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Lead(II) iodate (Pb(IO3)2) has a solubility product constant of 3.69 x 10-13. Calculate the molar solubility of Pb(IO3)2 in water. 4.30 x 10-7 M 6.07 x 10-7 M 4.52 x 10-5 M 7.17 x 10-5 M

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JeongHyeokAhn
The answer for this question is Option C. 4.52x10^-5

Answer : The correct option is, [tex]4.52\times 10^{-5}M[/tex]

Explanation :

The balanced equilibrium reaction will be,

[tex]Pb(IO_3)_2\rightleftharpoons Pb^{2+}+2IO_3^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Pb^{2+}][IO_3^-]^2[/tex]

Let the solubility will be, 's'

[tex]K_{sp}=(s)\times (2s)^2[/tex]

[tex]K_{sp}=(4s)^3[/tex]

Now put the value of [tex]K_[sp}[/tex] in this expression, we get the solubility of Lead(II) iodate.

[tex]3.69\times 10^{-13}=(4s)^3[/tex]

[tex]s=4.52\times 10^{-5}M[/tex]

Therefore, the solubility of Lead(II) iodate is, [tex]4.52\times 10^{-5}M[/tex]

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