Verified Solution • 2025–26 Session moderate

QJEEM22Chf2c9: JEE Main 2022 Chemistry - Class 11 JEEM22Chf2c9 Solution: At , the solubility of in water is . The solubility product of…

At 310 K310 \mathrm{~K} , the solubility of CaF2\mathrm{CaF}_{2} in water is 2.34×10−3 g/100 mL2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL} . The solubility product of CaF2\mathrm{CaF}_{2} is ____________ ×10−8( mol/L)3\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3} . (Give molar mass : CaF2=78 g mol−1\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1} )

Step-by-Step Verified Solution

CaF2⇌sCa2+s+2 F−2 s\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}

Kspamp;=s(2 s)2amp;=4 s3\begin{aligned}\mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\&=4 \mathrm{~s}^{3}\end{aligned}

Solubility (s)=2.34×10−3 g/100 mL(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}

=2⋅34×10−3×1078=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78} mole // lit

=3×10−4 mole/lit=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}

∴Ksp=4×(3×10−4)3\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}

amp;=108×10−12amp;=0.0108×10−8( mole/lit)3\begin{aligned}&=108 \times 10^{-12} \\\\&=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3}\end{aligned}

amp;∴x≈0\begin{aligned}& \therefore x \approx 0\end{aligned}

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