Question

Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

Answer 1

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition)  or 4.60 (starting from the original buffer solution)

Explanation:

Step 1: Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH = 4.73

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = 4.78

c) pH after further addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826

pH = 4.75 + log(0.826/1.164)

pH = 4.60