Question

"Washing soda" (sodium carbonate) may be used to "soften" water by 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) + CO32-(aq) --> CaCO3(s) What mass of sodium carbonate would be required for removal of essentially all of the calcium ion from 750 L of solution containing 43 mg Ca2+ per liter?

Answer 1

Answer:

85.46 grams of sodium carbonate would be required for removal of essentially all of the calcium ion from 750 L of solution containing 43 mg calcium ion per liter.

Explanation:

$Ca^{2+}(aq) + CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Mass of calcium ions in solution = 43 mg/L = 0.043 g/L (1 mg = 0.001 g)

Volume of solution = 750 L

Mass of calcium ions in 750 L solution = 0.043 g/L × 750 L =32.25 g

Moles of calcium ions = $\frac{32.25 g}{40 g/mol}=0.8062 mol$

According to reaction, 1 mol of calcium ion reacts with 1 mol of carbonate ions

Then 0.8062 mol of calcium ion will react with:

$\frac{1}{1}\times 0.8062 mol=0.8062 mol$ carbonate ions.

$CO_3^{2-}+2Na^+\rightarrow Na_2CO_3$

1 mol of carbonation ions combines with 2 moles of sodium ions to form 1 mol sodium carbonate.

Then moles of sodium carbonate formed 0.8062 mol of carbonate ions will:

$\frac{1}{1}\times 0.8062 mol=0.8062 mol$

Mass of 0.8062 moles of sodium carbonate :

0.8062 mol × 106 g/mol= 85.46 g

85.46 grams of sodium carbonate would be required for removal of essentially all of the calcium ion from 750 L of solution containing 43 mg calcium ion per liter.