Question

(K_w = 1.0 x 10^–14 ;    pH = –log[H₃O⁺];   pK_a = –logK_a;        pH = pK_a + log( base/acid)

1. The following equation shows the ionization of boric acid, H₃BO₃:

           H₃BO₃(aq) + H₂O ⇌ H₃O⁺(aq) + H₂BO₃^– (aq);       K_a = 5.4 x 10^–10

   (a) Calculate the pH of 0.100 M boric acid solution.

   (b) Write an equation for the ionization of borate ion, H₂BO₃^– , in aqueous solution (that is, its reaction with water) and calculate its K_b value, and determine the pH of solution of 0.10 M sodium borate, NaH₂BO₃(aq).

   (c) A borate buffer solution contains 0.15 M boric acid and 0.050 M sodium borate. What is the pH of this solution?

   (d) What is the molar ratio of borate ion, H₂BO₃^–, to boric acid, [H₂BO₃^–]/[H₃BO₃], in a buffer solution with pH = 8.70? If the concentration of boric acid in the buffer is 0.200 M, what should be the concentration of the borate ion?

   (e) If the solution in (d) is reacted with 0.020 M NaOH, how would this change the ratio:

[H₂BO₃^–] / [H₃BO₃], and what would be the pH of the resulting solution?

Answer 1

a)

pH of boric acid = 1/2(pKa - logC) ( c = concentration of boric acid)

pH = 1/2(-log(5.4 * 10^-10) - log(0.1))

pH = 5.134

b)

H2BO3- + H2O ------> H3BO3 + OH-

kb = 1/Ka = 1/(5.4*10^-10) = 1.8581 * 10^9

c)

pH = pKa + log[salt]/[acid]

pH = -log(5.4*10^-10) + log(0.05/0.15)

pH = 8.79

d)

pH = pKa + log[salt]/[acid]

8.7 = -log(5.4*10^-10) + log(salt/acid)

[salt]/[acid] = 0.2706

[salt]/0.2 = 0.2706

[salt] = borate ion concentration = 0.05412

[acid] = boric acid concentration

e)

when 0.02 NaOH is added [borate] = 0.05412 + 0.02 = 0.07412

[boric acid] = 0.2 - 0.02 = 0.18

pH = pKa + log [salt]/[acid]

pH = -log(5.4 * 10^-10) + log(0.07412/0.18)

pH = 8.882