Question

71\. Calcium in blood or urine can be determined by precipitation as calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4} .\) The precipitate is dissolved in strong acid and titrated with potassium permanganate. The equation for reaction is $$ \begin{aligned} 2 \mathrm{MnO}_{4}^{-}(a q)+6 \mathrm{H}^{+}(a q)+& 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Mn}^{2+}(a q)+10 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O} \end{aligned} $$ A 24 -hour urine sample is collected from an adult patient, reduced to a small volume, and titrated with \(26.2 \mathrm{~mL}\) of \(0.0946 \mathrm{M} \mathrm{KMnO}_{4}\). How many grams of calcium oxalate are in the sample? Normal range for \(\mathrm{Ca}^{2+}\) output for an adult is 100 to \(300 \mathrm{mg}\) per 24 hour. Is the sample within the normal range?

Short answer

Answer: Yes, the amount of calcium oxalate in the sample is within the normal range. The sample contains 248.6 mg of Ca²⁺ per 24 hours, which falls within the normal range of 100 to 300 mg.

Step-by-step solution

Write down the given information

We are given the following information: - Volume of potassium permanganate solution used: \(26.2 \ \text{mL}\) - Concentration of the potassium permanganate solution: \(0.0946 \ \text{M}\) - The equation for the reaction: $$2 \text{MnO}_4^- + 6 \text{H}^+ + 5 \text{H}_2\text{C}_2\text{O}_4 \longrightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O}$$

Calculate the moles of potassium permanganate

Using the volume and concentration of the potassium permanganate solution, we can calculate the moles of \(\text{MnO}_4^-\) present: $$\text{moles of MnO}_4^- = \text{volume} \times \text{concentration} = (26.2 \ \text{mL})(0.0946 \ \text{M})$$ $$\text{moles of MnO}_4^- = 0.0024792 \ \text{moles}$$

Use stoichiometry to determine moles of \(\text{CaC}_2\text{O}_4\)

From the balanced reaction equation, we can see that 2 moles of \(\text{MnO}_4^-\) react with 5 moles of \(\text{H}_2\text{C}_2\text{O}_4\). We can use stoichiometry to determine the moles of \(\text{H}_2\text{C}_2\text{O}_4\): $$\text{moles of H}_2\text{C}_2\text{O}_4 = \frac{\text{moles of MnO}_4^-}{2} \times 5$$ $$\text{moles of H}_2\text{C}_2\text{O}_4 = \frac{0.0024792 \ \text{moles}}{2} \times 5$$ $$\text{moles of H}_2\text{C}_2\text{O}_4 = 0.006198 \ \text{moles}$$ Note that 1 mole of \(\text{H}_2\text{C}_2\text{O}_4\) contains 1 mole of \(\text{CaC}_2\text{O}_4\), since the \(\text{CaC}_2\text{O}_4\) precipitate was dissolved in strong acid before titration. Therefore, the moles of \(\text{CaC}_2\text{O}_4\) equal the moles of \(\text{H}_2\text{C}_2\text{O}_4\).

Convert moles of \(\text{CaC}_2\text{O}_4\) to grams

Now we can calculate the grams of \(\text{CaC}_2\text{O}_4\) in the sample: $$\text{grams of CaC}_2\text{O}_4 = \text{moles of CaC}_2\text{O}_4 \times \text{molar mass of CaC}_2\text{O}_4$$ The molar mass of \(\text{CaC}_2\text{O}_4 = 40.08 \ \text{g/mol (for Ca)} + 2 \times (12.01 \ \text{g/mol (for C)} + 16.00 \ \text{g/mol (for O)}) = 128.10 \ \text{g/mol\) $$\text{grams of CaC}_2\text{O}_4 = (0.006198 \ \text{moles})(128.10 \ \text{g/mol})$$ $$\text{grams of CaC}_2\text{O}_4 = 0.794 \ \text{g}$$

Determine if the sample is within the normal range

The normal range for \(\text{Ca}^{2+}\) output for an adult is 100 to \(300 \ \text{mg}\) per 24 hours. To check if the sample is within this range, we need to convert the grams of \(\text{CaC}_2\text{O}_4\) to mg of \(\text{Ca}^{2+}\). We can obtain the ratio of the molar mass of \(\text{Ca}^{2+}\) to that of \(\text{CaC}_2\text{O}_4\): $$\frac{\text{molar mass of Ca}^{2+}}{\text{molar mass of CaC}_2\text{O}_4} = \frac{40.08 \ \text{g/mol}}{128.10 \ \text{g/mol} } = 0.313$$ Now we can find the mg of \(\text{Ca}^{2+}\) in the sample: $$\text{mg of Ca}^{2+} = (0.794 \ \text{g})(0.313)(1000 \ \text{mg/g})$$ $$\text{mg of Ca}^{2+} = 248.6 \ \text{mg}$$ The sample has \(248.6 \ \text{mg}\) of \(\text{Ca}^{2+}\) per 24 hours, which is within the normal range of 100 to \(300 \ \text{mg}\).