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Clark's method of water softening involves adding slaked lime (calcium hydroxide, $$Ca(OH)_2$$) to hard water that contains temporary hardness (bicarbonates of calcium and magnesium).
Reaction with calcium bicarbonate:
$$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$$Calcium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate ($$CaCO_3$$), which precipitates out.
Reaction with magnesium bicarbonate:
$$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$$Magnesium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate and insoluble magnesium hydroxide ($$Mg(OH)_2$$), both of which precipitate out.
Therefore, the metal salts formed during softening of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.
The correct answer is Option B: $$CaCO_3$$ and $$Mg(OH)_2$$.
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