The degree to which a base dissociates is given by the base ionization constant,

Question

The degree to which a base dissociates is given by the base ionization constant, K_b. For the generic week base B. B (aq) + H_2O(I) rightharpoonoverleftharpoon BH^+ (aq) + OH^- (aq) this constant is given by K_b = [BH^+][OH^-]/[B] Strong bases will have a higher K_b value. Similarly bases will have a higher percent ionization value. Percent ionization = [OH^-] equilibrium/[B] initial times 100% Strong bases, for which K_b is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization. Ammonia, NH_3, is a weak base with a K_b value of 1.8 times 10^-5. What is the pH of a 0.375 M ammonia solution? Express your answer numerically to two decimal places. pH = What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units. % ionization =

Explanation / Answer

A)

NH3 + H2O -----> NH4+ + OH-

0.375 0 0

0.375-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

since kb is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.375) = 2.598*10^-3

pOH = -log [OH-] = -log (2.598*10^-3) = 2.59

PH = 14 - pOH = 14 - 2.59 = 11.41

Answer: 11.41

B)

% ionisation = x*100/c

= (2.598*10^-3)*100 / (0.375)

= 0.693 %

Answer: 0.693 %

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