Question

One litre hard water contains 12.00 mg $Mg^{2+}$. Milliequivalents of washing soda required to remove its hardness is

• A. 1
• B. 12.16
• C. $1 \times 10^{-3}$
• D. $12.16 \times 10^{-3}$

Answer 1

The Correct Option is A

To determine the milliequivalents of washing soda required to remove hardness caused by Mg^{2+} ions in one litre of hard water, we must first understand the concept of hardness removal using washing soda (Na_2CO_3). Hardness in water due to Mg^{2+} ions is often removed by precipitation. Here's a step-by-step breakdown of the solution:

1. Identify the mass of Mg^{2+} ions in the given water sample:
   • Mass of Mg^{2+} ions = 12.00 mg = 12.00 x 10^-3 g
2. Calculate the moles of Mg^{2+}:
   • Molar Mass of Mg^{2+} = 24 g/mol
   • Moles of Mg^{2+} = \frac{12.00 \times 10^{-3}}{24}
   • Moles of Mg^{2+} = 0.5 \times 10^{-3}
3. Determine the equivalent factors:
   • One equivalent of Mg^{2+} is 0.5 moles because each mole of Mg^{2+} requires one mole of carbonate ions for precipitation.
4. Convert to milliequivalents:
   • Milliequivalents of Mg^{2+} = Moles of Mg^{2+} \times 1000
   • Milliequivalents of Mg^{2+} = 0.5 \times 10^{-3} \times 1000 = 0.5
5. Calculate the milliequivalents of washing soda needed:
   • Since 1 mole of Na_2CO_3 will react with 1 mole of Mg^{2+}, 0.5 milliequivalent of washing soda is required per 0.5 milliequivalent of Mg^{2+}.
   • The option closest to this calculated value is 1 milliequivalent.

Hence, the correct answer is 1 milliequivalent of washing soda is required to remove the hardness from 1 litre of hard water.