Problem 91 - Chemistry: Principles and Reactions

Question

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?

Step-by-Step Solution

Answer

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 1: 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

Step 2: 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

Step 3: 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

Step 4: 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.

Key Concepts

Titration

Titration is a laboratory technique used to determine the concentration of a particular substance in a solution. In the context of calcium oxalate determination, it involves gradually adding a titrant, such as potassium permanganate (\(\text{KMnO}_4\)), to a solution containing the dissolved analyte, in this case, calcium oxalate (\(\text{CaC}_2\text{O}_4\)).

The titrant has a known concentration, and by measuring the volume used to reach the endpoint—a marked change indicating the completion of the reaction—one can calculate the concentration of the analyte in the original solution. In our specific problem, 26.2 mL of potassium permanganate with a molarity of 0.0946 M was used to titrate the calcium oxalate solution.

Titration is essential in applications such as determining the concentration of ions in biological fluids to assess normal physiological ranges, ensuring the safety and effectiveness of pharmaceutical compounds, and more. It demonstrates the precision and reliability required in chemical analysis activities.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants transforming into products, maintaining a balance of elements before and after the reaction. The importance of a balanced equation cannot be overstated in stoichiometry for understanding how substances interact.

In the calcium oxalate titration problem, the balanced equation: \[5\ \text{CaC}_2\text{O}_4 + 24\ \text{KMnO}_4 + 32\ \text{H}_2\text{SO}_4 \rightarrow 5\ \text{CaSO}_4 + 24\ \text{MnSO}_4 + 12\ \text{K}_2\text{SO}_4 + 60\ \text{CO}_2 + 80\ \text{H}_2\text{O}\] is crucial. It tells us the proportions of calcium oxalate and potassium permanganate that react with sulfuric acid to form calcium sulfate, manganese sulfate, potassium sulfate, carbon dioxide, and water.

Understanding this equation helps in determining the quantities of reactants needed and the expected products from a given reaction. Writing and balancing equations are foundational skills for anyone studying chemistry.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is an essential concept for converting between the amount of substance in moles and its mass in grams.

In our exercise, the molar mass of calcium oxalate (\(\text{CaC}_2\text{O}_4\)) is needed to convert moles to grams. According to the step-by-step solution, the molar mass was calculated as follows:

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol, and there are two carbon atoms in each molecule
• Oxygen (O): 16.00 g/mol, and there are four oxygen atoms in each molecule

Summing these gives:\[40.08 + 2(12.01) + 4(16.00) = 128.1\ \text{g/mol}\]
The molar mass helps us convert the calculated moles of \(\text{CaC}_2\text{O}_4\) into its mass form, allowing for practical applications such as measuring the mass of this compound in urine to determine calcium levels.

Stoichiometry

Stoichiometry involves using relationships from balanced chemical equations to calculate the amount of reactants or products involved in a chemical reaction. It is a fundamental concept for quantifying interactions in chemistry.

In the context of calcium oxalate titration, stoichiometry helps understand the ratio of calcium oxalate to potassium permanganate needed to react fully. According to the balanced chemical equation, the interaction ratio is 5 moles of \(\text{CaC}_2\text{O}_4\) to 24 moles of \(\text{KMnO}_4\).

Once the moles of \(\text{KMnO}_4\) are known (0.00248012 mol in this case), stoichiometry allows the calculation of the corresponding moles of \(\text{CaC}_2\text{O}_4\) by applying the ratio:

• \[\text{Moles of } \text{CaC}_2\text{O}_4 = \frac{0.00248012 \times 5}{24} = 0.0005166875\text{ mol}\]

This stoichiometric analysis enables precise calculations necessary for verifying amounts, such as ensuring the calcium oxalate in a patient's urine is within normal health ranges.