Question

The hardness of water due to $HC{O_3}^ -$ is $122ppm$ . Select the correct
statements.
A. The hardness of water in terms of $CaC{O_3}$ is $200ppm$
B. The hardness of water in terms of $CaC{O_3}$ is $100ppm$
C. The hardness of water in terms of $CaC{l_2}$ is $222ppm$
D. The hardness of water in terms of $MgC{l_2}$ is $95ppm$

Answer 1

This is a multiple choice question where we need to find out the moles of the carbonates and then compare it with the options. Here, the hardness of water is the amount of minerals present in it.

Step by step answer: First we have to discuss what water hardness is. Hardness of water is the amount of dissolved calcium and magnesium in the water. Hard water is high in dissolved minerals like carbonates, bicarbonates and sulphates. When it is heated solid deposits of calcium carbonate can form. Water’s hardness depends on the amount or concentration of cations in water. There are two types of hardness, Temporary and Permanent Hardness. Temporary hardness is a type when bicarbonates are dissolved in water. They yield calcium and magnesium ions. Hardness can be removed by boiling. Permanent hardness is caused due to presence of sulphates and chlorides of calcium and magnesium. This hardness can be caused by the water softener and ion exchange column.
Hardness can be quantified as instrumental analysis. The total water hardness is the sum of molar concentrations of calcium and magnesium ions in $mol/L$ units. It measures the total concentrations of the elements.
According to the United States Geological survey, they classify water into four types which are soft, moderately hard, hard and very hard according to minerals dissolved, pH and its temperature.

Now, in the question it is said that the hardness of water caused by bicarbonates is $122ppm$. Since 2 moles of $HC{O_3}$ is present. So there should be one mole of each of $CaC{O_3}$ , $CaC{l_2}$ and $MgC{l_2}$ to have equal hardness. Now, the molecular weight of $HC{O_3}$ is $61g/mol$ . Therefore,
$ppm$ of $HC{O_3}$ $= 61 \times 2 = 122g$ in ${10^6}ml$ ${H_2}O$
Therefore,
$1mol$ of $CaC{O_3}$ = $100ppm$
$1mol$ of $CaC{l_2}$ = $111ppm$
$1mol$ of $MgC{l_2}$ = $95ppm$

So, the correct answers are B and D.

Note: Please take care of unit conversions and the exact conversion formula as it may lead to wrong answers, go through the details of parts per million(ppm) once.