Question

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

Answer 1

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

Answer 2

Answer : The correct option is, $4.52\times 10^{-5}M$

Explanation :

The balanced equilibrium reaction will be,

$Pb(IO_3)_2\rightleftharpoons Pb^{2+}+2IO_3^-$

The expression for solubility constant for this reaction will be,

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

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

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

$3.69\times 10^{-13}=(4s)^3$

$s=4.52\times 10^{-5}M$

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