The solubility product of `Pbl_(2)` is `7.2 xx 10^(-9)`. The maximum mass of `Nal` which may be added in `500ml` of `0.005M Pb (NO_(3))_(2)` solution without any precipitation of `Pbl_(2)` is `( = 127)`:

Options

A. `0.09g`
B. `1.2 xx 10^(-3)g`
C. `6 xx 10^(-4)g`
D. `1.08 xx 10^(-5)g`

Answer

A

Explanation

`[I^(-)] = sqrt(7.2 xx 10^(-9))/(0.005) = 1.2 xx 10^(-3) M`
Maximum moles of Nal that can be added to `500 ml` of `Pb(NO_(3))_(2)` are
`(1.2 xx 10^(-3))/(1000) xx 500 = 6 xx 10^(-4) "moles" rArr 0.09g`