Question

1) What concentration of ammonia is required to have a solution with a pH of 11.32? K_b = 1.8x10^-5

2) Calculate the K_a of the conjugate acid of a weak base with a K_b = 1.69 x 10⁻⁴.

3) A diprotic acid, H₂A has the following K_as: K_a1 = 4.24 x 10^-7 and K_a2 = 8.51 x 10^-11. What is the K_b of HA^-?

Answer 1

1. Here we have a amonia with pH of 11.32 and Kb = 1.8 * 10^-5

NH3 H+ + NH2-

We know pH of 11.32

[H+] = e^-pH

[H+] = e^-11.32

= 1.21 * 10^-5

Lets find out equilibrium concentration for ammonia

NH3    NH2- + H+

Initial x 0 0

Change -1.21 * 10-5 1.21 * 10-5 1.21 * 10-5

Wquilibrium x - 1.21 * 10-5 1.21 * 10-5   1.21 * 10-5

Kb = [NH2-][H+]/[NH3]

Kb = 1.8 * 10^-5

1.8*10^-5 = 1.21*10^-5*1.21*10^-5/(x - 1.21*10^-5)

0.55 * 10⁵ = x - 1.21 * 10^-5/1.4641 * 10^-10

0.805 * 10^-5 = x - 1.21 * 10^-5

x = 2.01 * 10^{-5.......................} Concentration.