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Q 7.48 Assuming complete dissociation, calculate the pH of the following solutions:
(d) 0.002 M KOH
Answers (1)
$\begin{aligned} & \mathrm{KOH}_{(\mathrm{aq})} \longleftrightarrow \mathrm{K}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-} \\ & {\left[\begin{array}{l}{\left[\mathrm{OH}^{-}\right]=[\mathrm{KOH}]} \\ \begin{array}{rl}\Rightarrow & {\left[\mathrm{OH}^{-}\right]=.002}\end{array} \\ \text { Now, } \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] \\ \quad=2.69\end{array}\right.} \\ & \begin{array}{c}\therefore \mathrm{pH}=14-2.69 \\ \quad=11.31\end{array}\end{aligned}$
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