Question

$100\mathrm{ml}$of tap water containing $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ was titrated with$\frac{N}{50}$ $\mathrm{HCl}$with methyl orange as indicator. If $30\mathrm{ml}$ of $\mathrm{HCl}$ were required, what is the degree of temporary hardness as ppm?

Answer 1

Step 1: Given data

Volume of tap water taken containing $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$=$100\mathrm{ml}$

Volume of $\frac{N}{50}$ $\mathrm{HCl}$required for titration = $30\mathrm{ml}$

The indicator used in the titration: Methyl orange

Step 2: Formula used

As the reaction is an acid-base reaction, the product of normality and volume of the reacting acid and base is equal.

Hence, the normality equation is used:

$N_1{V}_1=N_2{V}_2$

where

${\mathrm{N}}_1$= normality of acid

$V_1$= volume of acid

$N_2$= normality of base

$V_2$= volume of base

For calculating the normality of a solution, the following formula is used:

$\mathrm{Normality}=\frac{\mathrm{Number}\mathrm{of}\mathrm{equivalents}}{\mathrm{Volume}\mathrm{in}\mathrm{liters}}$

For calculating the number of equivalents, the following formula is used:

$\mathrm{Number}\mathrm{of}\mathrm{equivalents}=\frac{Givenmass}{\mathrm{Equivalent}\mathrm{mass}}=\frac{Givenmass}{{\displaystyle \frac{\mathrm{Molar}\mathrm{mass}}{\mathrm{n}-\mathrm{factor}}}}$

Step 3: Calculation of the normality $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$.

The molar mass of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$=$100\mathrm{g}$

n factor of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$=$2$

Let the mass of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$present in the tap water be $\mathrm{w}\mathrm{g}$.

The normality of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$can be calculated using the formula:

$\mathrm{Normality}=\frac{\frac{\mathrm{w}}{\frac{100}{2}}}{1}=\frac{\mathrm{w}}{\frac{100}{2}}=\frac{\mathrm{w}}{50}$

Step 4: Calculation of the weight of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ present in the tap water.

Putting the values in the normality equation to calculate the weight of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ in the tap water.

$$\begin{gathered} \frac{w}{{\displaystyle 50}}\times 100=\frac{30}{50} \\ \Rightarrow w=0.3 \end{gathered}$$

For $100\mathrm{ml}$ the amount of $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ = $0.3\mathrm{g}$
For ${10}^6=300\mathrm{ppm}$

Hence, the degree of temporary hardness is $300\mathrm{ppm}$.