91Ó°ÊÓ

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:

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

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

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:

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

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

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

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

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

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

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

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