Question

A certain weak base has a Kb of 7.50 × 10-7. What concentration of this base will produce a pH of 10.02?

Answer 1

B + H2O <=> CA + HO-
Equilibrium equation:
Kb= [HO-][CA]/[B]
[B]= concentration of base [CA]= concentration of conjugate acid
Also [HO-]= [CA] = x

14 - pH = pOH............................14-10.02 = 3.98
-log[HO-] = pOH................................... -log[HO-]= 3.98
[HO-] = 10^ (-pOH) .............................. [HO-] = 10^-3.98 = 1.047 x 10^-4

Since [HO-]= [CA] = x and we now know that x is 1.047 x 10^-4

So using the equilibrium equation and replacing x for [HO-] and [CA] we get:

Kb = x^2/[B]

Solve for [B]

[B] = x^2/Kb

[B] = (1.047 x 10^-4)^2 / (7.50 × 10-7)

[B]= 1.461 * 10^-2