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Question

Consider the following statements :
1. The temporary hardness in water is due to the presence of bicarbonates of calcium.
2. The permanent hardness in water is due to the presence of soluble chlorides and sulphates of calcium and magnesium.
3. The permanent hardness in water is removed by adding \(Na_2CO_3\) to water.
Which of the statements given above are correct ?

This question was previously asked in
CDS 1 2026 Maths Question Paper (12-Apr-2026)
The correct answer is
1, 2 and 3

Water Hardness Statements Analysis

Let's evaluate each statement regarding water hardness:

Statement 1: Temporary Hardness Cause

Temporary hardness is indeed caused by the presence of dissolved bicarbonates of calcium (\(Ca(HCO_3)_2\)) and magnesium (\(Mg(HCO_3)_2\)). This type of hardness can often be removed by boiling.

Statement 1 is correct.

Statement 2: Permanent Hardness Cause

Permanent hardness results from the presence of soluble chlorides (\(Cl^-\)) and sulphates (\(SO_4^{2-}\)) of calcium (\(Ca^{2+}\)) and magnesium (\(Mg^{2+}\)) in water. These salts do not precipitate upon boiling.

Statement 2 is correct.

Statement 3: Removing Permanent Hardness

Permanent hardness is characterized by the presence of \(Ca^{2+}\) and \(Mg^{2+}\) ions. Adding sodium carbonate (\(Na_2CO_3\)) introduces carbonate ions (\(CO_3^{2-}\)), which react with these metal ions to form insoluble precipitates:

  • \(Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)\)
  • \(Mg^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MgCO_3(s)\)

The removal of these ions effectively reduces permanent hardness. This method is part of the lime-soda process.

Statement 3 is correct.

Conclusion

Since all three statements (1, 2, and 3) are correct descriptions of water hardness causes and treatment, the correct option includes all of them.

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