Question

A sample of $AgCl$ was treated with $5.00 \,mL$ of $1.5\, M\,Na_2 CO_3 $ solution to give $Ag_2CO_3$ . The remaining solution contained $0.0026\, g$ of $ Cl^-$ ions per litre. Calculate the solubility product of $AgCl [ K_{ sp } \, ( Ag_2 \, CO_3 ) = 8.2 \times 10^{ - 12} ] $

Answer 1

$ 2 AgCl \, ( g) + CO_3^{ 2 - } \rightleftharpoons Ag_2CO_3 \, ( g) + 2Cl^- $
$K = \frac{ [ Cl^- ]^2 }{ [ CO_3^{ 2 - } ] } = \frac{ [ Cl^- ]^2 }{ [ CO_3^{ 2 - } ] } \times \frac{ [ Ag^+ ]^2 }{ [ Ag^+ ]^2 } = \frac{ [ K_{ sp } \, ( Ag Cl)]^2 }{ k_{ sp } \, ( Ag_2 CO_3) } $
$ [ Cl^- ] = \frac{ 0.0026 }{ 35.5 } M = 7.3 \times 10^{ - 5 } $ M
The above concentration of $Cl^-$ indicates that $[CO_3^{2-}$ ] remains almost unchanged.
$ \frac{ 7.3 \times 10^{ - 5 }}{ 1.5 } = \frac{ [ K_{ sp } \, ( AgCl)]^2}{ 8.2 \times 10^{ - 12 } } $
$ K_{ sp } = ( AgCl) = 2 \times 10^{ - 8 } $