First slide

Hardness of water

Question

In Clarke’s process, the salts responsible for hardness get precipitated as

Moderate

A

${MgCO}_{3}, Ca {(OH)}_{2}$
B

$Mg {(OH)}_{2}, Ca {(OH)}_{2}$
C

$Mg {(OH)}_{2}, {CaCO}_{3}$
D

${MgCO}_{3}, {CaCO}_{3}$

Solution

$Ca {({HCO}_{3})}_{2} and Mg {({HCO}_{3})}_{2}$ are responsible for temperorary hardness of water which are removed in Clarke's process;
$Ca {({HCO}_{3})}_{2} + Ca {(OH)}_{2} \to {CaCO}_{3} ↓ + H_{2} O + {CO}_{2}$
$Mg {({HCO}_{3})}_{2} + 2 Ca {(OH)}_{2} \to {CaCO}_{3} ↓ + Mg {(OH)}_{2} ↓ + H_{2} O + {CO}_{2}$
$K_{sp} of Mg {(OH)}_{2} < K_{sp} of {MgCO}_{3}$ $∴ Mg {(OH)}_{2} is precipitated but not {MgCO}_{3}$

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