Question

Calculate the pH of a solution of 0.10 M acetic acid after 100 mL of this solution is treated with 50.0 mL of 0.10 M NaOH.

[pKa of acetic acid= 4.74]

• A. $3.65$
• B. $4.74$
• C. $10.35$
• D. $11.71$

Answer 1

The correct option is B $4.74$

Number of moles of acetic acid $=\frac{100mL}{1000mL}\times 0.10M=0.010moles$

Number of moles of NaOH $=\frac{50mL}{1000mL}\times 0.10M=0.005moles$

$C{H}_{3}COOH+NaOH\to C{H}_{3}COONa+H_{2}O$

$0.005$ moles of $NaOH$ will react with $0.005$ moles of acetic acid to form $0.005$ moles of sodium acetate. $0.010-0.005=0.005$ moles of acetic acid will remain.

Use the Henderson-Hasselbalch equation to find its pH:

pH=pKa+log($\frac{\left[conjugatebase\right]}{\left[weakacid\right]}$)

Since the number of moles of sodium acetate (conjugate base) is equal to the number of moles of acetic acid(weak acid), the pH of the solution will be equal to pKa of the solution which is $4.74$

Option (B) is correct.