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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium3 Marks
Question
The molar solubility of lead iodate, $Pb(I0_3)_2$ is $4.0 \times 10^{-5} mol/ L at 25^\circ C$. Calculate its $K_{sp}$.
$\text{Pb}(\text{IO}_3)_2\rightleftharpoons\text{Pb}^{2+}+\text{2IO}_3$
✓

Answer

$\therefore \mathrm{K}_{\text {sp }}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_3^{-}\right]^2$
$\because \text { The molar solubility of } \mathrm{Pb}\left(\mathrm{IO}_3\right)_2 \text { is } 4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
$\therefore\left[\mathrm{~Pb}^{2+}\right]=4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
${\left[\mathrm{IO}_3^{-}\right]=2 \times 4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}}$
$\therefore \mathrm{~K}_{\text {sp }}=\left(4.0 \times 10^{-5}\right)\left(8.0 \times 10^{-5}\right)^2$
$=2.56 \times 10^{-13}$

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