The metal salts formed during softening of hardwater using Clark's method are :
- Ca(OH)2 and Mg(OH)2
- CaCO3 and Mg(OH)2
- Ca(OH)2 and MgCO3
- CaCO3 and MgCO3
The Correct Option is B
Solution and Explanation
To understand the softening of hard water using Clark's method, let's delve into the chemistry behind it. Clark's method involves the addition of slaked lime, chemically known as calcium hydroxide (Ca(OH)2), to hard water. Hard water typically contains dissolved bicarbonates of calcium and magnesium, which are responsible for its hardness.
The primary reactions involved in the Clark's method are:
- Addition of calcium hydroxide to the hard water:
Ca(OH)_2 + Ca(HCO_3)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O
This reaction causes the calcium bicarbonate, which is soluble, to be converted into calcium carbonate (CaCO3), which precipitates out of the solution. - Simultaneously, for the magnesium bicarbonate present:
Ca(OH)_2 + Mg(HCO_3)_2 \rightarrow Mg(OH)_2 \downarrow + CaCO_3 \downarrow + 2H_2O
Here, magnesium bicarbonate is converted into magnesium hydroxide (Mg(OH)2), which precipitates out. Calcium carbonate is also formed as a byproduct.
Thus, the precipitated compounds formed during the softening process, which are insoluble and can be removed by filtration, are calcium carbonate (CaCO3) and magnesium hydroxide (Mg(OH)2).
Given this explanation, the correct answer to the question is the option with the metal salts CaCO3 and Mg(OH)2, as calcium carbonate and magnesium hydroxide are the two compounds formed when hard water is softened using Clark's method.
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