Question

20. A 50.00-mL solution of 0.0350 M benzoic acid (Ka 6.4 x 10) is titrated with a 0.0134 M solution of sodium hydroxide as the titrant. What is the pH of the acid solution after 15.00 mL of titrant have been added? (Kw 1.00 x 10-14) A) 2.83 B) 9 C) 4.19 D) 3.37 E) 1.46 9.6

Answer 1

When benzoic acid is filtrated with sodium hydroxide, it will form sodium benzoate and water. When sodium hydroxide is used in lesser volume than that of benzoic acid, then benzoic acid salt i.e. sodium benzoate and excess benzoic acid form an acidic buffer solution C_6 H_5 COOH + NaOH rightarrow C_6 H_5 COONa + H_2 O \r\n 50 ml 0.0350 (M) solution of benzoic acid = (50 times 0.0350) millimole benzoic acid = 1.75 millimole benzoic acid \r\n 15 ml 0.0134 (M) NaOH solution = 0.201 millimole NaOH solution So salt form i.e. millimole of sodium benzoate = 0.201 millimole [N > B: -] Here 1 mole of NaOH reacts with 1 mole benzoic acid. So 0.201 moles of NaOH reacts with 0.201 moles of benzoic acid to form 0.201 moles of sodium benzoate salt] \r\n millimole of benzoic acid present after titration = (1.75 - 0.201) millimole = 1.549 millimole Here, k_a = 6.4 times 10^-5 (for C_6 H_5 COOH)