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## Buffer solutions

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# Buffer solution pH calculations

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## Video transcript

- [Voiceover] Let's do some
buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I
showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a
conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the
pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know
the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. So the pKa is the negative log of 5.6 times 10 to the negative 10. So let's get out the calculator
and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration
of A minus, our base. Our base is ammonia, NH three, and our concentration
in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the
concentration of our acid, that's NH four plus, and
our concentration is .20. So this is over .20 here
and we can do the math. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the
pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to
compare what happens to the pH when you add some acid and
when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what
that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH? So we're adding .005 moles of sodium hydroxide, and our total volume is .50. So if we divide moles by liters, that will give us the
concentration of sodium hydroxide. .005 divided by .50 is 0.01 molar. So that's our concentration
of sodium hydroxide. And since sodium hydroxide
is a strong base, that's also our concentration
of hydroxide ions in solution. So this is our concentration
of hydroxide ions, .01 molar. So we're adding a base and think about what that's going to react
with in our buffer solution. So our buffer solution has
NH three and NH four plus. The base is going to react with the acids. So hydroxide is going to
react with NH four plus. Let's go ahead and write out
the buffer reaction here. So NH four plus, ammonium is going to react with hydroxide and this is going to
go to completion here. So if NH four plus donates
a proton to OH minus, OH minus turns into H 2 O. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia. Alright, let's think
about our concentrations. So we just calculated
that we have now .01 molar concentration of sodium hydroxide. For ammonium, that would be .20 molars. So 0.20 molar for our concentration. And for ammonia it was .24. So let's go ahead and
write 0.24 over here. So if .01, if we have a concentration of hydroxide ions of .01 molar, all of that is going to
react with the ammonium. So we're gonna lose all of this concentration here for hydroxide. And that's going to neutralize the same amount of ammonium over here. So we're gonna be left with, this would give us 0.19 molar for our final concentration of ammonium. Hydroxide we would have
zero after it all reacts, And then the ammonium, since the ammonium turns into the ammonia,
if we lose this much, we're going to gain the same
concentration of ammonia. So over here we put plus 0.01. So the final concentration of ammonia would be 0.25 molar. And now we can use our
Henderson-Hasselbalch equation. So let's go ahead and plug everything in. So ph is equal to the pKa. We already calculated the pKa to be 9.25. And then plus, plus the log of the concentration of base, all right,
that would be NH three. So the concentration of .25. So this is .25 molar
for our concentration, over the concentration of
our acid and that's ammonium. So that's over .19. So this is all over .19 here. So if we do that math, let's go ahead and get
out the calculator here and let's do this calculation. Log of .25 divided by .19, and we get .12. So 9.25 plus .12 is equal to 9.37. So let's get a little
bit more room down here and we're done. The pH is equal to 9.25 plus .12 which is equal to 9.37. So let's compare that to the pH we got in the previous problem. For the buffer solution just
starting out it was 9.33. So we added a base and the
pH went up a little bit, but a very, very small amount. So this shows you mathematically how a buffer solution resists drastic changes in the pH. Next we're gonna look at what happens when you add some acid. So we're still dealing with
our same buffer solution with ammonia and ammonium, NH four plus. But this time, instead of adding base, we're gonna add acid. So we add .03 moles of HCl and let's just pretend like the total volume is .50 liters. And our goal is to calculate the pH of the final solution here. So the first thing we could do is calculate the concentration of HCl. So that would be moles over liters. So that's 0.03 moles divided by our total volume of .50 liters. And .03 divided by .5 gives us 0.06 molar. That's our concentration of HCl. And HCl is a strong
acid, so you could think about it as being H plus and Cl minus. And since this is all in
water, H plus and H two O would give you H three
O plus, or hydronium. So .06 molar is really the concentration of hydronium ions in solution. And so the acid that we
add is going to react with the base that's present
in our buffer solution. So this time our base is going to react and our base is, of course, ammonia. So let's write out the reaction between ammonia, NH3, and then we have hydronium ions in solution, H 3 O plus. So this reaction goes to completion. And if ammonia picks up a proton, it turns into ammonium, NH4 plus. And if H 3 O plus donates a proton, we're left with H 2 O. So we write H 2 O over here. For our concentrations,
we're gonna have .06 molar for our concentration of
hydronium ions, so 0.06 molar. And the concentration of ammonia
is .24 to start out with. So we have .24. And for ammonium, it's .20. So we write 0.20 here. So all of the hydronium
ion is going to react. So we're gonna lose all of it. So we're left with nothing
after it all reacts. So it's the same thing for ammonia. So that we're gonna lose the exact same concentration of ammonia here. So we're gonna lose 0.06 molar of ammonia, 'cause this is reacting with H 3 O plus. And so after neutralization,
we're left with 0.18 molar for the
concentration of ammonia. And whatever we lose for
ammonia, we gain for ammonium since ammonia turns into ammonium. So we're going to gain 0.06 molar for our concentration of
ammonium after neutralization. So we get 0.26 for our concentration. And now we're ready to use
the Henderson-Hasselbalch equation to calculate the final pH. So let's do that. Get
some more space down here. So the pH is equal to the pKa, which again we've already calculated in
the first problem is 9.25 plus the log of the concentration of the base and that's .18 so we put 0.18 here. Divided by the concentration of the acid, which is NH four plus. So that's 0.26, so 0.26. And we go ahead and take out the calculator and we plug that in. So log of .18 divided by .26 is equal to, is equal to negative .16. So let's go ahead and write that out here. So we have our pH is equal to 9.25 minus 0.16. And so that comes out to 9.09. So the pH is equal to 9.09. So remember for our original buffer solution we had a pH of 9.33. So we added a lot of acid,
the pH went down a little bit, but not an extremely large amount. So once again, our buffer
solution is able to resist drastic changes in pH.