Chapter 9.4 - Hard and Soft Water

In the previous section, we saw chemical properties of water. In this section, we will see hard water and soft water.

Some basics about hardness can be written in 6 steps:
1. Rain water is almost pure. That means, almost all molecules in a sample of rain water will be H₂O molecules.
• However, rain water may contain dissolved gases. Those gases were originally present in the atmosphere. They get dissolved in the rain water when the droplets fall through the atmosphere.
• Rain water may also contain dust particles and other harmful substances.
2. For our present discussion, we assume that, the rain water is pure.
• Water is a good solvent. So when this rain water flows on the surface of the earth, it dissolves many salts.
3. Let us see two types of those salts:
(i) Salts of calcium:
• bicarbonate (IUPAC name hydrogencarbonate) of calcium
   ♦ That is., calcium bicarbonate Ca(HCO₃)₂  
• chloride of calcium
   ♦ That is., calcium chloride CaCl₂       
• sulphate of calcium
   ♦ That is., calcium sulphate Ca(SO₄)₂       
(ii) Salts of magnesium:
• bicarbonate (IUPAC name hydrogencarbonate) of magnesium
   ♦ That is., magnesium bicarbonate Mg(HCO₃)₂
• chloride of magnesium
   ♦ That is., magnesium chloride MgCl₂
• sulphate of magnesium
   ♦ That is., magnesium sulphate Mg(SO₄)₂
4. When salts mentioned in (3) above are dissolved in water, it is called hard water. Hard water does not give lather with soap.
5. If the salts mentioned in (2) are not present in water, then soap will give lather. Then the water is called soft water.
6. Let us see how the salts causes hardness. It can be written in 4 steps:
(i) Soap contains sodium stearate C₁₇H₃₅COONa.
• When water is added, the sodium stearate dissociates as shown below:
C₁₇H₃₅COONa ⟶ C₁₇H₃₅COO^- + Na⁺
(ii) In the hard water, the salts are already present in dissociated form. So there will be Mg²⁺ or Ca²⁺ ions in the hard water.  
• Let us denote these metallic ions as M²⁺
(iii) The stearate ions react with metallic ions. The equation is:
2C₁₇H₃₅COO^- + M²⁺ ⟶ (C₁₇H₃₅COO)₂M
• (C₁₇H₃₅COO)₂M forms a precipitate.
• So instead of lather, we get a precipitate.
• Due to this reason, hard water is not suitable for laundry.
(iv) Hard water is not suitable for boilers also. This is because, the precipitate will accumulate and form scales on the inner surface of boilers. This will reduce the efficiency of those boilers.

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There are two types of hardness:
(i) Temporary hardness
(ii) Permanent hardness
 

• First we will see temporary hardness. It can be written in 4 steps:
1. We have seen that, bicarbonates, chlorides and sulphates of Ca and Mg cause hardness.
• If the hardness is due to the presence of bicarbonates, then it is called temporary hardness.
• So we can write:
Temporary hardness is caused due to the presence of any of the following two compounds:
   ♦ calcium bicarbonate Ca(HCO₃)₂
   ♦ magnesium bicarbonate Mg(HCO₃)₂.
2. There are two methods available to remove temporary hardness:
   ♦ Boiling
   ♦ Clark’s method  
3. First we will see boiling. It can be written in 3 steps:
(i) During boiling, the Ca(HCO₃)₂ if present, will be converted into CaCO₃
$\rm{Ca(HCO_3)_2~ \color {green}{\xrightarrow[{}]{\text{Heating}}} ~ CaCO_3 ↓~+~H_2O~+~ CO_2 ↑}$
• The CaCO₃ is insoluble. So it can be filtered out.
(ii) During boiling, the Mg(HCO3)₂ if present, will be converted into Mg(OH)₂
$\rm{Mg(HCO_3)_2~ \color {green}{\xrightarrow[{}]{\text{Heating}}} ~ Mg(OH)_2 ↓~+~ 2CO_2↑}$
• The Mg(OH)₂ is insoluble. So it can be filtered out.  
(iii) The filtrate will be soft water.
4. Now we will see Clark's method. It can be written in 3 steps:
(i) In this method, calculated amount of lime is added to the hard water.
(ii) When lime is added, the Ca(HCO₃)₂ if present, will be converted into CaCO₃
$\rm{Ca(HCO_3)_2~+~Ca(OH)_2~ \color {green}{\xrightarrow[{}]{}} ~ 2CaCO_3 ↓~+~2H_2O}$
• The CaCO₃ is insoluble. So it can be filtered out.
(ii) When lime is added, the Mg(HCO3)₂ if present, will be converted into Mg(OH)₂
$\rm{Mg(HCO_3)_2~+~2Ca(OH)_2~ \color {green}{\xrightarrow[{}]{}} ~ 2CaCO_3 ↓~+~Mg(OH)_2 ↓~+~2H_2O}$
• Mg(OH)₂ and CaCO₃ are insoluble. So they can be filtered out.  
(iii) The filtrate will be soft water.

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• Now we will see permanent hardness. It can be written in 4 steps:
1. We have seen that, bicarbonates, chlorides and sulphates of Ca and Mg cause hardness.
• If the hardness is due to the presence of chlorides or sulphates, then it is called permanent hardness.
• So we can write:
Permanent hardness is caused due to the presence of any of the following four compounds:
   ♦ calcium chloride CaCl₂
   ♦ calcium sulphate CaSO₄
   ♦ magnesium chloride MgCl₂.
   ♦ magnesium sulphate MgSO₄.
2. There are four methods available to remove permanent hardness:
   ♦ Treatment with washing soda (sodium carbonate)
   ♦ Calgon’s method 
   ♦ Ion exchange method 
   ♦ Synthetic resins method 
3. First we will see treatment with washing soda. It can be written in 5 steps:
(i) In this method, washing soda is added to the hard water. The metals get precipitated as the corresponding carbonates.
(ii) When washing soda is added, the CaCl₂ if present, will be converted into CaCO₃
CaCl₂ + Na₂CO₃ ⟶ CaCO₃ ↓ + 2NaCl
• The CaCO₃ is insoluble. So it can be filtered out.
(iii) When washing soda is added, the CaSO₄ if present, will be converted into CaCO₃
CaSO₄ + Na₂CO₃ ⟶ CaCO₃ ↓ + Na₂SO₄
• The CaCO₃ is insoluble. So it can be filtered out.
(iv) When washing soda is added, the MgCl₂ if present, will be converted into MgCO₃
MgCl₂ + Na₂CO₃ ⟶ MgCO₃ ↓ + 2NaCl
• The MgCO₃ is insoluble. So it can be filtered out.
(v) When washing soda is added, the MgSO₄ if present, will be converted into MgCO₃
MgSO₄ + Na₂CO₃ ⟶ MgCO₃ ↓ + Na₂SO₄
• The MgCO₃ is insoluble. So it can be filtered out.
4. Next we will see Calgon's method. It can be written in 4 steps:
(i) Calgon is the trade name of sodium hexameta phosphate (Na6qP6qO₁₈)
(ii) When Calgon is added to hard water, it dissociates into ions:
Na6qP6qO₁₈ ⟶ 2Na⁺ + Na₄P₆O₁₈^2-
(iii) If Mg²⁺ or Ca²⁺ (commonly denoted as M²⁺) are present, they will react with the Na₄P₆O₁₈^2-.
M²⁺ + Na₄P₆O₁₈^2- + ⟶ [Na₂MP₆O₁₈]^2- + 2Na⁺
• The M²⁺ thus gets locked inside the complex anion. This complex anion remains in the dissolved form.
(iv) So the M²⁺ will not be available to react with soap. Also, it will not precipitate to form scales in boilers. Thus hardness is removed.
5. Next we will see ion-exchange method. It can be written in 5 steps:
(i) Consider the compound NaAlSiO₄ (sodium aluminium silicate).
• For simplicity, this compound can be denoted as NaZ
(ii) This NaZ is a combination of two ions: Na⁺ and Z^-
• This combination is just like Na⁺Cl^-
(iii) Water is allowed to pass through a bed containing NaZ
• Suppose that, Mg²⁺ or Ca²⁺ (commonly denoted as M²⁺) are present in the water.
• Then the Na⁺ will get detached from the Na⁺Z^- combination. M²⁺ will take up the position of Na⁺
2NaZ (s) + M²⁺ (aq) ⟶ MZ₂ (s) + 2Na⁺ (aq)
• Thus the M²⁺ ions are removed from the water.
(iv) After some time, all the Na⁺ in the bed will be used up.
• When this happens, the treatment process is stopped and the bed is regenerated.
• For regeneration, the bed is washed with NaCl solution.
MZ₂ (s) + 2NaCl (aq) ⟶ 2NaZ (s) + MCl₂ (aq)
(v) The ion-exchange method is known by other names also:
   ♦ Zeolite process
   ♦ Permutit process
• NaAlSiO₄ (sodium aluminium silicate) is the zeolite/permutit.
6. Finally we will see synthetic resins method. It can be written in 5 steps:
(i) The word resin indicates an organic substance which is non-crystalline. It is usually a yellowish or brownish viscous liquid. It is insoluble in water.
(ii) The word synthetic indicates a substance which is made artificially using chemical processes.
(iii) The synthetic resin that we use for water softening is R-SO₃H
   ♦ 'R' indicates an organic group.
   ♦ So 'R-SO₃H' is a large organic molecule containing a -SO₃H group.
(iv) The R-SO₃H is first treated with NaCl solution.
• The SO₃H gets replaced by Na. Thus we get RNa
(v) Water is allowed to pass through a bed containing RNa
• Suppose that, Mg²⁺ or Ca²⁺ (commonly denoted as M²⁺) are present in the water.
• Then the Na will get detached from the RNa. Also, M²⁺ will take up the position of Na.
2RNa (s) + M²⁺ (aq) ⟶ R₂M (s) + 2Na⁺ (aq)
• Thus the M²⁺ ions are removed from the water.
(iv) After some time, all the Na in the bed will be used up.
• When this happens, the treatment process is stopped and the bed is regenerated.
• For regeneration, the bed is washed with NaCl solution.
R₂M (s) + 2NaCl (aq) ⟶ 2RNa (s) + MCl₂ (aq)

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• Until now, we were discussing about the methods to remove Mg²⁺ and Ca²⁺.
• What if we want to remove all soluble mineral salts from the water?
   ♦ Removing all soluble mineral salts means:
         ✰ removing all cations like Mg²⁺, Ca²⁺, Na⁺ etc.,
         ✰ removing all anions like Cl^-, HCO₃^-, SO₄^2- etc.,
• The process used for achieving this, can be explained in 8 steps:
(i) Consider the synthetic resins method that we saw above. In that method, we allowed the water to flow through a bed containing RNa.
(ii) Suppose that, instead of RNa, we use RH. Then the reaction will be as follows:
• If Mg²⁺ is present in the water:
2RH (s) + Mg²⁺ (aq) ⇌ MgR₂ (s) + 2H⁺ (aq)
• If Ca²⁺ is present in the water:
2RH (s) + Ca²⁺ (aq) ⇌ CaR₂ (s) + 2H⁺ (aq)
• If Na⁺ is present in the water:
RH (s) + Na⁺ (aq) ⇌ NaR (s) + H⁺ (aq)
• We see that:
Mg²⁺, Ca²⁺ and Na⁺ are effectively removed from the water. But at the same time, H⁺ is being released into the water.
(iii) As the quantity of H⁺ in the water increases, that water will become more and more acidic. But this problem is solved when this water is allowed to pass through a second bed.
(iv) The second bed contains RNH₂.
• This RNH₂ reacts with water to form RNH₃⁺.OH^-.
RNH₂ (s) + H₂O (l) ⇌ RNH₃⁺.OH^- (s)
(v) The RNH₃⁺.OH^- is able to remove anions.
• If Cl^- is present in the water:
RNH₃⁺.OH^- (s) + Cl^- (aq) ⇌ RNH₃⁺.Cl^- + OH^- (aq)
• If HCO₃^- is present in the water:
RNH₃⁺.OH^- (s) + HCO₃^- (aq) ⇌ RNH₃⁺.HCO₃^- + OH^- (aq)
• If SO₄^2- is present in the water:
2RNH₃⁺.OH^- (s) + SO₄^2- (aq) ⇌ (RNH₃⁺)₂.SO₄^2- + 2OH^- (aq)
• We see that:
Cl^-, HCO₃^- and SO₄2- are effectively removed from the water. But at the same time, OH^- is being released into the water.
(vi) The OH^- released in this way, will neutralize the H⁺ released in step (ii)
H⁺ (aq) + OH^- (aq) ⟶ H₂O (l)
(vii) As the process continues,
   ♦ All the H⁺ in the first bed will be used up.
   ♦ All the OH^- in the second bed will be used up.
• To regenerate the bed,
   ♦ The first bed is treated with dilute acid solution.
   ♦ The second bed is treated with dilute alkali solution.
(viii) Using the above seven steps, we can remove all the soluble mineral salts.
• When all soluble mineral salts are removed, it is called de-mineralised water. It is also known as de-ionized water.

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In the next section, we will see Hydrogen peroxide.

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