One liter of hard water contains 12.00 , mg o | vedclass

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(A) The hardness is removed by the reaction: $Mg^{2+} + Na_2CO_3 ightarrow MgCO_3 + 2Na^+$. Number of milliequivalents of $Mg^{2+} = \frac{	ext{mas

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$1$

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

One liter of hard water contains $12.00 \, mg$ of $Mg^{2+}$. How many milliequivalents of washing soda are required to remove its hardness?

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