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JEE MAIN 2023
29-1-23
Question
Millimoles of calcium hydroxyide required to produce 100 mL of the aqueous solution of pH 12 is x × 10–1. The value of x is _______ (Nearest integer). Assume complete dissociation.
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Solution
Sol. $$ \begin{array}{ll} \because & \mathrm{pH}=12 \\ \therefore & {\left[\mathrm{H}^{+}\right]=10^{-12} \mathrm{M}} \\ \therefore & {\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{M}} \\ \therefore & {\left[\mathrm{Ca}(\mathrm{OH})_2\right]=5 \times 10^{-3} \mathrm{M}} \\ 5 \times 10^{-3}=\frac{\text { milli moles of } \mathrm{Ca}(\mathrm{OH})_2}{100 \mathrm{~mL}} \\ \text { milli moles of } \mathrm{Ca}(\mathrm{OH})_2=5 \times 10^{-1} \\ \text { Ans. }=5 \end{array} $$
Question Tags
JEE Main
Chemistry
Easy
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