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* 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)* (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* 10^(-13)=(4s)^3`

`s=4.52* 10^(-5)M`

Therefore, the solubility of Lead(II) iodate is,
`4.52* 10^(-5)M`