A water is said to be soft water if it produces sufficient foam with the soap and waterthat 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+)toR_(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.
A 200 g sample of hard water is passed through the column of cation exchange resin, in which `H^(+)` is exchanged by `Ca^(2+)`.The outlet water of column required 50mL of 0.1 M NaOH for complete neutralization.What is the hardness of `Ca^(2+)` ion in PPm?

Options

A. 250 ppm
B. 500 ppm
C. 750 ppm
D. 1000 ppm

Answer

To solve the problem of determining the hardness of \( \text{Ca}^{2+} \) ions in parts per million (PPM), we can follow these steps:

### Step 1: Calculate the millimoles of \( \text{OH}^- \) used for neutralization
The volume of NaOH solution used is 50 mL with a concentration of 0.1 M.

\[ \text{Millimoles of } \text{OH}^- = \text{Volume (mL)} \times \text{Molarity (mol/L)} = 50 \, \text{mL} \times 0.1 \, \text{mol/L} = 5 \, \text{mmol} \]

### Step 2: Relate millimoles of \( \text{H}^+ \) to millimoles of \( \text{Ca}^{2+} \)
Since each \( \text{Ca}^{2+} \) ion replaces 2 \( \text{H}^+ \) ions in the cation exchange process, the millimoles of \( \text{Ca}^{2+} \) can be calculated as follows:

\[ \text{Millimoles of } \text{Ca}^{2+} = \frac{\text{Millimoles of } \text{OH}^-}{2} = \frac{5 \, \text{mmol}}{2} = 2.5 \, \text{mmol} \]

### Step 3: Calculate the mass of \( \text{Ca}^{2+} \)
To find the mass of \( \text{Ca}^{2+} \), we use the molar mass of calcium, which is approximately 40 g/mol:

\[ \text{Mass of } \text{Ca}^{2+} = \text{Millimoles} \times \text{Molar mass} = 2.5 \, \text{mmol} \times 40 \, \text{mg/mmol} = 100 \, \text{mg} \]

### Step 4: Calculate the hardness in PPM
The hardness in PPM is calculated based on the mass of \( \text{Ca}^{2+} \) in 200 g of water. To find the PPM in 1,000,000 g of water, we use the following formula:

\[ \text{Hardness (PPM)} = \left( \frac{\text{Mass of } \text{Ca}^{2+} \text{ in mg}}{\text{Sample mass in g}} \right) \times 10^6 \]

Substituting the values:

\[ \text{Hardness (PPM)} = \left( \frac{100 \, \text{mg}}{200 \, \text{g}} \right) \times 10^6 = 500 \, \text{PPM} \]

### Conclusion
The hardness of \( \text{Ca}^{2+} \) ions in the water sample is **500 PPM**.

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