The hydrogen ion concentration of 0.001 M Na | vedclass

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Question

(B) $NaOH$ is a strong base,so it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$. Given concentration of $NaOH = 0.001 \ M = 10^{-3} \ M$.

Vedclass · Accepted Answer

$1 	imes 10^{-11} \ 	ext{mole/litre}$

Vedclass · Answer

(B) $NaOH$ is a strong base,so it dissociates completely as: $NaOH ightarrow Na^+ + OH^-$. Given concentration of $NaOH = 0.001 \ M = 10^{-3} \ M$. Therefore,$[OH^-] = 10^{-3} \ M$. We know that the ionic product of water at $25^{\circ}C$ is $K_w = [H^+][OH^-] = 10^{-14}$. $[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ 	ext{mole/litre}$.

The hydrogen ion concentration of $0.001 \ M$ $NaOH$ solution is

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