Use the References to access Important values If needed for

Use the References to access Important values If needed for this question. The pH of an aqueous solution of 0.356 M benzoic acid, cHCOOH is

1) The ionization of benzoic acid is given as

C₆H₅COOH (aq) ---------> C₆H₅COO^- (aq) + H⁺ (aq)

The acid ionization constant K_a is written as

K_a = [C₆H₅COO^-][H⁺]/[C₆H₅COOH] = 6.5*10^-5

=====> 6.5*10^-5 = (x).(x)/(0.356 – x) [1:1 nature of ionization of benzoic acid]

Since K_a is small, we can assume (0.356 – x) M 0.356 M and write

6.5*10^-5 = x²/(0.356)

====> x² = 2.314*10^-5

====> x = 4.810*10^-3

Therefore, [H⁺] = 4.810*10^-3 M and pH = -log [H⁺] = -log (4.810*10^-3) = 2.3178 2.32 (ans).

2) The reaction of aniline with water can be shown as below.

C₆H₅NH₂ (aq) + H₂O (l) ---------> C₆H₅NH₃⁺ (aq) + OH^- (aq)

Since OH^- is formed, we will work with be base ionization constant, K_b defined as

K_b = [C₆H₅NH₃⁺][OH^-]/[C₆H₅NH₂] = 4.3*10^-10

====> 4.3*10^-10 = x²/(0.322 – x)

Since K_b is small, we can assume (0.322 – x) M 0.322 M and write

4.3*10^-10 = x²/(0.322)

====> x² = 1.3846*10^-10

====> x = 1.1767*10^-5

Therefore, [OH^-] = 1.1767*10^-5 M and pOH = -log [OH^-] = -log (1.1767*10^-5) = 4.9293.

The pH of the solution is pH = 14 – pOH = 14 – 4.9293 = 9.0707 9.07 (ans).