Question

Synthetic resin method is used to remove hardness of water by using synthetic cation exchangers. The pH of 1 litre of water when passed through an ion exchange resin $\sf RH_2$ is 3.3 . Find the hardness of water in terms of $\sf CaCO_3$ . ​

Answer 1

The hardness of water in terms of  $CaCo_{3}$ is 100 ppm

Explanation:

Given: The pH of 1 litre of water when passed through an ion exchange resin  is 3.3

To find: The hardness of water in terms of  $CaCo_{3}$

Solution:

Cation exchange resins are large organic molecules that contain the  - $So_{3}H$ group.

The resin is firstly changed to RNa by treating it with Nacl.

This resin then exchanges $Na^{+}$ ions with $Ca^{2+}$ and $mg^{2+}$ ions, thereby making the water soft.

$2RNa + M^{2+} - > R_{2} M(s) + 2Na^{+}$

There are cation exchange resins in $H^{+}$ form.

The resin exchange $H^{+}$ ions for $na^{+} ,ca^{2+} ,mg^{2+}$ ions

$2RH + M^{2+} - > R_{2} M +2H^{+}$

So,

$Ca^{2+} = 2H^{+}$

Given pH = 3.3

pH = -log[$H_{3} O^{+}$]

Hence,

pH = $10^{-3.3}$

= 5.0 × $10^{-4}$

Hence,

$ca^{2+} = 2$× 5.0×$10^{-4} = 10^{-3}$

No of moles $CaCo_{3} = 10^{-3}$

Given mass of $CaCo_{3} = 10^{-3}$ × 100

Hardness of water in terms of $CaCo_{3} = \frac{10^{-3} (10^{6} )(100)}{1000}$

= 100 ppm

Final answer:

The hardness of water in terms of  $CaCo_{3}$ is 100 ppm

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