One litre of hard water contains 12.00 mg of | vedclass

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(A) The hardness of water is removed by reacting $Mg^{2+}$ ions with washing soda $(Na_2CO_3)$.The reaction is: $Mg^{2+} + Na_2CO_3 \rightarrow MgCO_3

Vedclass · Accepted Answer

$1$

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(A) The hardness of water is removed by reacting $Mg^{2+}$ ions with washing soda $(Na_2CO_3)$.The reaction is: $Mg^{2+} + Na_2CO_3 ightarrow MgCO_3 + 2Na^+$.Number of milli-equivalents (meq) of $Mg^{2+} = \frac{	ext{mass in mg}}{	ext{equivalent weight}}$.Equivalent weight of $Mg^{2+} = \frac{	ext{Atomic mass}}{	ext{Valency}} = \frac{24}{2} = 12$.Number of meq of $Mg^{2+} = \frac{12.00 \ mg}{12} = 1 	ext{ meq}$.Since $1 	ext{ meq}$ of $Mg^{2+}$ reacts with $1 	ext{ meq}$ of $Na_2CO_3$,the required milli-equivalents of washing soda is $1$.

One litre of hard water contains $12.00 \ mg$ of $Mg^{2+}$. The milli-equivalents of washing soda $(Na_2CO_3)$ required to remove its hardness is:

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