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Question: 100 mL of a water sample contains 0.81g of calcium bicarbonate and 0.73g of magnesium bicarbonate. T...

100 mL of a water sample contains 0.81g of calcium bicarbonate and 0.73g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents ofCaCO3CaC{{O}_{3}} is: (molar mass of calcium bicarbonate is 162 g/mol and magnesium bicarbonate is 146 g/mol.
A) 1000 ppm
B) 10,000 ppm
C) 100 ppm
D) 5,000 ppm

Explanation

Solution

Here the answer is dependent on the concept of degree of hardness measurement which is given by the ratio of weight of hardness causing salt to the molecular weight which is multiplied by 100 and then converting it into ppm.

Complete step by step solution:
We have studied about the concepts of the formulas which are used for the calculation of molarity, equivalent weights, normality, hardness measurement etc.
Let us now see the calculation of hardness and then converting that value in terms of parts per million that is ppm.
Foe this, let us start with the reaction of calcium bicarbonate that is the dissociation ofCa(HCO3)2Ca{{(HC{{O}_{3}})}_{2}} into its constituent parts.
Ca(HCO3)2→CaCO3+H2O+CO2Ca{{(HC{{O}_{3}})}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}
Now, this total amount of calcium carbonate formed is to be measured by taking into consideration both calcium as well as magnesium bicarbonate.
Thus, according to the data given we have to find the total degree of hardness which is given by,neq.CaCO3=neq.Ca(HCO3)2+neq.Mg(HCO3)2{{n}_{eq}}.CaC{{O}_{3}}={{n}_{eq}}.Ca{{(HC{{O}_{3}})}_{2}}+{{n}_{eq}}.Mg{{(HC{{O}_{3}})}_{2}}
Now, neq.Ca(HCO3)2=0.81162=1200moles{{n}_{eq}}.Ca{{(HC{{O}_{3}})}_{2}}=\dfrac{0.81}{162}=\dfrac{1}{200}moles
Similarly, neq.Mg(HCO3)2=0.73146=1200moles{{n}_{eq}}.Mg{{(HC{{O}_{3}})}_{2}}=\dfrac{0.73}{146}=\dfrac{1}{200}moles
Now, total
neq.CaCO3=1200+1200=1100=0.01moles{{n}_{eq}}.CaC{{O}_{3}}=\dfrac{1}{200}+\dfrac{1}{200}=\dfrac{1}{100}=0.01moles
In terms of grams that is in 100mL we have,0.01×100=1g0.01\times 100=1g.
Thus, 1g of calcium carbonate is obtained and in terms of part per million in 100 mL, we have
⇒1100×106\Rightarrow \dfrac{1}{100}\times {{10}^{6}}
⇒104ppm=10,000ppm\Rightarrow {{10}^{4}}ppm=10,000ppm

Thus, the correct answer is option B) 10,000 ppm.

Note: Note that the hardness of water is mainly caused by the minerals that are because of the presence of calcium and magnesium ions and these ions are responsible for water to be hard and can be removed by the process that involves ion exchange, permutit and several others but not by physical process like boiling, filtration etc.