The planet Aragonose (which is made mostly of the mineral

aragonite, or ${\mathbf{CaCO}}_{\mathbf{3}}$) has an atmosphere containing methane and

carbon dioxide, each at a pressure of 0.10 bar. The oceans are

saturated with aragonite and have a concentration of ${\mathbf{H}}_{\mathbf{1}}$equal to

$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{27}}\mathbf{M}$. Given the following equilibria, calculate how many

grams of calcium are contained in 2.00 L of Aragonose seawater.

$\begin{array}{ccc}{\mathbf{CaCO}}_{\mathbf{3}}\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{aragonite}\mathbf{)}& \mathbf{\rightleftharpoons }{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}& {\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\\ {\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}& \mathbf{\rightleftharpoons }{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}& {\mathbf{K}}_{{\mathbf{CO}}_{\mathbf{2}}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\\ {\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}& \mathbf{\rightleftharpoons }{\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}& {\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\\ {\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}& \mathbf{\rightleftharpoons }{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}& {\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\end{array}$

Don’t panic! Reverse the first reaction, add all the reactions

together, and see what cancels.

A chemical reaction’s equilibrium constant is the value of its reaction quotient at chemical equilibrium, a state reached by a dynamic chemical system after a period of time has passed in which its composition shows no discernible tendency to change.

Consider the concentration of H:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=1.8\times {10}^{-7}\mathrm{M} \\ \mathrm{Volume}=2.00\mathrm{L} \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{pressure}\mathrm{of}{\mathrm{CO}}_2=0.10\mathrm{bar} \\ =0.098692\mathrm{atm} \end{gathered}$$

As $1\mathrm{bar}=0.98692\mathrm{atm}$

To calculate the concentration of ${\mathrm{Ca}}^2$in 2.00 L solution, reverse the first reaction of the given reactions and sum up other reactions to this reaction. Derive equation whose equilibrium constant gives the concentration of the ${\mathrm{Ca}}^{+2}$.

Add all the reactions as follows:

$$\begin{gathered} {\mathrm{Ca}}^{+2}\left(\mathrm{aq}\right)+{\mathrm{CO}}_3^{-}\left(\mathrm{aq}\right){\mathrm{CaCO}}_3\left(\mathrm{s}\right)K=\frac{1}{K_{sp}}=\frac{1}{6.0\times {10}^{-9}} \\ {\mathrm{CO}}_2\left(g\right){\mathrm{CO}}_2\left(\mathrm{aq}\right){\mathrm{K}}_{{\mathrm{CO}}_2}=3.4\times {10}^{-2} \\ {\mathrm{CO}}_2\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(I\right){\mathrm{HCO}}_3^{-}\left(aq\right)+{\mathrm{H}}^{+}\left(aq\right)K_1=4.5\times {10}^{-7} \\ {\mathrm{HCO}}_3^{-}\left(\mathrm{aq}\right){\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CO}}_3^{-}\left(\mathrm{aq}\right){\mathrm{K}}_2=4.7\times {10}^{-11} \\ {\mathrm{Ca}}^{+2}\left(\mathrm{aq}\right)+{\mathrm{CO}}_2\left(g\right)+{\mathrm{H}}_2\mathrm{O}\left(I\right){\mathrm{CaCO}}_3\left(s\right)+2H^{+}\left(aq\right)K=\frac{K_1{K}_2{K}_{{\mathrm{CO}}_2}}{K_{sp}} \end{gathered}$$

$$\begin{gathered} \mathrm{K}=\frac{3.4\times {10}^{-2}\times 4.5\times {10}^{-7}\times 4.7\times {10}^{-11}}{6.0\times {10}^9} \\ =1.1985\times {10}^{-10} \end{gathered}$$

Now, the equilibrium constant$$\begin{gathered} K=\frac{{\left[H^{+}\right]}^2}{\left[C{a}^{+2}\right]P_{C{o}_2}} \\ \Rightarrow 1.1985\times {10}^{-10}=\frac{{\left(1.8\times {10}^{-7}\right)}^2}{\left[C{a}^{+2}\right]0.09869} \\ \Rightarrow [C{a}^{+2}+2]=\frac{(1.8\times {10}^{-7}{)}^2}{1.1985\times {10}^{-10}\times 0.09869} \end{gathered}$$

$\Rightarrow \left[C{a}^{+2}\right]=2.74\times {10}^{-3}mol/L$

Thus, 1.0L of seawater has $2.74\times {10}^{-3}$moles of $C{a}^{+2}$, so 2.00L of seawater contains $2\times 2.74\times {10}^{-3}$moles of $C{a}^{+2}$.

Hence, the number of grams of calcium obtained from 2.00 L of seawater

role="math" localid="1663400375495" $$\begin{gathered} =2\times 2.74\times {10}^{-3}\frac{\mathrm{moles}\times 40.078\mathrm{gCa}}{1\mathrm{moleCa}} \\ =0.22g \end{gathered}$$

Therefore, 0.22 g of calcium is obtained from 2.00L of seawater.