Question

Given,$10L$ of hard water required $0.56g$ of lime ($CaO$) for removing hardness. Hence the temporary hardness in ppm of $CaC{O_3}$ is?
A) $100$
B) $200$
C) $10$
D) $20$

Answer 1

We have to remember that the units part per million is ${10^6}$ mL of any solution. We also remember that the hardness of water can be measured in parts per million calcium carbonate, which is the amount of calcium carbonate in gms present in ${10^6}$ ml of water. Hardness can be of two types: permanent and temporary.

Complete answer:
When we look at the chemical reaction it is inferred that one mole of calcium oxide is used to produce two moles of calcium carbonate. We have to know that the temporary hardness is caused due to bicarbonate ions of calcium and magnesium ions. The reaction of removing hardness from lime can be represented as:
$Ca{\left( {HC{O_3}} \right)_2} + CaO \to {\text{ }}2CaC{O_3} + {H_2}O$
We must have to know that the molar mass of lime is $56g$ and on the other hand molar mass of calcium carbonate is $100g$ since two moles of calcium carbonate is required thus $200g$. So from here we can conclude that $0.56g$ of calcium oxide i.e. lime requires $2g$ of calcium carbonate i.e. $CaC{O_3}$.
$0.56g$ of $CaO$= $2g$ of $CaC{O_3}$ in $10L$ of water
= $2g$ $CaC{O_3}$ in ${10^4}$ mL of ${H_2}O$
= $200g$ $CaC{O_3}$ in ${10^6}$ ml of ${H_2}O$

Note:
We have to remember that the hardness can be two types: permanent and temporary. Permanent hardness is caused by sulphate and chloride ions of calcium and magnesium whereas temporary hardness is due to bicarbonate ions of calcium and magnesium. We have to know that the Ppm is the unit's parts per million which is mass of solute divided by mass of solution in gms. We also know that the calcium oxide is also known as lime.