 25 degrees celsius so if you know the poh the ph is simply 14 minus the poh now when dealing with a weak acid solution the situation is different a weak acid doesn't dissociate completely so we're going to use a double arrow to indicate a situation of equilibrium so this reaction is reversible so when dealing with a weak acid or a weak base or let's say a weak acid you need to use something called ka the acid dissociation constant it's an equilibrium constant and it's equal to the concentration of the products divided by the concentration of the reactants water is a liquid so remember liquids and solids are not included in the equilibrium expression everything else is in the aqueous phase it's dissolved in water so you do include them in the equilibrium expression now once you calculate h3o plus using ice tables you can calculate the ph uh using the equation that we just considered now when dealing with bases in this case nh3 is a weak base you need to use the base dissociation constant which is kb kb like ka is equal to the concentration of the products divided by the concentration of the reactant now using ice tables you want to calculate the concentration of hydroxide because once you have that you can calculate the poh using this formula and then once you have the poh you could find the ph now sometimes you may need to calculate kb from ka or vice versa so remember ka times kb is equal to kw which is 1 times 10 to the minus 14 and this is temperature dependent so this equation holds true at 25 degrees celsius now sometimes you may need to calculate the h3o plus concentration from ph and this formula will help you to do that it's equal to 10 raised to the negative ph now if you need to calculate the hydroxide ion concentration it's 10 raised to the negative poh and also the hydronium ion concentration times the hydroxide ion concentration at 25 degrees celsius this is equal to kw the auto ionization constant for water and that's 1 times 10 to the negative 14. so those are some other formulas that you want to be familiar with now let's talk about acid strength i've covered this in some other videos but nevertheless it's a good concept to review acid strength increases with increasing ka values so if you have two acids with two different ka values the stronger acid is the one with the higher ka value now acid strength is inversely related to pka so the stronger acid is going to have the lower pka value pka is equal to negative log of ka and pkb is equal to negative log kb so you want to add these formulas to your notes if you haven't done so already now the pka plus the pkb these two they add up to 14 at 25 degrees celsius now the next thing we need to talk about is how to calculate percent association also referred to as percent ionization the set ionization is equal to x divided by the acid concentration that is the original acid concentration that was dissolved in the solution times 100 percent x represents the amount of acid that is dissociated and when dealing with acids for the most part x is usually the h2o plus concentration so you can find it from doing an ice table or if you know the ph you can find it that way as well when dealing with weak bases x is usually the hydroxide concentration so those are the formulas that we're going to be using throughout this video to calculate different things so let's go ahead and work on some practice problems let's put these equations to good use so what is the ph of a 0.75 molar acetic acid solution and we're given the ka so first let's write the dissociation reaction of acetic acid when it's combined with water so because it's a weak acid it's going to be a reversible reaction so we need to put two arrows whenever you add a weak acid in water it's going to generate the hydronium ion h3o plus and also the conjugate base which in this case is acetate so we're going to do is we're going to make an ice table the initial amount of acidic acid is 0.75 and this is going to be units of molarity and for the products we're going to start with zero now the reaction has no choice but to shift to right it can't shift to left so the products will increase by x and the reactant will decrease by x we don't have to worry about water because what is a pure liquid everything else is an aqueous phase so as you recall you can't have any liquids or solids in the equilibrium expression when you're dealing with ka or kb or any type of k so if we add the first two rows it's going to be x and 0.75 minus x now in order to calculate the ph of the solution we need to calculate the concentration of h3o plus which we could see that it's equal to x so we've got to find the value of x so we can write the expression for ka the acid dissociation constant it's equal to the concentration of the products which can be found on the right side that's h3o plus and the acetate ion divided by the reactants and we can't use water since it's a liquid now ka is equal to 1.8 times 10 to the minus 5. h3o plus and acetate they're both equal to x acetic acid at equilibrium it's 0.75 minus x now because 10 to the minus 5 is pretty small we could ignore this x so therefore we could say that ka or this number is equal to x squared divided by 0.75 now let's get rid of a few things we really don't need this anymore but keep in mind that h3o plus is equal to x now in order to find the value of x we need to cross multiply so 1 times x squared is x squared and 1.8 times 10 to the minus 5 multiplied by 0.75 that's equal to 1.35 times 10 to the minus 5. now let's take the square root of both sides so the square root of that number is going to be 3.674 times 10 to the minus 3. that's the concentration of h3o plus now that we have the concentration of h3o plus we can calculate the ph of the solution the ph is equal to negative log of the hydronium ion concentration so it's a negative log of 3.674 times 10 to the minus 3. so you should get about 2.43 so that's the answer for this problem that's the ph of the solution so if you want to quickly get the answer whenever you have a weak acid and if you're given the concentration and the ka value and you want to find the ph let's simply review the steps first use this equation it's going to be ka which is equal to x squared divided by the acid concentration minus x now if ka is very small if it's like 10 to negative 5 or less typically you can ignore this x value if ka is not very small then you may have to use the quadratic formula so once you get the equation in this form ax squared plus bx plus c if you can't ignore this x here's the quadratic formula that you need it's negative b plus or minus the square root of b squared minus 4ac divided by 2a so keep this in mind if you're going to use the quadratic formula you need to rearrange the equation until it's in this form so on one side of the equation you have to have a zero now when you're dealing with x i mean when you're dealing with ka x is equal to h2o plus so once you find the value of x you can calculate the ph using this equation number two what is the ph of a .25 molar ammonia solution and we're given the kb the base dissociation constant so since we have kb basically we have a weak base in the solution and a weak base will generate hydronium ions so we need to use kb to find x which x is going to represent the hydroxide ion concentration and once we have that we could find the poh of the solution and then the ph so first let's write a reaction when you put a weak base with water it's going to be a reversible reaction and you're going to get the conjugate acid in this case nh4 plus and hydroxide so let's make an ice table just as we did before the initial concentration for ammonia is 0.25 the products will begin at zero and the reaction will initially shift towards the right so add in the first two rows we're going to have 0.25 minus x and x and x so kb like any other equilibrium constant k it's going to be the concentration of the products divided by the reactants and don't include water now kb we know it's 1.8 times 10 to the minus 5. nh4 plus and oh minus they're both equal to x x times x is x squared and nh3 is 0.25 minus x but since kb is small we could ignore the x that's next to the 0.25 so we're going to get this equation so let's cross multiply so we're going to have x squared is equal to .25 times 1.8 times 10 to the minus 5. and that's 4.5 times 10 to the minus 6. now let's take the square root of both sides the square root of 4.5 times 10 to the minus 6 that's two point one two times ten raised to the minus three now that we have the value of x we have the concentration of hydroxide which means we could find the poh of the solution so the poh is negative log of the hydroxide concentration or of 2.12 times 10 to the minus 3. so this is equal to 2.67 and whenever you have the poh of a solution you can easily find the ph it's going to be 14 minus the poh so the ph is about 11.33 and so that's the answer for this problem so this is what you need to do if you need to find the ph of a solution that contains a weak base number three what is the ph of a 0.4 molar ammonium chloride solution so in this problem we want to find the ph of a salt solution ammonium chloride is an ionic compound and we're given the kb of an nh string so what should we do in this problem we need to realize that nh4 plus is a weak acid because nh3 the conjugate base is a weak base so let's write the reaction when it's added to water so because it's a weak acid it's going to donate a proton to water and so we're going to get the h3o plus ion and also nh3 so nh4 plus is the acid it's the proton donor water is the base is the proton acceptor astrio plus is the conjugate acid and h3 is the conjugate base now let's make an ice table so initially this is going to be 0.4 and as usual this is 0 0 the reaction is going to shift to the right increasing the products by x decreasing the reactants by x as well now in this problem should we use ka or kb what do we have in the solution do we have a weak acid or a weak base we don't have nh3 in the solution initially we have nh4 plus and that's a weak acid which means we need to use ka ka is always associated with h3o plus if you have something that produces hydroxide ions that's going to be associated with kb which is a base or which is four bases so we have a weak acid in a solution we need to use ka but we're given kb so how can we find ka from kb ka times kb is equal to kw which is the auto ionization constant of water and that's equal to 1 times 10 to the negative 14. so therefore ka is basically equal to that number divided by kb and we have kb it's 1.8 times 10 to the negative 5. so let's go ahead and divide these two numbers so this is about 5.56 times 10 to negative 10. so now that we have ka we can write the equilibrium expression which is going to be products over reactants so we know it's going to be x squared divided by 0.4 minus x now ka is very very small so therefore we can neglect this x value it's going to be insignificant now let's cross multiply so we're going to have x squared which is equal to 0.4 times 5.56 times 10 to the minus 10. as usual we're going to take the square root of both sides and calculate the value of x so x is one point four nine times ten to the minus five now because we're dealing with ka x represents the h3o plus concentration which means we can now find ph directly so the ph is going to be negative log of the hydronium ion concentration which is equal to x so therefore the ph is about 4.83 that's the answer number four what is the ph of a 1.5 molar sodium fluoride solution so we're given the ka of hydrofluoric acid let's go ahead and find the ph so we don't have to worry about sodium it's a spectator ion so let's ignore it by the way feel free to pause this video and work on this problem yourself see if you can get the answer so fluoride is the iono vitreous because it's associated with hf if hf is a weak acid fluoride is basic enough to change the ph of the solution so fluoride is a weak base it's going to react with water and it does so reversibly generating the conjugate acid hf and because fluoride is a weak base it's going to generate a small amount of hydroxides that's changing the ph of the solution so we're going to make our ice table as usual and the initial amount of fluoride is 1.5 the products are zero and they're going to increase by x fluoride will decrease by x and as you can see the ice table for the most part is pretty much the same now we need to find kb because fluoride is a weak base we have sodium fluoride in the solution not hf we're going to have a small amount of hf but for the most part we have a large amount of sodium fluoride so we need to use kb kb is going to equal the products divided by the reactant so let's go ahead and find kb kb is going to be equal to kw which is 1 times 10 to the negative 14 divided by ka which is uh this number so kb is 1.39 times 10 to the minus 11. hf and hydroxide they're both equal to x so x times x is x squared and f minus is going to be 1.5 minus x but we could ignore that x since kb is very small let's cross multiply so this is going to be x squared which is equal to let's get rid of this x 1.39 times 10 to the minus 11 times 1.5 and that's 2.085 times ten to the minus eleven now let's take the square root of both sides so x is equal to four point 4.566 times 10 to the negative 6. now keep in mind that x is equal to the hydroxide concentration so we could find the poh and then we could find the ph of the solution so the poh is equal to negative log of the hydroxide concentration which is this number so notice that you see a 6 here the ph i mean excuse me the poh is going to be somewhere between 5 and 6. so it's actually 5.34 now that we have the poh we can calculate the ph which is 14 minus the poh and it's going to be about 8.66 so that's the answer for this problem now you know how to calculate the ph of the solution if you're given a basic solution now let's focus on percent ionization problems i'm actually mixing in some old videos that i had in the past with this video so what is the formula in order to find the percent ionization it's equal to x divided by the acid concentration now if you're dealing with a weak acid x represents the amount of h plus that is in equilibrium in the solution and let's not forget to multiply this by a hundred percent now when you're dealing with bases x represents the amount of hydroxide and instead of h a it's going to be the concentration of the base that is the initial concentration so now let's go ahead and write a reaction so once a weak acid is in the solution let's say when it's dissolved in water it's going to ionize and because it's weak the reaction is going to be reversible so we need to put two arrows acids will always generate h3o plus the hydronium ion and we're going to get the conjugate base fluoride so now let's make an ice table the initial amount of hydrofluoric acid is 0.75 and initially we don't have any hydronium ions or fluoride ions so therefore the reaction has no choice but to shift to the right increasing the products by x decrease in the reactants by x so if we add the first two rows this is what we're going to get now ka the acid dissociation constant like any equilibrium constant is equal to the ratio of the products over the reactants so ka is equal to 7.2 times 10 to the minus 4. h3o plus and f minus they're both equal to x x times x is x squared and hf is 0.75 minus x so basically we're using the values at equilibrium when dealing with k now because ka is fairly small we can ignore this x x is going to be relatively small compared to 0.75 so what we can do at this point is cross multiply 1 times x squared is x squared and 0.75 times ka and that's equal to 5.4 times 10 to the minus 4. now let's take the square root of both sides so x is equal to .0232 so 0.75 minus 0.0232 that's about 0.07 i mean 0.7268 it doesn't change that much it changes a little but we're going to go with this estimation so now that we have the value of x we can calculate the percent dissociation now keep in mind this is not the exact answer but it's an estimation of the answer to get the exact answer you can use the quadratic formula but we're just going to go with the estimation so the percent ionization is going to be equal to x which is .0232 divided by the initial concentration of hf which is 0.75 so it's about 3.1 percent you