What is the $\mathrm{pH}$ of the $\mathrm{NaOH}$ solution when $0.04 \mathrm{~g}$ of it dissolved in water…
What is the $\mathrm{pH}$ of the $\mathrm{NaOH}$ solution when $0.04 \mathrm{~g}$ of it dissolved in water and made to $100 \mathrm{~mL}$ solution?
- $2$
- $1$
- $13$
- $12$
Solution
Molecular mass of $\mathrm{NaOH}=23+16+1=40$
$\begin{aligned}
& \text { Molarity }=\frac{\text { Mass of solute (in } \mathrm{g})}{\text { Molecular mass of the solute }} \\
& \times \text { Volume of solution (in L) } \\
& =\frac{0.04 \times 1000}{40 \times 100}=10^{-2} \\
& \mathrm{OH}^{-}=10^{-2} \mathrm{~mol} / \mathrm{L} \\
& \text { So, } \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\
& \mathrm{pOH}+\mathrm{pH}=14 \text { or } \mathrm{pH}=14-2=12 \\
&
\end{aligned}$
Asked in: AP EAMCET 2015
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