Consider a buffer made from 0.13 M sodium acetate and 0.10 M acetic acid in 1L. What amount of KOH must be added to cause an increase of the pH by 0.14?

Question

A) 0.048 mol
B) 0.038 mol
C) 0.028 mol
D) 0.018 mol
E) 0.008 mol

Answer 1

Let HA = CH3COOH
millimols HA = mL x M = 1000 x 0.1 = 100
mmols A^- = 1000 x 0.13 = 130

First determine what the pH of the solution is initially using the Henderson-Hasselbalch equation. I obtained 4.85 but you should ocnfirm that. Then you want to increase that by 0.14 by adding KOH so 4.85+0.14 = 4.99.

.........HA + OH^- --> A^- + H2O
I.......100...0.......130.........
add...........x....................
C.......-x...-x........+x
E.......100-x..0......130+x

Substitute the E line into the HH equation and solve for x. That will be millimols KOH to add to a liter. Convert to mols and that should be the answer. My answer was 17.3 millimols KOH and the closest answer is D. 17.3 mmols makes it come to 4.99; using 18 makes it come out to 4.996 or 5.00 and that is the closest answer which makes me think of the Zn/Cu cell we worked on earlier. How did that answer turn out?

Answer 2

This is a beautiful way of thinking, thank you for sharing.
The problem is that at the test I am not given the Ka value for the acid.
I solved it this way: I used HH equation, then since we are dealing with the same acid the value of Ka would not change (i thought of it as coarse tuning); to change the pH we would have to variate the concentration of base to acid ratio (fine tuning);
since the pH was increased by 0.14, I just multiplied by mol of base (Sodium acetate) 0.14* 0.13mol=0.0182

Not, sure if it's the right rationale but I got the answer.
Please, let me know what do you think?

Zn/Cu cell problem remained the same (1.09V) I couldn't come up with at better answer.

Answer 3

P.S. I just found out that our calculation were right all along for Zn/Cu cell problem... indeed the answer is 1.09V =)
Thank you!

Answer 4

Also, could you tell me please how you have solved for X?

Answer 5

First thanks for sharing that Zn/Cu answer. I was sure that was right and as I posted last night, I might be missing something but if so I didn't see it.
My equation for the buffer was
4.99 = 4.74 + log (130+x)/(100-x)
0.25 = log (130+x)/(100-x)see below
1.78 = (130-x)/(100-x)
1.78(100-x) = 130+x
178-1.78x = 130+x
48 = 2.78x
x = 17.3

I didn't realize you weren't given Ka. However, that doesn't change the way to work the problem. For example, make up a number for pKa. Let's call it 6 to make it easy. Then
pH = 6 + log(130/100)
pH = 6.11 for the buffer initially using Ka = 1E-6
Then add 0.14 to make that 6.25
Now 6.25 = 6 + log (130+x)/(100-x)
0.25 = log (130+x)/(100-x)
1.78 = (130+x)/(100-x)
and from here on the numbers are the same. So you really don't need a pKa because you're working on the difference. But wait, you really don't need a pKa at all if you do it this way.
pH = 0 + log(130+x)/(100-x)
This ratio is 0.11, you add 0.14 to it to make 0.25 so you have
0.25 = log(130+x)/(100-x)
1.78 = (130+x)/(100-x)
and from here on in you have the same numbers.

Answer 6

aha very interesting, thank you so much for this detailed explanation, now I get it.