in this video we're going to focus on acid-base titrations we're going to talk about titration curves and how to calculate the ph at various points of the acid-base titration process and some other problems as well so let's start with this one 28.9 milliliters of h2so4 was completely titrated with 38.4 milliliters of a 0.25 molar sodium hydroxide solution what is the concentration of h2so4 so how can we find the unknown concentration of this acid so there's two ways in which we can do this we could use stoichiometry to get the answer or we could use an equation let's do it both ways for the sake of learning so let's begin by writing a balanced chemical equation sulfuric acid h2so4 is going to react with sodium hydroxide so this is an acid-base neutralization reaction sodium is going to pair up with sulfate sodium have sodium has a positive one charge sulfate has a negative two charge so when you pair these two together the formula is going to be na2 so4 the hydrogen ions will react with the hydroxide ions to create h2o so whenever you add a strong acid with a strong base they will completely neutralize each other creating salt and water now let's balance the equation so we have two sodium ions on the right which means we need to put a two in front of naoh now notice that we have two hydrogen ions two hydroxide ions that's going to create two water molecules so now the equation is balanced so how can we use this equation to calculate the concentration of h2so4 well let's start with the other substance that is sodium hydroxide let's start with the volume of the naoh solution so we have 38.4 milliliters of sodium hydroxide and let's convert that to liters one liter is equal to one thousand milliliters now the next thing we need to do is we need to use the molarity of the solution to convert liters to moles a .25 molar solution means that we have .25 moles of solute in this case the solute being sodium hydroxide per 1 liter of solution so now that we have moles of sodium hydroxide we can convert that to the moles of h2so4 using the balanced chemical equation so the molar ratio is one and two for every one mole of h2so4 that reacts two moles of sodium hydroxide will be consumed so right now we have moles of h2so4 to get the concentration of h2so4 we need to take the moles and divide it by the liters of solution and we we are looking for the original concentration of h2so4 so we need to use the original uh volume of that solution let's convert 28.9 milliliters into liters to do that we need to divide it by a thousand so 28.9 milliliters is point zero two eight nine liters so now let's go ahead and plug this in it's going to be 38.4 divided by 1000 times 0.25 divided by 2 divided by .0289 so the concentration of h2so4 in the original unknown solution i mean it's not an unknown solution but the concentration was unknown that unknown concentration we now know what it is it's 0.1661 h2so4 now let's talk about another way in which we can get that same answer we could use this formula m1 v1 is equal to m2 v2 but we're going to need to modify a bit so m1v1 let's relate that to h2so4 and naoh is going to correspond to m2 v2 so in order for this equation to work we'll need to take into account the one to two molar ratio of h2so4 and sodium hydroxide and the way we could do that is by looking at the number of protons in sulfuric acid and the number of hydroxide ions in the naoh formula unit so in one formula unit of h2so4 there are two protons and in one formula unit of sodium hydroxide there's only one hydroxide ion so once you incorporate that this formula is going to work so remember m1v1 that's going to be the information for h2so4 m2v2 that's going to be it's going to correspond to the information for naoh if you mix it up it won't work so we're trying to calculate m1 the original concentration of h2so4 v1 the volume of h2so4 is 28.9 milliliters m2 the concentration of sodium hydroxide that's point 25 m v2 is 38.4 milliliters now v1 and v2 they have to be in the same unit if v1 is in liters v2 has to be in liters if v1 is in milliliters v2 has to be milliliters so these units will cancel now since m2 is a molarity m1 will be in the same unit molarity so on the left we have 2 times 28.9 which is 57.8 times m1 on the right we have 0.25 times 38.4 which is 9.6 so now let's divide both sides by 57.8 so 9.6 divided by 57.8 that will give us the original concentration of h2so4 which is 0.1661 those are the two ways in which you can solve this particular problem number two 23.6 milliliters of a 0.46 molar monoprotic acid solution was titrated with a 0.19 molar sodium hydroxide solution what is the volume of naoh that should be added to the solution in order to reach the equivalence point so we have a monoprotic acid that's an acid that can give away one proton per form a unit h2so4 this is what is known as a diprotic acid it can release two hydrogen ions per formal unit so we're combining a monoprotic acid with sodium hydroxide and we could use this formula m1v1 is equal to m2 v2 there's one proton per formal unit here and one hydroxide performing unit there at the equivalence point the number of moles of h plus is going to equal the moles of sodium hydroxide therefore we could use this formula to calculate the volume of naoh that will be needed to take us to the equivalence point because at the equivalence point these two must be equal and if you take the molarity of hx and multiply by the volume it's going to give you the moles of hx and that's going to equal the moles of h plus since these two are in a one-to-one ratio likewise here when you multiply m2 v2 you have the moles of naoh over the liters times the liters of the solution and that will give you the moles of hydroxide the bottom line is that we can use this formula to get the equivalence point so let's go ahead and do that the concentration of the monoprotic acid is 0.46 so that's m1 v1 is the volume of that monoprotic acid so it's 23.6 milliliters m2 is point 19 and we're trying to calculate the volume of base that we need to add to reach the equivalence point so we're looking for v2 so let's divide both sides by 0.19m so those units will cancel and we're going to get v2 in milliliters so it's 0.46 times 23.6 divided by 0.19 so the volume that's needed to reach the equivalence point is going to be 57.14 milliliters and so that's it for number two now let's focus on acid-base titration curves and let's begin our discussion with a strong acid strong base titration so based on the order in which it's written initially we have strong acid in the solution and we're adding strong base to it so let's draw the titration curve for this type of situation so initially we have strong acid in the solution so the ph is going to start out very low and then it's going to rise dramatically and then it's going to eventually taper off for a strong acid strong base titration the ph of the equivalence point is 7 and this is the volume at the equivalence point now anytime you add base to a solution the ph is going to increase now the reverse is true if we have a strong base strong acid titration anytime you add an acid the ph will decrease so this graph is basically just going to be inverted from this graph it's just the uh it's just that graph basically flipped so this time we're going to start off from a high ph and then it's going to decrease dramatically and then it's going to taper off the ph at the equivalence point is still 7 for strong base strong acid titration the only difference is here the ph is increasing and then for a strong base strong acid titration the ph is decreasing but besides that the graphs they look relatively similar just flipped so whenever you're dealing with a ph curve or titration curve you have ph on the y axis volume on the x-axis now for the titration curve on the left this corresponds to the volume of the strong base that was added to the solution here this corresponds to the volume of the strong acid that was added to the solution and the equivalence point occurs when the curve is basic essentially vertical so whenever you have like a slope that's very steep almost vertical that's where the equivalence point is going to be located now what about a weak acid strong base titration how can we draw a ph curve for that so because we're starting with an acid we're going to start with a low ph but it's going to be a little bit higher than the strong acid strong based titration assuming the concentration of the weak acid and the strong acid are the same so let's start with a value that's slightly above two so the ph is going to increase and then it's going to become somewhat vertical and then it's going to taper off again so it's somewhat similar to the strong acid strong base titration but it varies and the variation has to do with the ka of the acid now for a weak acid strong base titration the equivalence point is above seven it could be eight it could be nine it could be ten it really depends on the ka of the acid and uh the concentration as well but for the most part at the equivalence point the ph is greater than seven and if you think about this conceptually let's say if the weak acid is ammonia if we react ammonia with a strong base like hydroxide we're going to get water and we're going to get the weak base nh3 so imagine if we have one mole of ammonia one mole of hydroxide these two will react to completion they will decrease by one becoming zero this will increase by one so at the equivalence point we're not going to have any more weak acid or strong base what's gonna be left over is a weak base and the ph of a weak based solution is greater than seven so that is why at the equivalence point the ph has to be greater than seven because at that point all you'll have is the conjugate of the weak acid which is the weak base and the ph of any weak based solution will always be above seven so that's the situation that we have now another important thing that you need to keep in mind is that at one half of the volume of the equivalence point the ph is going to be equal to the pka so whatever the ph value here is let's assume it's four the pka for this acid will be equal to four so remember the ph is equal to the pka at one half of the equivalence point and here's the reason for this so if we go back to this bca table before change after let's say we have one mole of ammonia in a one liter solution so the concentration is one and we add 0.5 moles of hydroxide so this would be half of the equivalence point because we have half of the moles of ammonia with in terms of hydroxide and let's say this is zero so this is going to go to completion hydroxide will be the limiting reactant therefore we'll subtract both sides by 0.5 this is going to increase by 0.5 hydroxide will go to zero so at half the equivalence point notice that we have equal amounts of weak acid and weak base and according to the henderson-hasselbach equation ph is equal to pka plus log base over acid when the amount of base and acid are the same either in terms of concentration or in moles you get log of one log of one is zero so this disappears and the ph becomes equal to the pka so that's why at half the equivalence point the ph will be equal to the pka and you can identify the ka of the acid so once you get the pka from a titration curve the ka of the acid is going to be 10 raised to the negative pka so that's what we need to know when dealing with a weak acid strong base titration curve now let's consider a weak base strong acid titration curve so because we're starting with a weak base we're going to start at a relatively higher ph but not as high as if it would be a strong base so the graph will be somewhat horizontal and then eventually it's going to decline and then it's going to be horizontal again now the takeaway from this is that at the equivalence point the ph will be less than seven for a weak base strong acid titration so imagine if we have ammonia ammonia is a weak base and we react it with hcl we're going to get nh4 plus the weak acid and cl minus at the equivalence point these two will go to zero and we're only going to have the weak acid left over and the ph will be less than seven at that point so here this will be the volume at the equivalence point and then that one half of that volume like before we could determine the pka of the acid so let's say this value is 10. so the pka for this acid this hypothetical acid not necessarily ammonia or nh4 plus this is just some generic acid i mean genetic base but the pka of that conjugate acid is going to be 10 but because we're dealing with a weak base we want to find the pkb of the weak base the pkb is 14 minus the pka when water is a solvent at 25 degrees celsius so if the pka is 10 the pkb has to be 4. and once you know the pkb you can calculate the kb of the weak base using this now let's replace this equation with a more generic version so here we have the weak base we'll call the weak base a minus and we'll add an acid to it which we'll call h plus we're going to get the conjugate acid h so 10 would represent the pka not of the weak base but of the conjugate acid 4 is the pkb of the weak base which is a minus so when you get to pka it doesn't correspond to the weak base it corresponds to the conjugate weak acid so make sure you're aware of that now at the equivalence point you need to know that at this point you pretty much have mostly ha the predominant form of the predominant species at this point is h a at half of the equivalence point the concentration of a minus will be equal to the concentration of h a because at that point when you have equal amounts of basin acid the ph is going to be equal to the pka at this point between zero and half the equivalence point you have more of the base than the conjugate acid between half the equivalence point and the equivalence point you have more of the acid and less of the base and at the equivalence point the amount of weak base and strong acid that you have is almost zero i say almost because once all of these go to zero then you get h a a small amount of h a will go back into this form it's going to be very close to zero but a small amount will be dissociated into h plus and a minus so they don't go exactly to zero but close to it at the equivalence point nearly not exactly but almost 100 percent will be ha and then beyond the equivalence point you're going to have excess strong acid so this can be excess hcl or access h2so4 depend on the strong acid that you're using now the next thing i want to talk about is the buffer region so this region here would be considered the buffer region that's where we have a buffer solution anytime you have a weak acid and a conjugate base in a solution you have a buffer solution a buffer solution is designed to resist changes in the ph of the solution the buffer solution is ideal at this point when you have equal amounts of base and acid so at that point that is at half the equivalence point where the ph is equal to pka graph should be somewhat horizontal it's not going to be perfectly horizontal because as you add acid the ph will decrease but depending on the buffering capacity and how good of a buffer you have this should be somewhat horizontal because the buffer solution will resist changes in the ph it's going to try to maintain the ph to a relatively constant level now at the equivalence point the slope should be steep the equivalence point is going to be the point in the curve where the line is most vertical but at half the equivalence point it should be somewhat horizontal so those are some other things you want to be aware of when dealing with titration curves so remember at half the equivalence point you have an ideal buffer solution and the ph should be relatively constant at that point but at the equivalence point the ph changes drastically so the curve should be somewhat vertical at that point now let's work on some acid-base titration problems that ask us to calculate the ph at various points along the titration curve and let's start specifically with a strong acid strong base titration first 50 milliliters of a one molar hcl solution is titrated with a 0.50 molar sodium hydroxide solution calculate the volume of the sodium hydroxide solution needed to reach the equivalence point so we're combining hcl and naoh to calculate the volume at the equivalence point we could use the dilution formula m1 v1 is equal to m2 v2 so m1 that's going to be the concentration of the acid we have a one molar solution v1 is the volume of the acid so that's 50 milliliters m2 is the concentration of sodium hydroxide v2 is what we're looking for so let's go ahead and divide both sides by point five m one divided by point five is two two times fifty is a hundred so the volume at the equivalence point is 100 milliliters so that's the volume of sodium hydroxide that we need to add to reach the equivalence point now let's move on to part b what is the ph of the hydrochloric acid solution before any o h is added the concentration of hdl is going to be the same as the concentration of h2o plus because this acid dissociates completely so when hydrochloric acid reacts with water it's going to create h3o plus and cl minus so a one molar hcl solution will produce a one molar h2o plus solution so the ph is going to be negative log of the h3o plus concentration so that's going to be negative log of 1. negative log of 1 is zero so the ph of that solution will be zero now let's move on to part c what is the ph of the solution after 30 milliliters of sodium hydroxide has been added so we need to write a reaction so hydrochloric acid is going to react with sodium hydroxide and when hydrogen and hydroxide get together it's going to form water and when sodium and chlorine get together they will make sodium chloride now we need to do an icf table or a bca table icf initial change final bca before change after so you could use either one the results will be the same what we need to do is we need to get the moles of the initial reactants so let's start with hdl we have a volume of 50 milliliters we're going to convert that to liters by dividing it by a thousand and then we're going to get moles using the molarity so a one molar solution means that we have one mole of hcl in one liter of solution 50 divided by a thousand is .05 liters times one so we have .05 moles of hcl let's do the same thing with sodium hydroxide so the volume is 30 milliliters let's convert that to liters and then we're going to multiply that by the concentration of sodium hydroxide so we got .50 moles of naoh per 1 liter of solution so 30 divided by 1000 is point zero three times point five so that's going to give us point zero one five moles now the limiting reactant is sodium hydroxide so that's going to go to zero therefore the change will be negative 0.105 it's always going to be the smaller of those two so this is going to be zero and if we subtract .05 minus 0.015 we're going to get .035 and we really don't need to worry about anything on the right because that's not going to affect the ph of the solution what's important is that we have strong acid left over after mixing hcl and naoh so now let's calculate the concentration of hcl that we have so it's .035 moles and then we need to divide this by the total volume so we had 50 milliliters of the acid solution 30 milliliters of the base solution so that's 80 milliliters if we convert 80 milliliters to liters by dividing it by a thousand that's going to be point zero eight liters and keep in mind the concentration of hdl is the same as the concentration of h2o plus and that's going to be 0.4735 m so that's going to be equivalent to the concentration of the hydronium ion in solution now let's calculate the ph it's going to be negative log of the h2o plus concentration so negative log of that number will give us a ph of 0.359 so that's the answer for part c now for part d what is the ph at the equivalence point so what is it going to be if we were to add 100 milliliters of sodium hydroxide to the acid at the equivalence point the ph is going to be seven remember this is a strong acid strong base titration at the equivalence point the ph will always be seven for that type of titration and you could follow the same steps here what's going to happen is these two numbers will go to zero and so you just have pure water and the ph is going to be seven now what about part e what is the ph of the solution after 125 milliliters of sodium hydroxide has been added feel free to pause the video if you want to try that so the initial number of moles of hcl will be the same it's going to be 0.05 that's not going to change what's changing is the moles of sodium hydroxide that were introduced into the solution so we have to recalculate that let's start with the volume of naoh so it's 125 milliliters let's convert it to liters by dividing it by a thousand and then let's multiply it by the concentration of the solution so 125 divided by a thousand gives us the volume of 0.125 liters multiply that by 0.5 and we get 0.0625 moles of sodium hydroxide so notice that the volume is beyond the equivalence point the equivalence point occurred at a volume of 100 milliliters so we're going past the equivalence point we're going to have excess hydroxide in the solution so the ph is going to be above 7. the limiting reactant this time is hydrochloric acid so we're going to subtract by the smaller of those two values hcl is going to go to zero naoh is going to be .0125 moles after the titration now let's get the concentration of sodium hydroxide so let's divide this by the total volume of the solution so we mix 50 milliliters of acid with 125 milliliters of base therefore the total volume of the solution is 175 milliliters and we need to convert that to liters if we divide 175 by a thousand that's going to be 0.175 liters so that gives us the concentration of 0.07143 m that's the concentration of hydroxide so what we're going to do now is we're going to calculate the poh of the solution the poh is negative log of the hydroxide concentration so we're going to take this number plug it into that equation negative log of 0.07143 that gives us a poh of 1.146 once we have the poh we can calculate the ph the ph is 14 minus the poh so it's 14 minus 1.146 and we can round this answer to 12.85 so that's the ph of the solution after 125 milliliters of sodium hydroxide has been added now let's move on to our next example problem and that is a titration between a weak acid and a strong base you