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Difficulty: HardHydrogen: Preparation, Properties, Isotopes, and Water Hardness

A 250 dm3250\text{ dm}^3 sample of hard water contains 0.012 mol dm−30.012\text{ mol dm}^{-3} of dissolved magnesium tetraoxosulfate(VI), MgSO4\text{MgSO}_4. What mass, in grams, of anhydrous sodium trioxocarbonate(IV), Na2CO3\text{Na}_2\text{CO}_3, is required to completely precipitate all the magnesium ions as magnesium trioxocarbonate(IV) and soften the water? [Molar mass of Na2CO3=106 g mol−1][\text{Molar mass of Na}_2\text{CO}_3 = 106\text{ g mol}^{-1}]

Answer: 318 g

Answer

318 g of anhydrous sodium trioxocarbonate(IV) is required.
Permanent water hardness caused by soluble magnesium salts like magnesium tetraoxosulfate(VI) (MgSO4\text{MgSO}_4) is removed by reaction with sodium trioxocarbonate(IV) (Na2CO3\text{Na}_2\text{CO}_3). The balanced reaction MgSO4(aq)+Na2CO3(aq)→MgCO3(s)+Na2SO4(aq)\text{MgSO}_4\text{(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{MgCO}_3\text{(s)} + \text{Na}_2\text{SO}_4\text{(aq)} shows a 1:1 molar ratio. A 250 dm3250\text{ dm}^3 volume at 0.012 mol dm−30.012\text{ mol dm}^{-3} contains 3.0 moles3.0\text{ moles} of MgSO4\text{MgSO}_4, which requires 3.0 moles3.0\text{ moles} of Na2CO3\text{Na}_2\text{CO}_3. Multiplying by its molar mass (106 g mol−1106\text{ g mol}^{-1}) gives 318 g318\text{ g}.

Step-by-Step Solution

1
Calculate the total number of moles of magnesium tetraoxosulfate(VI) in the water sample.
n(MgSO4)=250 dm3×0.012 mol dm−3=3.0 moln(\text{MgSO}_4) = 250\text{ dm}^3 \times 0.012\text{ mol dm}^{-3} = 3.0\text{ mol}
Molar amount is calculated by multiplying the solution volume by its molar concentration.
2
Write the balanced chemical equation for softening permanent hardness with sodium trioxocarbonate(IV).
MgSO4(aq)+Na2CO3(aq)→MgCO3(s)+Na2SO4(aq)\text{MgSO}_4(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow \text{MgCO}_3(\text{s}) + \text{Na}_2\text{SO}_4(\text{aq})
Soluble magnesium ions causing permanent hardness are removed by precipitation as insoluble magnesium trioxocarbonate(IV).
3
Calculate the mass of anhydrous sodium trioxocarbonate(IV) needed.
Mass=3.0 mol×106 g mol−1=318 g\text{Mass} = 3.0\text{ mol} \times 106\text{ g mol}^{-1} = 318\text{ g}
From the 1:1 stoichiometric ratio, 3.0 moles of sodium trioxocarbonate(IV) is required.

Key Concept

Quantitative removal of permanent water hardness using sodium trioxocarbonate(IV) (washing soda)
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Topics

Hydrogen: Preparation, Properties, Isotopes, and Water Hardness

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