20 ml. of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid solution to give 70 mL of the solution. What is the pH of this solution? Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. (Ionisation constant of CH3COOH = 1.8 × 10 - 5 ).

Question

20 ml. of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid solution to give 70 mL of the solution. What is the pH of this solution? Calculate the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74. (Ionisation constant of CH3COOH = 1.8 × 10 - 5 ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

mmol of NaOH = 20 × 0.2 = 4 mmol of acetic acid = 50 × 0.2 = 10 After neutralisation, buffer solution is formed which contain 6 mmol CH3COOH and 4 mmol CH3COONa. pH = pKa + log ( [ CH3COONa ]/ CH3COOH ]) = - log ( 1.8 × 10 - 5 ) + log ( 4 / 6 ) = 4 . 56 Now, let x mmol of NaOH is further added so that pH of the resulting buffer solution is 4.74. Now, the buffer solution contains (4 + x) mmol CH3COONa and (6 - x) mmol of CH3COOH. 4.74 = - log ( 1.9 × 10 - 5 ) + log ( 4 + x/ 6 - x) ⇒ ( 4 + x/ 6 - x) = 1 ⇒ x = 1.0 mmol = 0.2 × V ⇒ V = 5.0 mmol NaOH.

Q. 20ml. of 0.2M sodium hydroxide is added to 50mL of 0.2M acetic acid solution to give 70mL of the solution. What is the pH of this solution? Calculate the additional volume of 0.2MNaOH required to make the pH of the solution 4.74. (Ionisation constant of CH3​COOH=1.8×10−5).