Question

The solubility of lead(II) chloride is 0.45 g/100 mL of solution. What is the Ksp of PbCl2?a. 8.5 ? 10^-6b. 4.2 ? 10^-6c. 1.7 ? 10^-5d. 4.9 ? 10^-2e. < 1.0 ? 10^-6

Answer 1

The value of  $K_{sp}$ of $PbCl_{2}$  is 1.7×$10^{-5}$ .

So, option C is correct one.

Calculation,

Number of moles of $PbCl_{2}$  = mass/molar mass = 0.45/278.1 = 1.618×$10^{-3}$ moles

Concentration of $Pb^{+2}$ and $Cl^{-}$ ion

Molarity  of $Pb^{+2}$ = Number of moles of solute /volume in lit

Molarity  of $Pb^{+2}$ = 1.618×$10^{-3}$ moles /0.1 lit = 0.0162 M

$PbCl_{2}$ →  $Pb^{+2}$ +  2$Cl^{-}$

For certain amount of  $Pb^{+2}$ , twice this amount of  $Cl^{-}$ ion formed.

The concentration of   $Cl^{-}$ ion formed = 2× 0.0162 M = 0.0324 M

Now,

$K_{sp}$ of $PbCl_{2}$ = [ $Pb^{+2}$ ]$[Cl^{-}]^{2}$ = 0.0162 ×$[0.0324 ]^{2}$ =  1.7×$10^{-5}$ .

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