What is the dissociation constant for NH_4OH | vedclass

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(C) Given: $pH = 11.27$,$C = 0.1 \ N$,$K_w = 7.1 	imes 10^{-15}$.First,calculate $[H^+]$: $[H^+] = 10^{-pH} = 10^{-11.27} = 10^{-12} 	imes 10^{0.73}

Vedclass · Accepted Answer

$1.75 	imes 10^{-5}$

Vedclass · Answer

(C) Given: $pH = 11.27$,$C = 0.1 \ N$,$K_w = 7.1 	imes 10^{-15}$.First,calculate $[H^+]$: $[H^+] = 10^{-pH} = 10^{-11.27} = 10^{-12} 	imes 10^{0.73} = 5.37 	imes 10^{-12} \ M$.Next,calculate $[OH^-]$: $[OH^-] = \frac{K_w}{[H^+]} = \frac{7.1 	imes 10^{-15}}{5.37 	imes 10^{-12}} = 1.322 	imes 10^{-3} \ M$.For a weak base,$[OH^-] = \sqrt{K_b 	imes C}$.Squaring both sides: $[OH^-]^2 = K_b 	imes C$.$K_b = \frac{[OH^-]^2}{C} = \frac{(1.322 	imes 10^{-3})^2}{0.1} = \frac{1.747 	imes 10^{-6}}{0.1} = 1.747 	imes 10^{-5}$.

What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1 \ N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$)?

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