Respuesta :
Answer:
The concentration of lead ion required to just initiate precipitation is -[tex]2.37\times10^-^5 M[/tex]
Explanation:
Lets calculate -:
Solubility equilibrium -: [tex]PbI_2(s)[/tex] ⇄ [tex]Pb^2^+ (aq) + 2I^- (aq)[/tex]
Solubility product of [tex]PbI_2[/tex] ,[tex]Q=[Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]=9.8\times10^-^9[/tex]
Concentration of [tex]I^-[/tex][tex]=[KI]=0.0492M[/tex]
When the ionic product exceeds the solubility product , precipitation of salt takes place .
[tex]Q_s_p\geq K_s_p[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][0.0492]^2[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq \frac{9.8\times10^-^9}{[0.0492]^2}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]\frac{9.8\times10^-^9}{2.42\times10^-^3}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]2.37\times10^-^5 M[/tex]
Thus , [tex]PbI_2[/tex] will start precipitating when [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq 2.37\times10^-^5 M[/tex].
