Question

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

Answer 1

Hint: ${10^{ - 3}}$ Mol/liter = $MgS{O^4}$ ${10^{ - 3}}$ M/liter $CaC{O_3}$ (given)
Hardness in ppm (in terms of equivalents $CaC{O_3}$ ) $ = \dfrac{{(100 \times {{10}^{ - 3}})}}{{1000}}$

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

Note: This question can also be solved by an alternate method,
1 liter has ${10^{ - 3}}$ moles $MgS{O^4}$
So, 1000 liter has 1 mole $MgS{O^4}$
= 1 mole $CaC{O_3}$ = 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 $(CaC{O_3})$ 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 $CaC{O_3}$ , we calculate as if magnesium etc. were there as calcium.