Question

If the Kb of a weak base is 2.8 × 10-6, what is the pH of a 0.30 M solution of this base?

Answer #1

**let the weak base is BOH**

**BOH dissociates as:**

**BOH +H2O -----> B+ + OH-**

**0.3 0 0**

**0.3-x x x**

**Kb = [B+][OH-]/[BOH]**

**Kb = x*x/(c-x)**

**Assuming x can be ignored as compared to c**

**So, above expression becomes**

**Kb = x*x/(c)**

**so, x = sqrt (Kb*c)**

**x = sqrt ((2.8*10^-6)*0.3) = 9.165*10^-4**

**since c is much greater than x, our assumption is
correct**

**so, x = 9.165*10^-4 M**

**so.[OH-] = x = 9.165*10^-4 M**

**use:**

**pOH = -log [OH-]**

**= -log (9.165*10^-4)**

**= 3.04**

**use:**

**PH = 14 - pOH**

**= 14 - 3.04**

**= 10.96**

**Answer: 10.96**

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