Question

The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be : (pKa for CH3COOH = 4.74)

• A. 4.74
• B. 8.87
• C. 9.10
• D. 8.57

Answer 1

Answer: (B)

8.87

Answer 2

CH3COOH + OH– $\longrightarrow$ CH3COO– + H2O

t = 0 20 20

t = eq – – 20

So, [CH3COO–] = $\frac { 20 } { 200 }$ = 0.1 M

pH = 7 + $\frac { 1 } { 2 }$ log C = 7 + 2.37 + $\frac { 1 } { 2 }$ log 10–1 = 7 + 2.37 – 0.5 = 8.87