Introduction

Temporary hardness in water is caused by the presence of dissolved bicarbonates of calcium ($Ca(HCO_3)_2$) and magnesium ($Mg(HCO_3)_2$). Softening this water involves converting these soluble bicarbonates into insoluble precipitates, which can then be filtered out.

Step-by-step Evaluation of Options

1. Option A: $Ca_3(PO_4)_2$ (Calcium phosphate)

   • Calcium phosphate is an insoluble salt. Adding it to water will not cause a reaction with the dissolved bicarbonates.
   • If it were to dissolve even slightly, it would introduce more $Ca^{2+}$ ions, potentially increasing the hardness.
   • Therefore, $Ca_3(PO_4)_2$ cannot be used to soften water. Option A is incorrect.

2. Option B: $Ca(OH)_2$ (Calcium hydroxide or Slaked lime)

   • This is a well-known method for removing temporary hardness, called Clark's method.
   • Calcium hydroxide reacts with calcium bicarbonate to form insoluble calcium carbonate: $Ca(HCO_3)_2(aq) + Ca(OH)_2(aq) \rightarrow 2CaCO_3(s) \downarrow + 2H_2O(l)$
   • It also reacts with magnesium bicarbonate to precipitate both magnesium hydroxide and calcium carbonate. The required amount of lime is carefully calculated to avoid adding excess $Ca^{2+}$ ions. $Mg(HCO_3)_2(aq) + 2Ca(OH)_2(aq) \rightarrow Mg(OH)_2(s) \downarrow + 2CaCO_3(s) \downarrow + 2H_2O(l)$
   • Since $Ca(OH)_2$ effectively removes temporary hardness, Option B is correct.

3. Option C: $Na_2CO_3$ (Sodium carbonate or Washing soda)

   • Sodium carbonate is used to soften both temporary and permanent hardness.
   • It reacts with calcium and magnesium bicarbonates to form insoluble carbonates: $Ca(HCO_3)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) \downarrow + 2NaHCO_3(aq)$ $Mg(HCO_3)_2(aq) + Na_2CO_3(aq) \rightarrow MgCO_3(s) \downarrow + 2NaHCO_3(aq)$
   • The products, calcium carbonate and magnesium carbonate, are precipitates and can be removed. Sodium bicarbonate ($NaHCO_3$) remains dissolved but does not contribute to hardness.
   • Therefore, $Na_2CO_3$ is a suitable reagent. Option C is correct.

4. Option D: NaOCl (Sodium hypochlorite)

   • Sodium hypochlorite is the salt of a strong base (NaOH) and a weak acid (HOCl). When dissolved in water, the hypochlorite ion ($OCl^-$) hydrolyzes to produce hydroxide ions ($OH^-$), making the solution alkaline: $OCl^-(aq) + H_2O(l) \rightleftharpoons HOCl(aq) + OH^-(aq)$
   • The hydroxide ions ($OH^-$) react with the bicarbonate ions ($HCO_3^-$) present from the temporary hardness, converting them to carbonate ions ($CO_3^{2-}$): $HCO_3^-(aq) + OH^-(aq) \rightarrow CO_3^{2-}(aq) + H_2O(l)$
   • These newly formed carbonate ions then react with the $Ca^{2+}$ and $Mg^{2+}$ ions to form insoluble precipitates: $Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s) \downarrow$ $Mg^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MgCO_3(s) \downarrow$
   • Thus, by increasing the pH and promoting the formation of carbonate precipitates, NaOCl can soften temporarily hard water, although it is primarily used as a disinfectant.
   • Therefore, Option D is correct.

Conclusion

The reagents that can be used for softening the temporary hardness of water are Calcium hydroxide ($Ca(OH)_2$), Sodium carbonate ($Na_2CO_3$), and Sodium hypochlorite (NaOCl).