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

$_{1g - eq}^{MG^{2+}} + _{\, \, \, 1g - eq}^{Na_2CO_3} \rightarrow MgCo_3 + 2Na^+$ 1 g-equivalent of $Mg^{2+} = 12 \, g \, of \, Mg^{2+}$ $= 12000 \, mg \, of \, Mg^{2+}$ Now, 12000 mg of $Mg^{2+} \equiv 100$ milliequivalent of $ \, \, \, \, \, \, \, \, \, \, \, Na_2 CO_3$ $\therefore 12 mg \, of \, Mg^{2+} \equiv milliequivalent \, of \, Na_2Co_3$