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Question: 1 L of pond water contains 20 mg of \(C{a^{2 + }}\) and 12 mg of \(M{g^{2 + }}\) ions. What is the v...

1 L of pond water contains 20 mg of Ca2+C{a^{2 + }} and 12 mg of Mg2+M{g^{2 + }} ions. What is the volume of a 2N Na2CO3N{a_2}C{O_3} solution required to soften 5000 L of pond water?
A. 500 L
B. 50 L
C. 5L
D. None of these

Explanation

Solution

To calculate the volume of Na2CO3N{a_2}C{O_3}, the formula for calculating normality and milliequivalent is used. The millie equivalent is calculated in milligrams. The normality is defined as the gram equivalent mass of compound present in one litre solution.

Complete answer: Given
Volume of pond water is 1 L.
Mass of Ca2+C{a^{2 + }} ion is 20 mg.
Mass of Mg2+M{g^{2 + }} ion is 12 mg.
Normality of Na2CO3N{a_2}C{O_3} is 2N.
Ca2+C{a^{2 + }} ion forms calcium carbonate and Mg2+M{g^{2 + }} ion forms magnesium carbonate.
Milliequivalent of Ca2+C{a^{2 + }} ion and milliequivalent of Mg2+M{g^{2 + }} ion is equal to the milli equivalent of Na2CO3N{a_2}C{O_3}.
The equation is given as shown below
Milli eq. of Ca2+C{a^{2 + }}+ Milli eq. of Mg2+M{g^{2 + }}= Milli.eq. of Na2CO3N{a_2}C{O_3}……(i)
Milliequivalent is defined as the mass in milligrams of the solute component equal to 1/1000 of its gram equivalent mass by considering the valency of the ions.
The formula for calculating the milli equivalent is shown below.
mEq=m×VMmEq = \dfrac{{m \times V}}{M}
Where,
mEq is the millie equivalent.
m is the mass in mg.
V is the valency
M is the molecular weight.
Normality is defined as the concentration of solution measured in terms of gram equivalent of solute in one litre solution or milliequivalent of solute is one litre solution.
The formula for normality is shown below.
N=meqVN = \dfrac{{meq}}{V}
Where,
N is the normality.
meq is milliequivalent
V is the volume.
To calculate the volume Na2CO3N{a_2}C{O_3} solution, substitute the values, in equation (i).
⇒2040×2+1224×2=2×V\Rightarrow \dfrac{{20}}{{40}} \times 2 + \dfrac{{12}}{{24}} \times 2 = 2 \times V
⇒0.5×2+0.5×2=2×V\Rightarrow 0.5 \times 2 + 0.5 \times 2 = 2 \times V
⇒1+1=2×V\Rightarrow 1 + 1 = 2 \times V
⇒V=1mL\Rightarrow V = 1mL
1mL of Na2CO3N{a_2}C{O_3} solution is used to soften 1 L of pond water.
To soften 5000 L of pond water, the volume needed is
⇒1×10−3×5000\Rightarrow 1 \times {10^{ - 3}} \times 5000
⇒5L\Rightarrow 5L
Therefore, the correct option is C.

Note:
Make sure to convert volume in millilitre to litre as the volume is calculated in terms of litre. Calcium carbonate and magnesium carbonate present in the pond water is insoluble in water, thus it is removed from the water to form soft water.