{K_{sp}} of Pb I_{2} is 1.4 × {10^{ - 8}} . molecular mass of Pb I_{2} is 461 g mol^{-1} Then

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6-2.Equilibrium-II (Ionic Equilibrium)

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${K_{sp}}$ of $Pb$ $I_{2}$ is $ 1.4 \times {10^{ - 8}}$. The molecular mass of $Pb$ $I_{2}$ is $461$ $g$ $mol^{-1}$ Then molecular mass of $Pb{\left( {N{O_3}} \right)_2}$ is $331.9$ $mol^{-1}$ So, $(a)$ In $500$ $mL$ water $(b)$ $500$ $mL$ $0.10$ $M$ $KI$ $(c)$ What is the weight of $PbI_{2}$ when soluble in $1.33$ $g$ $Pb{\left( {N{O_3}} \right)_2}$ containing $500$ $mL$ solution ?

Option A

Option B

Option C

Option D

Solution

$(a)$ $0.346 \mathrm{~g}$ $(b)$ $0.322 \times 10^{-3} \mathrm{~g}$ $(b)$ $0.15 \mathrm{~g}$

Standard 11

Chemistry

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