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If $0.4 \mathrm{gm~} \mathrm{NaOH}$ is present in 1 litre solution, then its $\mathrm{pH}$ will be
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The correct answer is:
12
$[\mathrm{NaOH}]=\frac{0.4}{40}=0.01 \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{M}$
$\left[H^{+}\right]=10^{-12}, p H=-\log \left[H^{+}\right]=12$
$\left[H^{+}\right]=10^{-12}, p H=-\log \left[H^{+}\right]=12$
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