Permanent hardness due to Mg^{2+} ions is bes | vedclass

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(C) Permanent hardness due to $Mg^{2+}$ ions is best removed by using a mixture of $Ca(OH)_2$ and $Na_2CO_3$ (Clark's process combined with soda ash t

Vedclass · Accepted Answer

$Na_2CO_3 + Ca(OH)_2$

Vedclass · Answer

(C) Permanent hardness due to $Mg^{2+}$ ions is best removed by using a mixture of $Ca(OH)_2$ and $Na_2CO_3$ (Clark's process combined with soda ash treatment).$Ca(OH)_2$ reacts with $Mg^{2+}$ ions to precipitate $Mg(OH)_2$:$Mg^{2+} + Ca(OH)_2 	o Mg(OH)_2 \downarrow + Ca^{2+}$The resulting $Ca^{2+}$ ions (along with existing permanent hardness) are then removed by $Na_2CO_3$:$Ca^{2+} + Na_2CO_3 	o CaCO_3 \downarrow + 2Na^{+}$Thus,the combination of $Na_2CO_3 + Ca(OH)_2$ is the most effective method.

Permanent hardness due to $Mg^{2+}$ ions is best removed by

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