Given that the dissociation constant for H_2O | vedclass

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(D) $KOH$ is a strong base,so it dissociates completely as: $KOH ightarrow K^+ + OH^-$.Given $[KOH] = 0.001 \ M = 1 	imes 10^{-3} \ M$,therefore $[

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$11$

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(D) $KOH$ is a strong base,so it dissociates completely as: $KOH ightarrow K^+ + OH^-$.Given $[KOH] = 0.001 \ M = 1 	imes 10^{-3} \ M$,therefore $[OH^-] = 1 	imes 10^{-3} \ M$.Using the ionic product of water: $[H^+] 	imes [OH^-] = K_w = 1 	imes 10^{-14}$.$[H^+] = \frac{1 	imes 10^{-14}}{[OH^-]} = \frac{1 	imes 10^{-14}}{1 	imes 10^{-3}} = 1 	imes 10^{-11} \ M$.$pH = -\log[H^+] = -\log(10^{-11}) = 11$.

Given that the dissociation constant for $H_2O$ is $K_w = 1 \times 10^{-14} \ mol^2 \ L^{-2}$,what is the $pH$ of a $0.001 \ M \ KOH$ solution?

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