Answer:
Approximately [tex]4.2 \times 10^{-7}\; \rm g \cdot L^{-1}[/tex].
Explanation:
Start by finding the concentration of [tex]\rm Ag_2CO_3[/tex] at equilibrium. The solubility equilibrium for
[tex]\rm Ag_2CO_3 \; (s) \rightleftharpoons 2\, Ag^{+}\; (aq) + {CO_3}^{2-}\; (aq)[/tex].
The ratio between the coefficient of [tex]\rm Ag_2CO_3[/tex] and that of [tex]\rm Ag^{+}[/tex] is [tex]1:2[/tex]. For
Let the increase in [tex]\rm {CO_3}^{2-}[/tex] concentration be [tex]+x\; \rm mol \cdot L^{-1}[/tex]. The increase in [tex]\rm Ag^{+}[/tex] concentration would be [tex]+2\,x\; \rm mol \cdot L^{-1}[/tex]. Note, that because of the [tex]0.057\; \rm mol \cdot L^{-1}[/tex]of [tex]\rm AgNO_3[/tex], the concentration of
Apply the solubility product expression (again, note that in the equilibrium, the coefficient of [tex]\rm Ag^{+}[/tex] is two) to obtain:
[tex]\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}[/tex].
Note, that the solubility product of [tex]\rm Ag_2CO_3[/tex], [tex]K_{\text{sp}} = 8.1 \times 10^{-12}[/tex] is considerably small. Therefore, at equilibrium, the concentration of
Apply this approximation to simplify [tex](0.057 + x)^2\cdot x = 8.1 \times 10^{-12}[/tex]:
[tex]0.057^2\, x \approx (0.057 + x)^2 \cdot x = 8.1 \times 10^{-12}[/tex].
[tex]\begin{aligned} x &\approx \frac{8.1 \times 10^{-12}}{0.057^2}\end{aligned}[/tex].
Calculate solubility (in grams per liter solution) from the concentration. The concentration of [tex]\rm Ag_2CO_3[/tex] is approximately [tex]\displaystyle \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol\cdot L^{-1}[/tex], meaning that there are approximately [tex]\displaystyle n = \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol[/tex] of
[tex]\begin{aligned}m &= n \cdot M \\ &\approx \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2} \; \rm mol\times 167.91\; g \cdot mol^{-1} \\ &\approx 4.2 \times 10^{-7}\; \rm g \end{aligned}[/tex].
As a result, the maximum solubility of [tex]\rm Ag_2CO_3[/tex] in this solution would be approximately [tex]4.2 \times 10^{-7}\; \rm g \cdot L^{-1}[/tex].
