Question

Be sure to answer all parts. Determine the K, of a weak base if a 0.15 M solution of the base has a pH of 10.77 at 25°C. х 10 Къ (Enter your answer in scientific notation.)

Answer 1

pH of the solution is 10.77.

pH is defined as the negative logarithm of [H⁺].

Hence, we can calculate the [H⁺] at equilibrium as follows:

$pH = -\log[H^+] \\ \Rightarrow [H^+] = 10^{-pH} = 10^{-10.77} \approx 1.698 \times 10^{-11}$

Note that the [OH^-] and [H⁺] of an aqueous solution at 25 ⁰C are related as

H* OH ]= 1.00x 10

Hence, we can calculate the OH^-  concentration at equilibrium as follows:

$[H^+][OH^-] = 1.00 \times 10^{-14} \\ \Rightarrow [OH^-] = \frac{1.00 \times 10^{-14}}{1.698 \times 10^{-11}} = 5.888 \times 10^{-4} \ M$

Now, we can write the equilibrium reaction of weak base with water as follows:

$B_{(aq)} + H_2O_{(l)} \rightarrow BH^+_{(aq)} + OH^-_{(aq)}$

Given that the initial concentration of B is 0.15 M,  we can create the ICE table for the above equilibrium as follows:

	$B_{(aq)}$	$BH^+_{(aq)}$	$[OH^-]_{(aq)}$
Initial, M	0.15	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.15-x	x	x

Now, we already know the value of x in the ICE table from the pH data as

x = [OH^-]=$5.888 \times 10^{-4}$

Hence, we can calculate the base equilibrium constant K_b as follows:

$K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{x \times x}{0.15-x} = \frac{5.888 \times 10^{-4} \times 5.888 \times 10^{-4}}{0.15 - 5.888 \times 10^{-4}} \\ \Rightarrow K_b\approx 2.32 \times 10^{-6}$

Hence, the Kb of the base is about $2.32 \times 10^{-6}$ . (Rounded to three significant figures).