For sea water one may take \(\mathrm{pH}=8\). Use the equilibrium constants
\(K_{1}=\) \(\left[\mathrm{H}_{3}
\mathrm{O}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right] /\left[\mathrm{H}_{2}
\mathrm{CO}_{3}\right]=4.3 \times 10^{-7}\) and \(K_{2}=\left[\mathrm{H}_{3}
\mathrm{O}^{+}\right]\left[\mathrm{CO}_{3}^{--}\right]
/\left[\mathrm{HCO}_{3}^{-}\right]=\) \(4.8 \times 10^{-11}\) to calculate the
fractions in sea water of \(\left[\mathrm{CO}_{3}{
}^{2-}\right],\left[\mathrm{HCO}_{3}{ }^{-}\right]\)and \(\left[\mathrm{H}_{2}
\mathrm{CO}_{3}\right]\). Students proficient with computers may plot the three
fractions as a function of \(\mathrm{pH}\).
