Question

Calculate the maximum solubility of silver carbonate, Ag2CO3 in g/L when in the presence of 0.057 M AgNO3. The solubility product of Ag2CO3 is 8.1x10-12 and Ag2CO3 has a molar mass of 167.91 g/mol. Express your answer to the correct number of significant figures, in scientific notation and include units with your answer.

Answer 1

Answer:

Approximately $4.2 \times 10^{-7}\; \rm g \cdot L^{-1}$.

Explanation:

Start by finding the concentration of $\rm Ag_2CO_3$ at equilibrium. The solubility equilibrium for

$\rm Ag_2CO_3 \; (s) \rightleftharpoons 2\, Ag^{+}\; (aq) + {CO_3}^{2-}\; (aq)$.

The ratio between the coefficient of $\rm Ag_2CO_3$ and that of $\rm Ag^{+}$ is $1:2$. For

Let the increase in $\rm {CO_3}^{2-}$ concentration be $+x\; \rm mol \cdot L^{-1}$. The increase in $\rm Ag^{+}$ concentration would be $+2\,x\; \rm mol \cdot L^{-1}$. Note, that because of the $0.057\; \rm mol \cdot L^{-1}$of $\rm AgNO_3$, the concentration of

• The concentration of $\rm Ag^{+}$ would be $(0.057 + 2\, x) \; \rm mol\cdot L^{-1}$.
• The concentration of $\rm {CO_3}^{2-}$ would be $x\; \rm mol \cdot L^{-1}$.

Apply the solubility product expression (again, note that in the equilibrium, the coefficient of $\rm Ag^{+}$ is two) to obtain:

$\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}$.

Note, that the solubility product of $\rm Ag_2CO_3$, $K_{\text{sp}} = 8.1 \times 10^{-12}$ is considerably small. Therefore, at equilibrium, the concentration of

Apply this approximation to simplify $(0.057 + x)^2\cdot x = 8.1 \times 10^{-12}$:

$0.057^2\, x \approx (0.057 + x)^2 \cdot x = 8.1 \times 10^{-12}$.

$\begin{aligned} x &\approx \frac{8.1 \times 10^{-12}}{0.057^2}\end{aligned}$.

Calculate solubility (in grams per liter solution) from the concentration. The concentration of $\rm Ag_2CO_3$ is approximately $\displaystyle \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol\cdot L^{-1}$, meaning that there are approximately $\displaystyle n = \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol$ of

$\begin{aligned}m &= n \cdot M \\ &\approx \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2} \; \rm mol\times 167.91\; g \cdot mol^{-1} \\ &\approx 4.2 \times 10^{-7}\; \rm g \end{aligned}$.

As a result, the maximum solubility of $\rm Ag_2CO_3$ in this solution would be approximately $4.2 \times 10^{-7}\; \rm g \cdot L^{-1}$.