Question #133447

Show how to calculate the molar solubility of PbI2 in 0.1M KI (aq)? Ksp=7.1x10-9

Expert's answer

The dissociation of Pbl, is as follows:

Pbl_{2 }gives Pb^{2+ }+ 2I^{-}

K_{sp }of Pbl_{2} =7.1x10^{-9}

The expression for the K of the reaction is as follows:

Construct an ICE table for the reaction.

Pbl_{2 gives} Pb^{+2}+^{ }2I^{-}

Initial: 0 0.1

Change +x +2x

Equilibrium: x (0.1+2x)

Substitute these values in the expression as follows:

K_{sp}=_{ }(Pb^{+2})(I^{-})^{2}

7.1x10 =(x){0.1+2x)^{2}

Solve for x

x= 7.1x10^{-7} M

Therefore, the molar solubility of Pbl_{2 }in KI is 7.1x10^{-7} M.

The molar solubility of Pbl_{2} is 1.22x10^{-3} M

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