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

Answer:

The concentration of lead ion required to just initiate precipitation is -$2.37\times10^-^5 M$

Explanation:

Lets calculate -:

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

Solubility product of $PbI_2$ ,$Q=[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $=9.8\times10^-^9$

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

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

                                   $Q_s_p\geq K_s_p$

        $[Pb^2^+]_i_n_i_t_i_a_l$ $[I^-]^2_i_n_i_t_i_a_l$ $\geq 9.8\times10^-^9$

      $[Pb^2^+]_i_n_i_t_i_a_l$  $[0.0492]^2$ $\geq 9.8\times10^-^9$

                       $[Pb^2^+]_i_n_i_t_i_a_l$ $\geq \frac{9.8\times10^-^9}{[0.0492]^2}$

                        $[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $\frac{9.8\times10^-^9}{2.42\times10^-^3}$

                        $[Pb^2^+]_i_n_i_t_i_a_l$ $\geq$ $2.37\times10^-^5 M$

Thus , $PbI_2$ will start precipitating when $[Pb^2^+]_i_n_i_t_i_a_l$   $\geq 2.37\times10^-^5 M$.