Question: The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, \({K_2}{C_2}{O_4}\), is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of\(Ca{C_2}{O_4} \times {H_2}O\). It is necessary to remove all but 1.0% of the \(C{a^{2 + }}\) in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of \(C{a^{2 + }}\) per 100 mL of serum, what mass of \({K_2}{C_2}{O_4}\)is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the \({K_{sp}}\)value for \(Ca{C_2}{O_4}\)in serum is the same as in water.)

In 100 ml of serum, there are 9.5 mg of calcium.

In 5.5.ml of serum, we get \(\frac{{9.5 \cdot 5.5}}{{100}}\)mg of calcium, which is 0.52 mg

molar mass of calcium is \({\rm{Mr}} = 40.08{\rm{g}}/{\rm{mol}}\)

moles of calcium \(\begin{array}{l}{\rm{n}} = \frac{{\rm{m}}}{{{\rm{Mr}}}}\\ = \frac{{0.52 \cdot {{10}^{ - 3}}{\rm{g}}}}{{40.08{\rm{g}}/{\rm{mol}}}}\\ = 1.3 \cdot {10^{ - 5}}{\rm{\;moles\;}}\end{array}\)

To prevent coagulation, we precipitate 99% of calcium with oxalate. Thus, we precipitate,

\(1.3 \cdot {10^{ - 5}} \cdot 0.99 = \)\(1.29 \cdot {10^{ - 5}}\)mole of calcium

Remaining \(1{\rm{\% }} = {10^{ - 7}}\)moles, and is in equilibrium with oxalate in a solution

\(\begin{array}{l}{\rm{c}}\left( {{\rm{C}}{{\rm{a}}^{2 + }}} \right) = \frac{{\rm{n}}}{{\rm{V}}}\\ = \frac{{{{10}^{ - 7}}{\rm{moles}}}}{{0.01{\rm{L}}}}\\ = {10^{ - 5}}{\rm{M}}\end{array}\)

Expression for \({{\rm{K}}_{sp}}{\rm{\;for\;Ca}}{{\rm{C}}_2}{{\rm{O}}_4}\),

\({{\rm{K}}_{sp}} = \left( {{\rm{C}}{{\rm{a}}^{2 + }}} \right)\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)\)

Calculate \(\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)\) in equilibrium with calcium ions that we calculated above, so concentration that does not precipitate,

\(\begin{array}{*{20}{c}}{1.96 \cdot {{10}^{ - 8}} = \left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right) \cdot {{10}^{ - 5}}}\\{\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right) = \frac{{1.96 \cdot {{10}^{ - 8}}}}{{{{10}^{ - 5}}}}}\end{array}\)

\(\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right) = 1.96 \cdot {10^{ - 3}}{\rm{M}}\)

calculate number of moles by multiplying molar concentration by volume \(\left( {{{10}^{ - 2}}{\rm{L}}} \right)\), \(n = 1.96 \cdot {10^{ - 5}}{\rm{\;moles\;}}\)

One ion of calcium precipitates with one oxalate ion. Likely, one mole of calcium reacts with one mole of oxalate.

Calculate total moles used in these reactions are as follow,\({\rm{\;Total oxalate = oxalate used to precipitate\;}}99{\rm{\% \;of\;C}}{{\rm{a}}^{2 + }} + {\rm{\;oxalate in equilibrium with\;}}1{\rm{\% C}}{{\rm{a}}^{2 + }}\)

\({\rm{\;Total oxalate\;}} = 1.29 \cdot {10^{ - 5}} + 1.96 \cdot {10^{ - 5}} = 3.25 \cdot {10^{ - 5}}{\rm{\;moles\;}}\)

Since 1 mol of oxalate ions is obtained from 1 mol of \({{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4},3.25 \cdot {10^{ - 5}}\) moles of \({{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4}\) are needed to avoid coagulation.

\(Mr\left( {{{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4}} \right) = 166{\rm{g}}/{\rm{mol}}\)

\({\rm{\;Mass of\;}}{{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4} = {\rm{n}} \cdot {\rm{Mr}}\)

\({\rm{\;Mass of\;}}{{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4} = 3.25 \cdot {10^{ - 5}}{\rm{mol}} \cdot 166{\rm{g}}/{\rm{mol}}\)

\({\rm{\;Mass of\;}}{{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4} = 5.39 \cdot {10^{ - 3}}{\rm{g}} = 5.39{\rm{mg}}\)

The solution for mass of \({{\rm{K}}_2}{{\rm{C}}_2}{{\rm{O}}_4}\) is \(5.39{\rm{mg}}\).