Question

A buffer solution contains 0.30 mol of hypochlorous acid (HClO) and 0.83 mol of sodium hypochlorite (NaOCl) in 7.50 L. The Ka of hypochlorous acid (HClO) is Ka = 3e-08.

(a) What is the pH of this buffer?

pH =

(b) What is the pH of the buffer after the addition of 0.24 mol of NaOH? (assume no volume change)

pH =

(c) What is the pH of the original buffer after the addition of 0.25 mol of HI? (assume no volume change)

pH =

Answer 1

pH pf any buffer can be calculated according to:

pH=pKa+log([salt]/[acid])

pKa=-logKa

(a) [salt]=0.83mol/7.5L=0.111 M

[acid]=0.30mol/7.5L=0.04

pH=7.52+log(0.111/0.04)=7.52+0.44=7.96

(b) What is the pH of the buffer after the addition of 0.24 mol of NaOH? (assume no volume change)

HClO+NaOH=NaClO + H2O

You lose 0.24 mol of acid and obtain 0.24 mol salt

New concentration are:

[salt]=(0.83mol+0.24mol))/7.5L=0.143 M

[acid]=(0.30mol-0.24mol)/7.5L=0.008

pH=7.52+log(0.143/0.008)=7.52+1.25=8.77

(c) What is the pH of the original buffer after the addition of 0.25 mol of HI? (assume no volume change)

NaClO+HI=NaI+HClO

You lose 0.25 mol of salt and obtain 0.25 mol of acid

New concentration are:

[salt]=(0.83mol-0.25mol))/7.5L=0.077 M

[acid]=(0.30mol+0.25mol)/7.5L=0.073

pH=7.52+log(0.077/0.073)=7.52+0.02=7.54