Question

20mL of 0.2M sodium hydroxide is added to 50mL of 0.2M acetic acid to give 7C)mL of the solution. What is the pH of this solution? Calculate the additional volume of 0.2M NaOH required to make the pH of the solution 4.74. (Ionisation constant of CH₃COOH = 1.8 x 10^-5)

Answer 1

NaOH + CH₃COOH → CH₃COONa + H₂O 

20 mL x 0.2M 50mL x 0.2M 

Milliequivalents of NaOH reacts with CH₃COOH 

= 20 x 0.2 

Milliequivalent of CH₃COONa formed = 20 x 0.2 

Milliequivalent of acetic acid remains

= (50 x 0.2 - 20 x 0.2)

= 30 x 0.2

When NaOH is added, volume of CH₃COONa increases and that of acid decreases. Let the volume of NaOH required to make pH 4.74 is x, then after addition of NaOH, 

Milliequivalent of CH₃COONa = 0.2 (20 + x) 

Milliequivalent of acetic acid= 0.2 (30 - x)