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$0.023 \mathrm{~g}$ of sodium metal is reacted with $100 \mathrm{~cm}^{3}$ of water. The $\mathrm{pH}$ of the resulting solution is
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Verified Answer
The correct answer is:
12
$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$
$(2 \times 23=46) \quad(2 \times 40=80)$
$0.023 \quad 0.04$
${\left[\mathrm{OH}^{-}\right]=\frac{0.04}{40} \times 10=10^{-2} }$
$\because \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]$
$\mathrm{pOH}=2$
$\because \quad \mathrm{pH}+\mathrm{pOH}=14$
$\mathrm{pH}=14-\mathrm{pOH}$
$\mathrm{pH}=14-2$
$\mathrm{pH}=12$
$(2 \times 23=46) \quad(2 \times 40=80)$
$0.023 \quad 0.04$
${\left[\mathrm{OH}^{-}\right]=\frac{0.04}{40} \times 10=10^{-2} }$
$\because \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]$
$\mathrm{pOH}=2$
$\because \quad \mathrm{pH}+\mathrm{pOH}=14$
$\mathrm{pH}=14-\mathrm{pOH}$
$\mathrm{pH}=14-2$
$\mathrm{pH}=12$
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