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Question: The hardness of a water sample containing \({10^{ - 3}}\) M \(MgS{O^4}\) expressed as \(CaC{O_3}\) e...

The hardness of a water sample containing 10−3{10^{ - 3}} M MgSO4MgS{O^4} expressed as CaCO3CaC{O_3} equivalents (in ppm) is.

Explanation

Solution

10−3{10^{ - 3}} Mol/liter = MgSO4MgS{O^4} 10−3{10^{ - 3}} M/liter CaCO3CaC{O_3} (given)
Hardness in ppm (in terms of equivalents CaCO3CaC{O_3} ) =(100×10−3)1000 = \dfrac{{(100 \times {{10}^{ - 3}})}}{{1000}}

Complete step by step answer:
The molar mass of MgSO4MgS{O^4} = 120 g/mol.
The molar mass of
CaCO3CaC{O_3} = 100 g/mol
Given,
Hardness is expressed in terms of CaCO3CaC{O_3}
Moles of MgSO4=10−3MgS{O^4} = {10^{ - 3}} M
10−3{10^{ - 3}} Mol/litre MgSO4MgS{O^4} = 10−3{10^{ - 3}} Mol/litre CaCO3CaC{O_3}
∴ Moles of CaCO3CaC{O_3} in water sample = 10−3{10^{ - 3}} M
And, Mass of water (H2O)({H_2}O) = 1000 g
Hardness in PPM (in terms of equivalents of CaCO3CaC{O_3} )= 10−3×100g×1000{10^{ - 3}} \times 100g \times 1000 (mg/g)
100 mg/lit =(100×10−3)1000 = \dfrac{{(100 \times {{10}^{ - 3}})}}{{1000}}
100 PPM
Therefore, the hardness of a water sample containing 10−3{10^{ - 3}} M MgSO4MgS{O^4} expressed as CaCO3CaC{O_3}
equivalents are 100 ppm.

Note: This question can also be solved by an alternate method,
1 liter has 10−3{10^{ - 3}} moles MgSO4MgS{O^4}
So, 1000 liter has 1 mole MgSO4MgS{O^4}
= 1 mole CaCO3CaC{O_3} = 100 PPM
Hard water is a mixture of calcium and magnesium together with sulphate, bicarbonate, chloride etc. We express the hardness of water in terms of ppm because the molecular weight of Calcium carbonate (CaCO3)(CaC{O_3}) is 100 g/mol and thus it becomes easy to calculate. This is one of the main reasons for calculating the hardness of water in terms of ppm. The hardness is not completely due to the calcium but magnesium is also present. When we express the hardness as CaCO3CaC{O_3} , we calculate as if magnesium etc. were there as calcium.