If K_ w =49 10^ -14 , what will be neutral pH of H_2 O ? - Vidyadip

Question

If $\mathrm{K}_{\mathrm{w}}=49 \times 10^{-14}$, what will be neutral pH of $\mathrm{H}_2 \mathrm{O}$ ?

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Answer

$\text{K}_{\text{w}}=49\times10^{}-14[\text{H}_3\text{O}^+][\text{OH}^-]$
$=49\times10^{-14}$
$[\text{H}_3\text{O}^+]=[\text{OH}^-]$
$\Rightarrow[\text{H}_3\text{O}^+]^2=49\times10^{-14}$
$\Rightarrow[\text{H}_3\text{O}^+]=7\times10^{-7}\text{ mol L}^{-1}$
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log7\times10^{-7}$
$=-\log7-\log10^{-7}$
$=-\log7+7\log10$
$\text{pH}=-0.8451+7.000$
$=6.1549$

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