Question

20 ml of 0.2 M NaOH is added to 50 ml of 0.2 M acetic acid. What is the pH of the resulting solution? Calculate the additional volume of 0.2 M NaOH required for making the solution of pH 4.74. The ionization constant of acetic acid is $1.8\times {10}^{-5}.$ $(log1.8=0.2552,log0.66=-0.18)$

• A. pH = 3.56
  Additional volume = 6 mL
• B. pH = 3.56
  Additional volume = 5 mL
• C. pH = 4.56
  Additional volume = 5 mL
• D. pH = 4.56
  Additional volume = 4 mL

Answer 1

The correct option is C pH = 4.56

Additional volume = 5 mL
20 ml of 0.2 M NaOH will react with 20 ml of 0.2 M acetic acid.
Unreacted [Acid] = 30 ml of 0.2 M present in 70 ml
$=\frac{30\times 0.2}{70}=\frac{6}{70}$ M
[Salt] formed = 20 ml of 0.2 M present in 70 ml
$=\frac{20\times 0.2}{70}=\frac{4}{70}$ M
$p{K}_a=-log1.8\times {10}^{-5}$
$p{K}_a=5-log1.8=4.74$
$pH=p{K}_a+log\frac{\left[\text{salt}\right]}{\left[\text{Acid}\right]}$
$pH=4.74+log\frac{4}{70}\times \frac{70}{6}$
$pH=4.74+log0.66=4.56$
$pH=pKa=4.74$ if [Acid] = [Salt]
Addition of 25 mL of 0.2 M NaOH to 50 ml of 0.2 M acetic acid leaves same concentration of salt and unreacted acid.
As 20 mL is already added addition of an extra volume of 5 mL NaOH (25 - 20 = 5 ml) will result in pH = 4.74