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).
B. 4.75
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 CH_3COOH. 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: CH_3COOH + NaOH CH_3COONa + H_2O Millimoles of CH_3COOH remaining after the reaction = 2.84 - 1.42 = 1.42 mmol Millimoles of CH_3COONa formed = 1.42 mmol Since the solution contains a weak acid and its conjugate base, it forms an acidic buffer. Using the Henderson-Hasselbalch equation: pH = pK_a + ( [ Salt ] [ Acid ] ) Substituting the values: pH = 4.75 + ( 1.42 1.42 ) pH = 4.75 + (1) pH = 4.75 Answer: 4.75
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