20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? ( pK_a value of acetic acid is 4.75 ).

Step-by-step solution

For complete neutralization of 20 mL of acetic acid, the millimoles of NaOH required is: 28.4 0.1 = 2.84 mmol Thus, 20 mL of the acetic acid solution contains 2.84 mmol of CH₃COOH . For the preparation of solution (X), the millimoles of NaOH added is: 14.2 0.1 = 1.42 mmol The reaction between acetic acid and sodium hydroxide is: CH₃COOH + NaOH CH₃COONa + H₂O Millimoles of CH₃COOH remaining after the reaction = 2.84 - 1.42 = 1.42 mmol Millimoles of CH₃COONa formed = 1.42 mmol Since the solution contains a weak acid