Question

100 kg hard water contains 5g $MgS{O}_4$ The hardness of water is:

Answer 1

Total hardness= $Ca$ hardness $+$ $Mg$ hardness

The calcium and $Mg$ hardness is the concentration of $Ca$ & $Mg$ ions expressed as equivalents of $CaC{O}_3$

For $Mg=\frac{M_{CaC{O}_3}}{M_{Mg}}=\frac{100.1}{24.3}=4.1$

So total hardness= $4.1\left[M{g}^{2+}\right]ppm$ of $CaC{O}_3$

Given $100kg$ of water contains $5g$ $MgS{O}_4$

i.e, $100L$ water contains $500mg$ of $MgS{O}_4$

$=\frac{\left(500mg\right)24.3}{M_{MgS{O}_4}}$

$=\frac{\left(500mg\right)\left(24.3\right)}{120.7}=100mg$ of $Mg$ ims

$\therefore \left[M{g}^{2+}\right]=100mg/100L=1mg/L$

So hardness of water= $4.1\left[1mg/L\right]$

$=4.1ppm$ of $CaC{O}_3$