Question

Suppose that $\text{150mL}$of $\text{0.200M}$${\text{K}}_{2}C{O}_3$and $100mL$of $0.400M$$Ca(N{O}_3{)}_2$are mixed together. Assume that the volumes are additive, thatis completely insoluble, and that all other substances that might be formed are soluble. Calculate the mass of$CaC{O}_3$precipitated, and calculate the concentrations in the final solution of the four ions that were present initially.

Short answer

The mass of $CaC{O}_3$is$2.2\times {10}^{-5}g$ .

The concentrations are:

$$\begin{gathered} \left[K^{+}\right]=2\times 2.2\times {10}^{-7}M \\ =4.4\times {10}^{-7}M \end{gathered}$$

$$\begin{gathered} \left[N{O}_3^{-}\right]=2\times 0.040M \\ =0.080M \end{gathered}$$

Step-by-step solution

Initial concentration of K+ and  Ca2+.

${\text{[k}}^{\text{+}}\text{]=}$$$\begin{gathered} \left(\frac{0.200mol{K}_{2}C{O}_3}{L}\times \frac{2mol{K}^{+}}{1mol{K}_{2}C{O}_3}\right)\times \frac{150mL}{250mL} \\ =0.240M \end{gathered}$$

$$\begin{gathered} \left[C{a}^{+2}\right]=0.400\times \frac{100mL}{250mL} \\ =0.160M \end{gathered}$$

$$\begin{gathered} Here{K}^{+}islimiting, \\ accordingtoreaction: \end{gathered}$$

$$\begin{gathered} K_{2}C{O}_3\left(aq\right)+Ca(N{O}_3{)}_2\left(aq\right) \\ CaC{O}_3\left(s\right)+2KN{O}_3\left(aq\right) \end{gathered}$$

$0.120$Mol of $K_{2}C{O}_3$will combine with $0.120$mol of $Ca(N{O}_3{)}_2$to produce $0.120$mol of $CaC{O}_3$.

The moles of $Ca(N{O}_3{)}_2$remains unreacted

$$\begin{gathered} =0.160M-0.120M \\ =0.040M \end{gathered}$$

Equilibrium calculation.

$CaC{O}_3\left(s\right)\to C{a}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$

If $0.040+x$is approximated as 0.040, then

$$\begin{gathered} \left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=K_{sp} \\ (0.040+x)\left(x\right)=8.7\times {10}^{-9} \\ (0.040)x=8.7\times {10}^{-9} \\ x=\frac{8.7\times {10}^{-9}}{0.040} \\ =2.2\times {10}^{-7}M \end{gathered}$$

Clearly,

$x=2.2\times {10}^{-7}<<<0.040$

So, the assumption was justified.

Finding gram solubility

Now the concentration of $C{a}^{2+}\left(aq\right)$and $C{O}_3^{2-}\left(aq\right)$ are:

$$\begin{gathered} \left[C{a}^{2+}\right]=0.0400M \\ \left[C{O}_3^{2-}\right]=2.2\times {10}^{-7}M \end{gathered}$$

The solubility$\left(x\right)$of$CaC{O}_3$ is$2.2\times {10}^{-7}mol$$L^{-1}$ .

Because the molar mass of$CaC{O}_3$is,$100.09g$$mo{l}^{-1}$ the gram solubility$CaC{O}_3$ is:

Gram solubility$$\begin{gathered} =(2.2\times {10}^{-7}mol{L}^{-1})(100.09gmo{l}^{-1}) \\ =2.2\times {10}^{-5}g{L}^{-1} \end{gathered}$$

That is, the solubility in$L$ is$2.2\times {10}^{-5}g$,

Thus, the mass of $CaC{O}_3$is$2.2\times {10}^{-5}g$.

Solubility equation

The solubility equations for ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$and ${\text{Ca(NO}}_3{\text{)}}_2$are:

$$\begin{gathered} K_{2}C{O}_3\left(aq\right)\to 2K^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right) \\ Ca(N{O}_3{)}_2\left(aq\right)\to C{a}^{2+}+2N{O}_3^{-}\left(aq\right) \end{gathered}$$

Since the concentrations of $C{a}^{2+}\left(aq\right)$ and $C{O}_3^{2-}\left(aq\right)$are $0.040M$ and, $2.2\times {10}^{-7}M$so the concentrations of $K^{+}\left(aq\right)$ and $N{O}_3^{-}\left(aq\right)$are:

$$\begin{gathered} \left[K^{+}\right]=2\times 2.2\times {10}^{-7}M \\ =4.4\times {10}^{-7}M \end{gathered}$$

$$\begin{gathered} \left[N{O}_3^{-}\right]=2\times 0.040M \\ =0.080M \end{gathered}$$