Question

Generally only the carbonates of the Group 1 elements and the ammonium ion are soluble in water; most other carbonates are insoluble. How many milliliters of \(0.125 \mathrm{M}\) sodium carbonate solution would be needed to precipitate the calcium ion from \(37.2 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{CaCl}_{2}\) solution? \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NaCl}(a q)\)

Short answer

To precipitate the calcium ion from 37.2 mL of 0.105 M calcium chloride solution, approximately 31.25 mL of 0.125 M sodium carbonate solution is required.

Step-by-step solution

Calculate moles of calcium chloride in the given solution.

To calculate the moles of calcium chloride in the solution, use the formula: moles = Molarity × Volume In this case, the molarity of calcium chloride solution is 0.105M and the volume is \(37.2 \mathrm{~mL}\). So, moles of \(\mathrm{CaCl_2}\) = \(0.105 \mathrm{M}\) × \(37.2 \mathrm{~mL}\) × \( \frac{1 \mathrm{L}}{1000 \mathrm{~mL}} \) = 0.003906 moles

Calculate moles of sodium carbonate needed to react with calcium chloride completely.

From the balanced chemical equation: $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NaCl}(a q)$ 1 mole of \(\mathrm{Na_2CO_3}\) reacts with 1 mole of \(\mathrm{CaCl_2}\). Therefore, the moles of sodium carbonate required will be the same as the moles of calcium chloride. moles of \(\mathrm{Na_2CO_3}\) = 0.003906 moles

Calculate the volume of sodium carbonate solution required to react with calcium chloride completely.

Now that we know the moles of sodium carbonate required, we can calculate the volume of the 0.125M sodium carbonate solution needed. Use the formula: Volume = moles / Molarity For sodium carbonate solution: Volume = \( \frac{0.003906 \mathrm{ moles}}{0.125 \mathrm{M}} \) Volume = 0.031248 L Convert this volume to milliliters: Volume = \(0.031248 \mathrm{ L} \) × \( \frac{1000 \mathrm{~mL}}{1 \mathrm{ L}}\) Volume ≈ 31.25 mL

Conclusion

To precipitate calcium ion from 37.2 mL of 0.105 M calcium chloride solution, approximately 31.25 mL of 0.125 M sodium carbonate solution is required.

Key concepts

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two soluble salts react in an aqueous solution to form one or more insoluble products, known as precipitates. In this particular reaction, we are dealing with sodium carbonate and calcium chloride. These two react to form calcium carbonate, which is the precipitate, and sodium chloride, which remains in solution. The formation of a solid from two aqueous solutions is the hallmark of a precipitation reaction. The creation of the solid, called a precipitate, occurs because one of the products of the reaction is insoluble in water. In general, knowing the solubility rules, such as those for carbonates, can help predict the formation of a precipitate. For example, most carbonates apart from those of ammonium and Group 1 elements are insoluble in water, enabling such predictions.

Molarity Calculations

Molarity is a measure of the concentration of a solution, expressed as moles of solute per liter of solution. Calculating molarity involves using the formula: \[ \text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \] For our problem, calculating the molarity helped us determine how much sodium carbonate is needed to react with the given calcium chloride to effect the complete precipitation of calcium ions. Starting with the moles of calcium chloride \[ \text{moles of } \mathrm{CaCl_2} = 0.105 \text{ M} \times 37.2 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.003906 \text{ moles} \] we determine the same amount of moles of sodium carbonate is needed since the reaction between sodium carbonate and calcium chloride is 1:1 in ratio according to the balanced equation. Finally, using the molarity of the sodium carbonate solution \[ \text{Volume} = \frac{0.003906 \text{ moles}}{0.125 \text{ M}} = 0.031248 \text{ L} \] we find this corresponds to approximately 31.25 mL.

Chemical Equations

A chemical equation represents a chemical reaction where the symbols and formulas of the reactants and products are used. Correct balancing of a chemical equation is crucial as it shows the stoichiometry of the reaction, that is, the law of conservation of mass is satisfied because the number of atoms of each element is the same on both sides of the equation. For instance, the equation given in the problem is balanced: \[ \mathrm{Na}_2\mathrm{CO}_3 (aq) + \mathrm{CaCl}_2 (aq) \rightarrow \mathrm{CaCO}_3 (s) + 2 \mathrm{NaCl} (aq) \] This equation tells us that one mole of sodium carbonate reacts with one mole of calcium chloride to yield one mole of calcium carbonate and two moles of sodium chloride. Having balanced chemical equations allows us to perform stoichiometric calculations effectively, determining the exact amounts of reactants required to fully undergo the reaction and the products formed thereof.