Question

The dibasic compound B forms BH+ and BH22+ with Kb1 = 1.00 x 10-5 and Kb2 = 1.00x10-9. Find the pH and concenrations of B, BH+,and BH22+ in each of the following solutions: (a) 0.100 M B; (b) 0.100 M BH+Br-; (c) 0.100 M BH22+ . For (b) use the approximation [BH+] ~ 0.100 M.

Answer 1

(a): 0.100 M B:

For the first protonation of B:

---B(aq) + H2O(l) < ------> BH⁺(aq) + OH^-(aq) : Kb1 = 1.00*10^-5

I: 0.100, ----------------------- 0 ---------- 0  

C: - X, ----------------------- +X ---------- +X  

E: (0.100 - X) --------------- X ------------ X

1.00*10^-5 = [BH⁺(aq)]*[OH^-(aq)] / [B(aq)] = X² / (0.100 - X)

=> X = 9.95*10^-4 M

=> [BH⁺(aq)] = [OH^-(aq)] = X = 9.95*10^-4 M  

[B(aq)] = (0.100 - X) = 0.100 - 9.95*10^-4 M = 0.099005 M (answer)

For the 2nd protonation of B:

---BH⁺(aq) + H2O(l) < ------> BH₂²⁺(aq) + OH^-(aq) : Kb1 = 1.00*10^-9

I: 9.95*10^-4 M---------------------- 0 ------------ 9.95*10^-4 M

C: - Y, ----------------------------- +Y ---------- +Y  

E: (9.95*10^-4 - Y) ----------------- Y ------------ (9.95*10^-4 +Y)

1.00*10^-9 = [BH₂²⁺(aq)]*[OH^-(aq)] / [BH⁺(aq)] = Y*(9.95*10^-4 +Y) / (9.95*10^-4 - Y)

=> Y = 1.00*10^-9 M

=> [BH₂²⁺(aq)] = Y = 1.00*10^-9 M (Answer)

[BH⁺(aq)] = (9.95*10^-4 - Y) = (9.95*10^-4 - 1.00*10^-9) = 9.95*10^-4 M (answer)

(b): 0.100 M BH⁺(aq):

For the protonation of BH⁺(aq):

---BH⁺(aq) + H2O(l) < ------> BH₂²⁺(aq) + OH^-(aq) : Kb1 = 1.00*10^-9

I: 0.100 M-------------------------- 0 -------------- 10^-7 M  

C: - Y, ----------------------------- +Y ------------ +Y  

E: (0.100 - Y) ---------------------- Y ------------ (10^-7 +Y)

1.00*10^-9 = [BH₂²⁺(aq)]*[OH^-(aq)] / [BH⁺(aq)] = Y*(10^-7 +Y) / (0.100 - Y)

=> Y = 9.95*10^-6 M  

=> [BH₂²⁺(aq)] = Y = 9.95*10^-6 M  (Answer)

[BH⁺(aq)] = (0.100 - Y) = (0.100- 9.95*10^-6) = 0.100 M (answer)

Deprotonation of BH⁺(aq) gives B.

For the deprotonation of BH⁺(aq):

BH⁺(aq) + H2O(l) < ------> B(aq) + H₃O⁺(aq) : Ka1 = Kw / Kb1 = 1.00*10^-9

I: 0.100 M----------------------- 0 ------- 10^-7 M  

C: - Y, -------------------------- +Z ------- +Z  

E: (0.100 - Z) ------------------ Z ------- (10^-7 +Z)

1.00*10^-9 = [B(aq)]*[H₃O⁺(aq)] / [BH⁺(aq)] = Z*(10^-7 +Z) / (0.100 - Z)

=> Z = 9.95*10^-6 M  

=> [B(aq)] = Y = 9.95*10^-6 M  (Answer)