If 0.4 g of NaOH is present in 1 L of solut | vedclass

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(D) The molar mass of $NaOH$ is $40 \ g/mol$.The number of moles of $NaOH = \frac{0.4 \ g}{40 \ g/mol} = 0.01 \ mol$.Since the volume of the solution

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$12$

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(D) The molar mass of $NaOH$ is $40 \ g/mol$.The number of moles of $NaOH = \frac{0.4 \ g}{40 \ g/mol} = 0.01 \ mol$.Since the volume of the solution is $1 \ L$,the molarity $[NaOH] = 0.01 \ M = 10^{-2} \ M$.Since $NaOH$ is a strong base,$[OH^-] = [NaOH] = 10^{-2} \ M$.Using the ionic product of water,$[H^+][OH^-] = 10^{-14}$,we get $[H^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ M$.$pH = -\log[H^+] = -\log(10^{-12}) = 12$.

If $0.4 \ g$ of $NaOH$ is present in $1 \ L$ of solution,then its $pH$ will be:

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