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) to 2CaCO_(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) to CaCO_(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+)toR_(2)Ca+2H^(+)`
`2ROH+SO_(4)^(2-) to R_(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.
One litre of a sample of hard water (d=1 g/mL) cotains 136 mg of `CaSO_(4)` and 190 mg of `MgCl_(2)`. What is the total hardness of water in terms of `CaCO_(3)` ?

Options

A. 100 ppm
B. 200 ppm
C. 300 ppm
D. None of these

Answer

To find the total hardness of the water sample in terms of CaCO₃, we will follow these steps:

### Step 1: Calculate the moles of CaSO₄ in the sample
The mass of CaSO₄ in the sample is given as 136 mg. We need to convert this mass to grams and then calculate the number of moles using the molar mass of CaSO₄.

- **Molar mass of CaSO₄**:
- Ca = 40.08 g/mol
- S = 32.07 g/mol
- O₄ = 4 × 16.00 g/mol = 64.00 g/mol
- Total = 40.08 + 32.07 + 64.00 = 136.15 g/mol

- **Convert mg to g**:
\[ 136 \text{ mg} = 0.136 \text{ g} \]

- **Calculate moles of CaSO₄**:
\[ \text{Moles of CaSO₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.136 \text{ g}}{136.15 \text{ g/mol}} \approx 0.00100 \text{ moles} \]

### Step 2: Convert moles of CaSO₄ to equivalent moles of CaCO₃
The hardness contribution of CaSO₄ can be converted to CaCO₃ using the molar ratio. Since both compounds contribute equally to hardness, we can say:

- **Moles of CaCO₃ equivalent to moles of CaSO₄**:
\[ \text{Moles of CaCO₃} = \text{Moles of CaSO₄} = 0.00100 \text{ moles} \]

### Step 3: Calculate the moles of MgCl₂ in the sample
The mass of MgCl₂ in the sample is given as 190 mg. We will convert this mass to grams and then calculate the number of moles using the molar mass of MgCl₂.

- **Molar mass of MgCl₂**:
- Mg = 24.31 g/mol
- Cl₂ = 2 × 35.45 g/mol = 70.90 g/mol
- Total = 24.31 + 70.90 = 95.21 g/mol

- **Convert mg to g**:
\[ 190 \text{ mg} = 0.190 \text{ g} \]

- **Calculate moles of MgCl₂**:
\[ \text{Moles of MgCl₂} = \frac{0.190 \text{ g}}{95.21 \text{ g/mol}} \approx 0.00199 \text{ moles} \]

### Step 4: Convert moles of MgCl₂ to equivalent moles of CaCO₃
Since Mg²⁺ also contributes to hardness, we convert the moles of MgCl₂ to moles of CaCO₃. The molar ratio is equivalent (1:1).

- **Moles of CaCO₃ equivalent to moles of MgCl₂**:
\[ \text{Moles of CaCO₃} = \text{Moles of MgCl₂} = 0.00199 \text{ moles} \]

### Step 5: Calculate total moles of hardness in terms of CaCO₃
Now, we add the moles of CaCO₃ contributed by both CaSO₄ and MgCl₂:

\[ \text{Total moles of CaCO₃} = 0.00100 + 0.00199 = 0.00300 \text{ moles} \]

### Step 6: Convert moles of CaCO₃ to ppm
To find the hardness in ppm, we convert moles of CaCO₃ to grams and then to ppm.

- **Mass of CaCO₃**:
\[ \text{Mass of CaCO₃} = \text{moles} \times \text{molar mass of CaCO₃} \]
- **Molar mass of CaCO₃**:
- Ca = 40.08 g/mol
- C = 12.01 g/mol
- O₃ = 3 × 16.00 g/mol = 48.00 g/mol
- Total = 40.08 + 12.01 + 48.00 = 100.09 g/mol

\[ \text{Mass of CaCO₃} = 0.00300 \text{ moles} \times 100.09 \text{ g/mol} \approx 0.30027 \text{ g} \]

- **Convert to ppm**:
\[ \text{Hardness in ppm} = \frac{\text{mass of CaCO₃ (g)}}{\text{volume of water (L)}} \times 10^6 = \frac{0.30027 \text{ g}}{1 \text{ L}} \times 10^6 \approx 300 \text{ ppm} \]

### Final Answer:
The total hardness of the water in terms of CaCO₃ is approximately **300 ppm**.

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