A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00L

Question

A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00L

a) What is the pH of the buffer? b) What is the pH of the buffer after the addition of 0.02 mol of KOH? c) What is the pH of the buffer after the addition of 0.02 mol of HNO3?

Answer 1

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76

(a). pH = 4.76 + log [0.13]/[0.10] = 4.76 + 0.11 = 4.87

(b) KOH + CH3COOH =>H2O + CH3COOK so [acid] goes down and [salt] goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt

pH = 4.76 + log [0.15]/[0.08] = 4.76 + 0.27 = 5.03

(c) HNO3 + CH3COO- => CH3COOH + NO3-, so [acid] goes up and [salt] goes down. Assuming no change in volume, you have 0.10 acid + 0.02 = 0.12 acid and 0.13 salt - 0.02 = 0.11 salt.

pH = 476 + log [0.11]/[0.12] = 4.76 + (-0.038) = 4.72

Answer 2

This is a question about buffers, and requires use of the Henderson-Hasselbach equation

pH = pKa + log([A-]/[HA])

The pKA is a property of the acid, that will be given in the problem or can be looked up. The value is 4.75 for acetic acid. [A-] will be the concentration OR number of moles of the conjugate base (just the anion). [HA] refers to concentration or number of moles of the acid.

An interesting feature of buffers is that the pH doesn't depend on absolute concentrations, but instead on the ratio of the two components.

Pluggin in the values for part a:

pH = 4.75 + log(.13/.10) = 4.86

b) Addition of a strong base will convert an equivalent amount of the acid into a base, changing the ratio. This is best done with an ICE table

HA + OH^- --> A^- + H₂O

I .10 .02 .13

C -.02 -.02 +.02

E .08 0 .15

Plug in the new values to the equation

pH = 4.75 + log(.15/.08) = 5.02

c) Similar to b, addition of a strong acid will convert some of the base form to the acid form.

A^- + H⁺ --> HA

I .10 .02 .13

C -.02 -.02 +.02

E .08 0 .15

pH = 4.75 + log(.08/.15) = 4.48

Notice that these pH changes are much smaller than you would get with the addition of acid or base to an unbuffered solution.