Question

Grade Problem 5. (10pts) A weak diprotic base B, has pKo=3 and pkp2=6 Which is / are the most abundant species (B, BH', or BH,2") at pH=11? Final Answer

Answer 1

For a diprotic base, the reaction proceeds in two steps.

$B(aq)+H_2O\rightleftharpoons BH^+(aq)+OH^-(aq)\qquad [K_b_1]$

$BH^+(aq)+H_2O\rightleftharpoons BH_2^{2+}(aq)+OH^-(aq) \qquad [K_b_2]$

Thus, we have,

$K_b_1=\frac{[BH^+][OH^-]}{[B]}\\ \Rightarrow -\log{K_b_1}=-\log{\frac{[BH^+][OH^-]}{[B]}}\\ \Rightarrow pK_b_1=pOH-\log{\frac{[BH^+]}{[B]}}\qquad(1)$

Similarly, for the other equilibrium, we obtain,

$K_b_2=\frac{[BH_2^{2+}][OH^-]}{[BH^+]}\\ \Rightarrow -\log{K_b_2}=-\log{\frac{[BH_2^{2+}][OH^-]}{[BH^+]}}\\ \Rightarrow pK_b_2=pOH-\log{\frac{[BH_2^{2+}]}{[BH^+]}} \qquad (2)$

We know that, pOH= 14 - pH = 14- 11 = 3

pK_b1 = 3 and pK_b3 = 6

Using these values, in equation (1), we have,

$pK_b_1=pOH-\log{\frac{[BH^+]}{[B]}}\\ \Rightarrow 3=3-\log{\frac{[BH^+]}{[B]}}\\ \Rightarrow \log{\frac{[BH^+]}{[B]}}=0\\ \Rightarrow {\frac{[BH^+]}{[B]}}=1\\ \Rightarrow [BH^+]=[B]$

Thus, concentrations of B and BH⁺ are equal.

From equation (2), we have,

$pK_b_2=pOH-\log{\frac{[BH_2^{2+}]}{[BH^+]}} \\ \Rightarrow 6 =3-\log{\frac{[BH_2^{2+}]}{[BH^+]}}\\ \Rightarrow\log{\frac{[BH_2^{2+}]}{[BH^+]}}=-3\\ \Rightarrow {\frac{[BH_2^{2+}]}{[BH^+]}}=10^{-3}\\ \Rightarrow [BH_2^{2+}]= 0.001[BH^+]$

Thus, [BH₂²⁺] is only 0.001 times [BH⁺]. Hence it is least abundant.

Most abundant species are [B] and [BH⁺] which are present in equal concentration.