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Chapter 4: Problem 91

Calcium in blood or urine can be determined by precipitation as call cium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\). The precipitate is dissolved in strong acid and titrated with potassium permanganate. The products of the reaction are carbon dioxide and manganese(II) ion. 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

Expert verified
Answer: Yes, the mass of calcium oxalate in the sample is 66.18 mg, which is within the normal range for an adult (100-300 mg per 24 hours).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between calcium oxalate and potassium permanganate is: 5 CaC2O4 + 24 KMnO4 + 32 H2SO4 → 5 CaSO4 + 24 MnSO4 + 12 K2SO4 + 60 CO2 + 80 H2O
02

Calculate the moles of potassium permanganate used

We are given the volume and the molarity of the potassium permanganate solution used in the titration. We can calculate the moles of KMnO4 used as follows: Moles of KMnO4 = volume * molarity Moles of KMnO4 = 26.2 mL * 0.0946 M = 0.00248012 mol
03

Calculate the moles of calcium oxalate present

From the balanced chemical equation, we can see that the stoichiometric ratio between calcium oxalate (CaC2O4) and potassium permanganate (KMnO4) is 5 : 24. Hence, we can determine the moles of CaC2O4 present using the moles of KMnO4 and the stoichiometric ratio as follows: Moles of CaC2O4 = (moles of KMnO4 * stoichiometric coefficient of CaC2O4) / stoichiometric coefficient of KMnO4 Moles of CaC2O4 = (0.00248012 mol * 5) / 24 = 0.0005166875 mol
04

Calculate the mass of calcium oxalate and verify if it is within the normal range

Now, we will convert the moles of calcium oxalate to mass using its molar mass. The molar mass of calcium oxalate is as follows: Molar mass of CaC2O4 = (40.08 (Ca) + 2*(12.01 (C) + 4*(16.00 (O)))) g/mol = 128.1 g/mol Mass of CaC2O4 = moles of CaC2O4 * molar mass of CaC2O4 Mass of CaC2O4 = 0.0005166875 mol * 128.1 g/mol = 0.066178425 g = 66.18 mg Since the mass of calcium oxalate in the sample is 66.18 mg, it is within the normal range for an adult, which is between 100 and 300 mg per 24 hours.

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