JEE MAIN - Chemistry (2023 - 29th January Morning Shift - No. 13)
Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is $$x\times10^{-1}$$. The value of $$x$$ is ___________ (Nearest integer).
Assume complete dissociation.
Answer
5
Explanation
$$
\begin{aligned}
& \mathrm{pH}=12, \\\\
& \mathrm{pH}+\mathrm{pOH}=14 \\\\
& \mathrm{pOH}=14-12=2 \\\\
& {\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{~mol} \mathrm{~L}} \\\\
& \underset{5 \times 10^{-3}}{\mathrm{Ca}(\mathrm{OH})_2} \longrightarrow \underset{5 \times 10^{-3}}{\mathrm{Ca}^{2+}}+\underset{10^{-2}}{2 \mathrm{OH}^{-}}
\end{aligned}
$$
Moles of $\mathrm{Ca}(\mathrm{OH})_2$ in $1000 \mathrm{~mL}=5 \times 10^{-3}$
Millimoles in $100 \mathrm{~mL}=5 \times 10^{-1}$
Moles of $\mathrm{Ca}(\mathrm{OH})_2$ in $1000 \mathrm{~mL}=5 \times 10^{-3}$
Millimoles in $100 \mathrm{~mL}=5 \times 10^{-1}$
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