Question

Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion required to just initiate precipitation?

Answer 1

The concentration of lead ion required to just initiate precipitation is -
`2.37*10^-^5 M`

Step-by-step explanation:

Lets calculate -:

Solubility equilibrium -:
`PbI_2(s)` ⇄
`Pb^2^+ (aq) + 2I^- (aq)`

Solubility product of
`PbI_2` ,
`Q=[Pb^2^+]_i_n_i_t_i_a_l`
`[I^-]^2_i_n_i_t_i_a_l`
`=9.8*10^-^9`

Concentration of
`I^-`
`=[KI]=0.0492M`

When the ionic product exceeds the solubility product , precipitation of salt takes place .

`Q_s_p\geq K_s_p`

`[Pb^2^+]_i_n_i_t_i_a_l`
`[I^-]^2_i_n_i_t_i_a_l`
`\geq 9.8*10^-^9`

`[Pb^2^+]_i_n_i_t_i_a_l`
`[0.0492]^2`
`\geq 9.8*10^-^9`

`[Pb^2^+]_i_n_i_t_i_a_l`
`\geq (9.8*10^-^9)/([0.0492]^2)`

`[Pb^2^+]_i_n_i_t_i_a_l`
`\geq`
`(9.8*10^-^9)/(2.42*10^-^3)`

`[Pb^2^+]_i_n_i_t_i_a_l`
`\geq`
`2.37*10^-^5 M`

Thus ,
`PbI_2` will start precipitating when
`[Pb^2^+]_i_n_i_t_i_a_l`
`\geq 2.37*10^-^5 M`.