Question

The pH of an aqueous solution of 0.593 M quinoline (a weak base with the formula C9H7N) is

Answer #1

**kb of quinoline = 6.3*10^-10**

**C9H7N dissociates as:**

**C9H7N +H2O -----> C9H7NH+ + OH-**

**0.593 0 0**

**0.593-x x x**

**Kb = [C9H7NH+][OH-]/[C9H7N]**

**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 ((6.3*10^-10)*0.593) = 1.933*10^-5**

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

**so, x = 1.933*10^-5 M**

**so.[OH-] = x = 1.933*10^-5 M**

**use:**

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

**= -log (1.933*10^-5)**

**= 4.7138**

**use:**

**PH = 14 - pOH**

**= 14 - 4.7138**

**= 9.29**

**Answer: 9.29**

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pH
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