Many common weak bases are derivatives of NH3, where one or more of the hydrogen

Question

Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent. Such reactions can be generically symbolized as NX_3 (aq) + H_2 O(1) HNX_3 + (aq) + OH^- (aq) where NX_3 is the base and HNX_3 is the conjugate acid. The equilibrium-constant expression for this reaction is K_b = [HNX_3^+][OH^-]/[NX_3] where Kb is the base ionization constant. The extent of ionization, and thus the strength of the base, increases as the value of K_b increases. K_a and K_b are related through the equation K_a times K_b = K_w As the strength of an acid increases, its KA value increase and the strength of the conjugate base decreases (smaller K_b value). Part B If K_b for NX_3 is 9.0times10^-6, what is the percent ionization of a 0.325 M aqueous solution of NX_3? Express your answer numerically to three significant figures. Part C If K_b for NX_3 is 9.0times10^-6, what is the the pK_a for the following reaction? HNX_3^+ (aq) + H_2O(1) NX_3(aq) + H_3O^+ (aq) Express your answer numerically to two decimal places.

Explanation / Answer

%ion = [OH-] / M * 100%

[OH-] = ?

[OH-] = sqrt(Kb*M) = sqrt(0.325*9*10^-6)= 0.00171

then

% ion = (0.00171)/(0.325) * 100 = 0.5261 %

C)

then

14 = pKa + pKb

pKb = -log(kb) = -log(9*10^-6) = 5.04575

pKa = 14-pKB = 14-5.04575

pKa = 8.95425

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