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QJEEM23Chdcba: JEE Main 2023 Chemistry - Class 11 JEEM23Chdcba Solution: Millimoles of calcium hydroxide required to produce 100 mL of…

Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is x×10−1x\times10^{-1} . The value of xx is ___________ (Nearest integer).

Assume complete dissociation.

Step-by-Step Verified Solution

amp;pH=12,amp;pH+pOH=14amp;pOH=14−12=2amp;[OH−]=10−2 mol Lamp;Ca(OH)25×10−3⟶Ca2+5×10−3+2OH−10−2\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 Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2 in 1000 mL=5×10−31000 \mathrm{~mL}=5 \times 10^{-3}

Millimoles in 100 mL=5×10−1100 \mathrm{~mL}=5 \times 10^{-1}

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