Question

What mass of NaCO3 is required to remove 90% of these ions from10.0L of laundry water?

Since soap and detergent action is hindered by hard water, laundryformulations usually include water softeners-calledbuilders-designed to remove hard water ions (especially Ca2+ andMg2+) from the water. A common builder used in North America issodium carbonate. Suppose that the hard water used to do laundrycontains 75 ppm CaCO3 and 55 ppm MgCO3 (by mass).

Answer 1

The sodium carbonate provide sufficient carbonate ions to precipitate Ca2+ and Mg2+ ions as their respective insoluble carbonates. It thus reduces the hardness of wwater

             Ca2+(aq) + CO32–(aq) → CaCO3(s)

And,     Mg2+(aq) + CO32–(aq) → MgCO3(s)

Note that : 1 ppm = 1 mg /l

75 ppm CaCO3 75 mg of CaCO3 is present in 1 litre of water,

                        So, 750 mg would be present in 10.0 litres of water.

Milli Moles (mmoles) of CaCO3 in 10.0 L = (750 mg )/ (100 g/mol) = 7.50 mmoles

Note that molar mass of CaCO3 = 100 g/mol

Similarly,

55 ppm MgCO3 55 mg of MgCO3 is present in 1 litre of water,

                        So, 550 mg would be present in 10.0 litres of water.

mmoles of MgCO3 in 10.0 L = (550 mg )/ (84.3 g/mol) = 6.52 mmoles

Note that molar mass of MgCO3 = 84.3 g/mol

Total mmoles = 6.52 + 7.50 = 14.02

Now 90 % of 14.02 mmoles are removed that is 0.90 x 14.02 = 12.62

So, 12.62 mmoles of carbonate ion or Na2CO3 (as one mole of Na2CO3 conatins one mole of carbonate), are required to remove the hardness.

Molar mass of Na2CO3 = 106 g/mol

Therefore, mass of Na2CO3 required = (12.62 mmoles) x (106 g/mol) = 1337.72 mg = 1.34 g ...Answer