Problem 116 - General Chemistry

Question

Bone was dissolved in hydrochloric acid, giving \(50.0 \mathrm{~mL}\) of solution containing calcium chloride, \(\mathrm{CaCl}_{2} .\) To precipitate the calcium ion from the resulting solution, an excess of potassium oxalate was added. The precipitate of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), weighed \(1.437 \mathrm{~g} .\) What was the molarity of \(\mathrm{CaCl}_{2}\) in the solution?

Step-by-Step Solution

Answer

The molarity of \( \text{CaCl}_2 \) in the solution is approximately 0.224 M.

Step 1: Write the Balanced Chemical Equation

The reaction to consider is the precipitation of calcium ions as calcium oxalate. The balanced equation is \[ \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{CaC}_2\text{O}_4 \] This equation shows that one mole of calcium ions reacts with one mole of oxalate ions to form one mole of calcium oxalate.

Step 2: Calculate Moles of Calcium Oxalate

Convert the mass of calcium oxalate to moles. The molar mass of \( \text{CaC}_2\text{O}_4 \) is approximately \( 40.08 + 2 \times 12.01 + 4 \times 16.00 = 128.10 \, \text{g/mol} \). \[ \text{moles of } \text{CaC}_2\text{O}_4 = \frac{1.437 \text{ g}}{128.10 \text{ g/mol}} \approx 0.01122 \text{ moles} \]

Step 3: Relate Moles to Calcium Chloride

From the balanced reaction, the moles of \( \text{Ca}^{2+} \) are equal to the moles of \( \text{CaC}_2\text{O}_4 \). Therefore, the moles of \( \text{CaCl}_2 \), which provides \( \text{Ca}^{2+} \), is also \( 0.01122 \text{ moles} \).

Step 4: Calculate Molarity of Calcium Chloride

Use the definition of molarity: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] The volume of the solution is \( 50.0 \text{ mL} \) or \( 0.0500 \text{ L} \). Hence, \[ M = \frac{0.01122 \text{ moles}}{0.0500 \text{ L}} = 0.2244 \text{ M} \] This is the molarity of \( \text{CaCl}_2 \) in the solution.

Key Concepts

Molarity Calculation

Molarity is a way of expressing the concentration of a solution. It tells us how many moles of a substance are present in one liter of solution. Calculating molarity involves a simple formula:

• \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

In this formula:

• The moles of solute refer to the amount of the chemical you have dissolved in your solution.
• The volume of the solution is always measured in liters.

This means to find molarity, you need to know two things: how much of the substance you've got in moles, and how much liquid you're dealing with inside the solution.
In the exercise, we determined the moles of calcium oxalate first, which equalled the moles of calcium ions since the reaction between calcium ions and oxalate is a 1:1 molar ratio. Then, using the volume of the solution provided (50 mL or 0.050 L), we plugged those numbers into the formula to find the molarity of \( \text{CaCl}_2 \). The calculated concentration was 0.2244 M.

Chemical Reactions

Chemical reactions occur when substances interact to form new products. In the context of the exercise, the reaction involved the interaction between calcium ions and oxalate ions. The equation is:

• \( \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{CaC}_2\text{O}_4 \)

This is an example of a precipitation reaction, where two soluble ions in solution react to form an insoluble solid, called a precipitate.

In this case, calcium ions (\( \text{Ca}^{2+} \)) from calcium chloride and oxalate ions (\( \text{C}_2\text{O}_4^{2-} \)) from potassium oxalate combined to form calcium oxalate, which is insoluble and precipitates out of the solution. These observations form a critical aspect of understanding chemical reactions by identifying the reactants, potential products, and the reaction type.

Stoichiometry

Stoichiometry is like a recipe for reactions. It uses balance and ratio to predict the amounts of reactants needed or products formed. It relies on the balanced chemical equation to provide these mole ratios.
In the exercise, the balanced equation tells us that one mole of calcium ions reacts with one mole of oxalate ions. This is crucial because stoichiometry uses the relationships between quantities of reactants and products in a reaction:

• By knowing the amount of one substance, you can calculate the amount of another.
• In our case, the moles of calcium ions produced from \( \text{CaCl}_2 \) were the same as the moles of calcium oxalate formed.

Understanding stoichiometry allows you to convert between mass, moles, and molecules so you can predict how much of each chemical you need or will produce in a reaction.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, which we call a precipitate. For a precipitation reaction to occur, one of the products must be insoluble (or only sparingly soluble) in water.

Consider the exercise, where adding potassium oxalate to a calcium chloride solution caused the formation of a solid calcium oxalate. We performed the reaction:

• The calcium ions (\( \text{Ca}^{2+} \)) met up with the oxalate ions (\( \text{C}_2\text{O}_4^{2-} \)).
• The result was calcium oxalate (\( \text{CaC}_2\text{O}_4 \)), a compound that is not soluble in the solution.

This demonstrates how precipitation reactions can be used to isolate specific compounds from a mixture. Watching for color changes or the appearance of solids in liquid can indicate such a reaction.