Question 0bff5

Jun 28, 2016

$= 200 m g {\mathrm{dm}}^{-} 3$

Explanation:

The sample of water contains

$C {a}^{\text{++",NO_3^-,HCO_3^"-"and Cl^"-" " ions}}$ . We are to find out the temporary hardness of water.
The temporary hardness in this sample will be caused by $C a {\left(H C {O}_{3}\right)}_{2}$ which reacts with acid solution used for titration as follows.

$C a {\left(H C {O}_{3}\right)}_{2} + {H}_{2} S {O}_{4} \to C a S {O}_{4} + 2 {H}_{2} O + 2 C {O}_{2}$

This reaction reveals that 1 mol of ${H}_{2} S {O}_{4}$ is required for nutralisation of 1 mol

$C a {\left(H C {O}_{3}\right)}_{2}$which is equivalent to 1 mol $C a C {O}_{3}$

Now the titration result is that

5cm^3 or 5xx10^-3 dm^3 0.01moldm^-3

acid nutralises $25 c {m}^{3} \mathmr{and} 25 \times {10}^{-} 3 {\mathrm{dm}}^{3}$water sample.

So the no of moles of acid required
$= \text{volume of acid"xx"its strength}$

$= 5 \cdot {10}^{-} 3 {\mathrm{dm}}^{3} \times 0.01 m o l {\mathrm{dm}}^{-} 3$

$= 5 \times {10}^{-} 5 m o l$

As per above noted equation the amount of $C a {\left(H C {O}_{3}\right)}_{2}$ in water sample will also be

$= 5 \times {10}^{-} 5 m o l$

$\text{In "CaCO_3" equivalent the no.of moles of "CaCO_3" will be}$

$= 5 \times {10}^{-} 5 m o l$

Now molar mass of $C a C {O}_{3}$ is

$= \left(40 + 12 + 3 \cdot 16\right) \frac{g}{\text{mol"=100g/"mol}}$

So the mass of $C a C {O}_{3}$

$= 5 \times {10}^{-} 5 m o l \times 100 \frac{g}{\text{mol}}$

$= 5 \times {10}^{-} 3 g = 5 m g$

Now

$25 \times {10}^{-} 3 {\mathrm{dm}}^{3}$ water sample contains 5 mg $C a C {O}_{3}$ equivalent $C a {\left(H C {O}_{3}\right)}_{2}$

So temporary hardness of water is
=("Mass expressed in " CaCO_3" equivalent")/"Volume of water"

$= \frac{5}{25 \times {10}^{-} 3} m g {\mathrm{dm}}^{-} 3$

$= 200 m g {\mathrm{dm}}^{-} 3$

A Short Cut

$\text{Hardness} = \frac{S \times V \times 50 \times 1000}{V} _ w m g {\mathrm{dm}}^{-} 3$

Where
S="Strength of acid in Normality"#

$\text{Normality"="Molarity"(M)xx"Basicity} \left(b\right)$

$V = \text{Volume of acid in L} \mathmr{and} {\mathrm{dm}}^{3}$

${V}_{w} = \text{Volume of water sample in L} \mathmr{and} {\mathrm{dm}}^{3}$