## Chemistry 12th Edition

$[NaOH]=0.62M$ We know that $[H^+]\times [OH^-]=1\times 10^{-14}$ Thus $[H^+]=\frac{1\times 10^{-14}}{[OH]^-}=\frac{1\times 10^{-14}}{ 0.62}=1.61\times 10^{-14}$