What is the concentration of [OH^-] in distil | vedclass

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(C) At $298 \; K$,the ionic product of water $(K_w)$ is $1 	imes 10^{-14}$.In pure distilled water,the concentration of hydrogen ions $[H^+]$ is equa

Vedclass · Accepted Answer

$1 	imes 10^{-7} \; M$

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(C) At $298 \; K$,the ionic product of water $(K_w)$ is $1 	imes 10^{-14}$.In pure distilled water,the concentration of hydrogen ions $[H^+]$ is equal to the concentration of hydroxide ions $[OH^-]$.Therefore,$[H^+] = [OH^-]$.Since $K_w = [H^+][OH^-]$,we can write $K_w = [OH^-]^2$.Substituting the value of $K_w$: $1 	imes 10^{-14} = [OH^-]^2$.Taking the square root on both sides: $[OH^-] = \sqrt{1 	imes 10^{-14}} = 1 	imes 10^{-7} \; M$.

What is the concentration of $[OH^-]$ in distilled water at $298 \; K$?

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