A water is said to be soft water if it produces sufficient foam with the soap and water that does not produce foam with soap is known as hard water. Hardness has been classified into two types (i)Temporary hardness (ii) Permanent hardness.
Temporary hardness is due to presence of calcium and magnesium bicarbonate. It is simply removed by boiling as
`Ca(HCO_(3))_(2)overset(Delta)rarr CaCO_(3)darr+CO_(2)uarr+H_(2)O`
`Mg(HCO_(3))_(2)overset(Delta)rarr MgCO_(3)darr+CO_(2)uarr+H_(2)O`
temporary hardness can also be removed by addition of slaked lime, `Ca(OH)_(2)`
`Ca(HCO_(3))_(2)+Ca(OH)_(2)to2CaCO_(3)darr+2H_(2)O`
permanent hardsness is due to presencce of sulphates and chlorides of Ca,Mg,etc. It is removed by washing soda as
`CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)darr+2NaCl`
`CaSO(4)+Na_(2)CO_(3)to CaCO_(3)darr+Na_(2)SO_(4)`
Permanent hardness also removed by ion exchange resin process as
`2RH+Ca^(2+) to R_(2)Ca+2H^(+)`
`2ROH+SO_(4)^(2-)toR_(2)SO_(4)+2OH^(-)`
The degree of hardness of water is measured in terms of PPm of `CaCO_(3)` 100 PPm means 100 g of `CaCO_(3)` is present in `10^(6)` g of `H_(2)O`. If any other water sample which contain 120 PPm of `MgSO_(4)`, hardness in terms of `CaCO_(3)` is equal to =100 PPm.
What is the mass of `Ca(OH)_(2)` required for 10 litre of water remove temporary hardness of 100 PPm due to `Ca(HCO_(3))_(2)` ?

Options

A. 1.62 g
B. 0.74 g
C. 7.4 g
D. None of these

Answer

To solve the problem of determining the mass of \( \text{Ca(OH)}_2 \) required to remove the temporary hardness of water due to \( \text{Ca(HCO}_3)_2 \) in 10 liters of water with a hardness of 100 ppm, we can follow these steps:

### Step 1: Understand the concept of ppm
PPM (parts per million) is a way to express very dilute concentrations of substances. 100 ppm means that there are 100 grams of \( \text{CaCO}_3 \) in 1,000,000 grams (or 1,000 liters) of water.

### Step 2: Calculate the amount of \( \text{CaCO}_3 \) in 10 liters
Since 100 ppm corresponds to 100 grams of \( \text{CaCO}_3 \) in 1,000,000 grams of water, we can calculate the amount of \( \text{CaCO}_3 \) in 10 liters (10,000 grams of water):
\[ \text{Amount of } \text{CaCO}_3 = \frac{100 \text{ grams}}{1,000,000 \text{ grams}} \times 10,000 \text{ grams} = 1 \text{ gram} \]

### Step 3: Determine the stoichiometry of the reaction
The reaction for the removal of temporary hardness due to \( \text{Ca(HCO}_3)_2 \) by \( \text{Ca(OH)}_2 \) is:
\[ \text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3 + 2\text{H}_2\text{O} \]
From the balanced equation, we see that 1 mole of \( \text{Ca(OH)}_2 \) reacts with 1 mole of \( \text{Ca(HCO}_3)_2 \) to produce 2 moles of \( \text{CaCO}_3 \).

### Step 4: Calculate the molar masses
- Molar mass of \( \text{Ca(OH)}_2 \):
- \( \text{Ca} = 40 \, \text{g/mol} \)
- \( \text{O} = 16 \, \text{g/mol} \)
- \( \text{H} = 1 \, \text{g/mol} \)
- Total = \( 40 + 2(16) + 2(1) = 74 \, \text{g/mol} \)

- Molar mass of \( \text{CaCO}_3 \):
- \( \text{Ca} = 40 \, \text{g/mol} \)
- \( \text{C} = 12 \, \text{g/mol} \)
- \( \text{O} = 16 \, \text{g/mol} \)
- Total = \( 40 + 12 + 3(16) = 100 \, \text{g/mol} \)

### Step 5: Calculate the mass of \( \text{Ca(OH)}_2 \) needed
From the stoichiometry, we know that 1 mole of \( \text{Ca(OH)}_2 \) is required to remove 1 mole of \( \text{CaCO}_3 \). Since we have 1 gram of \( \text{CaCO}_3 \):
\[ \text{Moles of } \text{CaCO}_3 = \frac{1 \text{ gram}}{100 \text{ g/mol}} = 0.01 \text{ moles} \]
Thus, the moles of \( \text{Ca(OH)}_2 \) required will also be 0.01 moles.

Now, we can find the mass of \( \text{Ca(OH)}_2 \) needed:
\[ \text{Mass of } \text{Ca(OH)}_2 = \text{Moles} \times \text{Molar Mass} = 0.01 \text{ moles} \times 74 \text{ g/mol} = 0.74 \text{ grams} \]

### Step 6: Conclusion
The mass of \( \text{Ca(OH)}_2 \) required to remove the temporary hardness of 100 ppm due to \( \text{Ca(HCO}_3)_2 \) in 10 liters of water is **0.74 grams**.

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