Chapter 3: Problem 15
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 \(\mathrm{pH}\).
Short Answer
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Key Concepts
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