"Washing soda" (sodium carbonate)may be used to "soften" water by the removal of certain ions that would otherwise react with common soaps.When the "hardness" is due to calcium ion,the "softening" process may be represented as: Ca²⁺(aq)+ CO₃^2-(aq)→ CaCO₃(s) What mass of sodium carbonate would be required to remove essentially all of the calcium ion from 750 L of solution containing 43 mg Ca²⁺ per liter?

Question

Quizplus · Accepted Answer

The reaction between calcium ions (Ca^2+) and sodium carbonate (Na2CO3) to form calcium carbonate (CaCO3) is a 1:1 molar ratio. First, calculate the moles of Ca^2+ in 750 L of solution: $750 \, \text{L} \times \frac{43 \, \text{mg Ca}^{2+}}{1 \, \text{L}} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} \times \frac{1 \, \text{mol Ca}^{2+}}{40.08 \, \text{g}} = 0.804 \, \text{mol Ca}^{2+}$. Since the reaction is 1:1, 0.804 mol of Na2CO3 is needed. The molar mass of Na2CO3 is approximately 106 g/mol, so $0.804 \, \text{mol} \times 106 \, \text{g/mol} = 85.224 \, \text{g}$. Therefore, 85 g of sodium carbonate is required.

"Washing Soda" (Sodium Carbonate)may Be Used to "Soften" Water by the Removal