Question

The products obtained during treatment of hard water using Clark’s method are:

• A. \(CaCO_3 and MgCO_3\)
• B. \(Ca(OH)_2 and Mg(OH)_2\)
• C. \(CaCO_3 and Mg(OH)_2\)
• D. \(Ca(OH)_2 and MgCO_3\)

Answer 1

The Correct Option is C

Clark's method is a common technique used to soften hard water. Hard water typically contains dissolved calcium (\(Ca^{2+}\)) and magnesium (\(Mg^{2+}\)) ions, which can be problematic as they tend to precipitate as salts that form scale in pipes and boilers. The principle of Clark's method involves the addition of lime (\(Ca(OH)_2\)) to the hard water. Let's explore the chemical processes involved and understand the resulting products.

1. When lime (\(Ca(OH)_2\)) is added to hard water, it reacts with the temporary hardness salts such as Calcium Bicarbonate (\(Ca(HCO_3)_2\)) and Magnesium Bicarbonate (\(Mg(HCO_3)_2\)).
2. The reactions can be shown as:
   • For Calcium Bicarbonate: \(Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O\)
   • For Magnesium Bicarbonate: \(Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 + Mg(OH)_2 + 2H_2O\)
3. From the reactions above, we observe that the products formed are Calcium Carbonate (\(CaCO_3\)) and Magnesium Hydroxide (\(Mg(OH)_2\)).
4. Calcium Carbonate and Magnesium Hydroxide are insoluble in water and precipitate out, allowing them to be filtered off, thus removing the hardness from water.

Thus, in Clark's method, the products of the chemical reactions involved in softening hard water are indeed \(CaCO_3\) and \(Mg(OH)_2\).

The correct answer is: \(CaCO_3\) and \(Mg(OH)_2\).