Calcium ions in blood trigger clotting. To prevent that in donated blood, sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is added to remove calcium ions according to the following equation. $$ \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s) $$ Blood contains about \(0.10 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{mL}\). If a 250.0-mL sample of donated blood is treated with an equal volume of \(0.160 \mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), estimate \(\left[\mathrm{Ca}^{2+}\right]\) after precipitation. \(\left(K_{s p} \mathrm{CaC}_{2} \mathrm{O}_{4}=4 \times 10^{-9}\right)\)

Step by step solution

01

Convert calcium concentration to moles

First, we need to convert the given concentration of \(\mathrm{Ca}^{2+}\) from mg/mL to moles per liter (M). The molar mass of \(\mathrm{Ca}^{2+}\) is 40.08 g/mol. $$ \frac{0.10 \, \text{mg}}{1 \, \text{mL}} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} \times \frac{\text{1 mol} \, \mathrm{Ca}^{2+}}{40.08 \, \text{g}} \times \frac{1000 \, \text{mL}}{1 \, \text{L}} \approx 2.50 \times 10^{-3} \text{M} $$

02

Find moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\)

Now, we will calculate moles of \(\mathrm{Ca}^{2+}\) in the 250 mL blood sample and moles of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) in the 250 mL \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) solution. $$ \text{moles of } \mathrm{Ca}^{2+} = (2.50 \times 10^{-3}\text{M})\times(250\text{mL})\times \frac{1\text{L}}{1000\text{mL}} \approx 6.25\times10^{-4}\text{ mol} $$ $$ \text{moles of }\mathrm{C}_{2}\mathrm{O}_{4}^{2-} = (0.160\text{M})\times(250\text{mL})\times \frac{1\text{L}}{1000\text{mL}} =0.040\text{mol} $$

03

Find the limiting reactant and moles of remaining \(\mathrm{Ca}^{2+}\)

As there is far more \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) than \(\mathrm{Ca}^{2+}\), \(\mathrm{Ca}^{2+}\) will be the limiting reactant. This means all \(\mathrm{Ca}^{2+}\) ions will react with \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions and some \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions will remain in the solution. $$ \text{moles of remaining } \mathrm{C}_{2}\mathrm{O}_{4}^{2-} = 0.040\text{mol} - 6.25\times10^{-4}\text{mol} \approx 0.0394\text{mol} $$

04

Calculate solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\)

Now, we will use the \(K_{sp}\) value of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) to determine its solubility. $$ K_{sp} = 4 \times 10^{-9} = [\mathrm{Ca}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] $$ Since initial \(\mathrm{Ca}^{2+}\) ions precipitated completely, we can assume the solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) will be equal to the concentration of the dissolved \(\mathrm{Ca}^{2+}\) ions. Thus, new \([\mathrm{Ca}^{2+}] = [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}] = S\).

05

Find the final concentration of \(\mathrm{Ca}^{2+}\) ions

Using the equation from Step 4, we can find the solubility, S, which is also the final concentration of \(\mathrm{Ca}^{2+}\) ions. $$ [\mathrm{Ca}^{2+}] = S = \sqrt{K_{sp}} = \sqrt{4 \times 10^{-9}} \approx 2.0 \times 10^{-5}\text{M} $$ Therefore, the final concentration of \(\mathrm{Ca}^{2+}\) ions will be approximately \(2.0 \times 10^{-5}\text{M}\) after precipitation.

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