Question

A sample of water is known to contain \(Mg(HCO_3)_2=7.3\ \text{mg/L\), \(Ca(HCO_3)_2=8.1\ \text{mg/L}\), and \(27.2\ \text{mg/L}\) of \(CaSO_4\). The total hardness associated with water sample in ppm in equivalents of \(CaCO_3\) is \[ (\text{At.wt. }H=1,\ C=12,\ O=16,\ Mg=24,\ Ca=40,\ S=32) \] }

• A. \(20\)
• B. \(25\)
• C. \(30\)
• D. \(40\)

Hint:

To convert hardness into \(CaCO_3\) equivalent, use \(\frac{\text{salt amount}\times 50}{\text{equivalent weight of salt}}\).

Answer 1

The Correct Option is C

Concept: Hardness is usually expressed in terms of \(CaCO_3\). The conversion formula is: \[ \text{Hardness as }CaCO_3 = \frac{\text{amount of salt} \times 50}{\text{equivalent weight of salt}} \]

Step 1: Calculate equivalent weight of \(Mg(HCO_3)_2\).
Molecular weight: \[ Mg(HCO_3)_2=24+2(1+12+48) \] \[ =24+2(61)=146 \] Since \(Mg^{2+}\) has valency \(2\), \[ \text{Eq. wt.}=\frac{146}{2}=73 \] Hardness due to \(Mg(HCO_3)_2\): \[ =\frac{7.3\times 50}{73}=5\ \text{ppm} \]

Step 2: Calculate equivalent weight of \(Ca(HCO_3)_2\).
Molecular weight: \[ Ca(HCO_3)_2=40+2(61)=162 \] \[ \text{Eq. wt.}=\frac{162}{2}=81 \] Hardness due to \(Ca(HCO_3)_2\): \[ =\frac{8.1\times 50}{81}=5\ \text{ppm} \]

Step 3: Calculate equivalent weight of \(CaSO_4\).
Molecular weight: \[ CaSO_4=40+32+64=136 \] \[ \text{Eq. wt.}=\frac{136}{2}=68 \] Hardness due to \(CaSO_4\): \[ =\frac{27.2\times 50}{68}=20\ \text{ppm} \]

Step 4: Total hardness: \[ =5+5+20 \] \[ =30\ \text{ppm} \] Therefore, \[ \boxed{30} \]