Question:
Which of the following statements are correct regarding methods used to remove hardness of water?
I. Temporary hardness due to Ca and Mg bicarbonates can be removed by boiling.
II. In Clark's process, lime is used to precipitate calcium carbonate.
III. Calgon softens water by forming soluble complexes with Ca$^{2+}$ and Mg$^{2+}$.
Which of the following statements are correct regarding methods used to remove hardness of water?
I. Temporary hardness due to Ca and Mg bicarbonates can be removed by boiling.
II. In Clark's process, lime is used to precipitate calcium carbonate.
III. Calgon softens water by forming soluble complexes with Ca$^{2+}$ and Mg$^{2+}$.
Show Hint
Calgon = “Calcium Gone” → forms complexes with Ca$^{2+}$ and Mg$^{2+}$ preventing scaling.
Updated On: Jun 10, 2026
- I, II, III
- I, II only
- II, III only
- I, III only
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Hardness of water is mainly due to dissolved calcium and magnesium salts. Different chemical methods are used to remove temporary and permanent hardness.
Step 1: Statement I Boiling decomposes bicarbonates: \[ Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \downarrow + CO_2 + H_2O \] Thus, temporary hardness is removed →
Correct
Step 2: Statement II Clark's process uses Ca(OH)$_2$: \[ Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O \] Thus, calcium carbonate precipitates →
Correct
Step 3: Statement III Calgon (sodium hexametaphosphate) forms soluble complexes: \[ Ca^{2+}, Mg^{2+} \rightarrow \text{complex ions (no precipitation)} \] Thus, it removes hardness →
Correct
Final Conclusion: All three statements are correct.
Step 1: Statement I Boiling decomposes bicarbonates: \[ Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \downarrow + CO_2 + H_2O \] Thus, temporary hardness is removed →
Correct
Step 2: Statement II Clark's process uses Ca(OH)$_2$: \[ Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O \] Thus, calcium carbonate precipitates →
Correct
Step 3: Statement III Calgon (sodium hexametaphosphate) forms soluble complexes: \[ Ca^{2+}, Mg^{2+} \rightarrow \text{complex ions (no precipitation)} \] Thus, it removes hardness →
Correct
Final Conclusion: All three statements are correct.
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