The pH of a 0.1 M solution of NH_{4}OH (havi | vedclass

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Question

(C) For a weak base,the concentration of hydroxyl ions is given by $[OH^{-}] = \sqrt{K_{b} 	imes C}$.$[OH^{-}] = \sqrt{1.0 	imes 10^{-5} 	imes 0.1}

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) For a weak base,the concentration of hydroxyl ions is given by $[OH^{-}] = \sqrt{K_{b} 	imes C}$.$[OH^{-}] = \sqrt{1.0 	imes 10^{-5} 	imes 0.1} = \sqrt{1.0 	imes 10^{-6}} = 1.0 	imes 10^{-3} \ M$.Using the ionic product of water,$K_{w} = [H^{+}] [OH^{-}] = 1.0 	imes 10^{-14}$.$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{1.0 	imes 10^{-14}}{1.0 	imes 10^{-3}} = 1.0 	imes 10^{-11} \ M$.The $pH$ is calculated as $pH = -\log [H^{+}]$.$pH = -\log (1.0 	imes 10^{-11}) = 11$.

The $pH$ of a $0.1 \ M$ solution of $NH_{4}OH$ (having $K_{b} = 1.0 \times 10^{-5}$) is equal to

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