Question
Asked Jan 9, 2020
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 Since soap and detergent action is hindered by hard water, laundry formulations usually include water softeners—called builders—designed to remove hard water ions (especially Ca2+ and Mg2+from the water. A common builder used in North America is sodium carbonate. Suppose that the hard water used to do laundry contains 75 ppm CaCO3   and 55 ppm MgCO3 (by mass). What mass of Na2CO3 will remove 90.0% of these ions from 10.0 L of laundry water? 

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Expert Answer

Step 1

Given, the concentration of CaCO3 = 75 ppm means that 75 mg of CaCO3 is present in 1 L of hard water.

The concentration of MgCO3 = 55 ppm means that 55 mg of MgCO3 is present in 1 L of hard water.

 Conversion of mg to g of CaCO3 and MgCO3 respectively

1mg = 10-3 g

75 mg of CaCO3 = 75 x 10-3 g = 0.075 g

55 mg of MgCO3 = 55 x 10-3 g = 0.055 g

The number of moles of CaCO3 and MgCO3 is calculated by –

Chemistry homework question answer, step 1, image 1
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Step 2

Therefore, the number of moles of CaCO3 present in 10 L of water = 10 x 7.5 x 10-4 mol = 7.5 x 10-3 mol

And the number of moles of MgCO3 present in 10 L of water = 10 x 6.52 x 10-4 mol = 6.52 x 10-3 mol

Now the total number of moles –

Chemistry homework question answer, step 2, image 1
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Step 3

It means that 14.02 x 10-3 moles of Na2CO3 is required to remove all the Ca2+ and Mg2+ ions.

Now mass of Na2CO3 is calculated by the formula –

Mass = Moles x Molar mass

Molar m...

Chemistry homework question answer, step 3, image 1
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