Question

0.00012% `MgSO_(4)` and `0.000111% CaCl_(2)` is present in water. What is the measured hardness of water and millimoles of washing soda required to purigy water 1000 L water ?

Answer 1

Basis of calculation = 100 g hard water
`MgSO_(4) = 0.00012g = (0.00012)/(120) ` mole
`CaCl_(2)=0.00012 g = (0.000111)/(111) ` mole `:. ` equivalent moles of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))` mole
`:.` mass of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))xx100=2xx10^(-4)g`
Hardness (in terms of ppm of `CaCO_(3)`)=`(2xx10^(-4))/(100)xx10^(6)=2`ppm
`CaCl_(2)+Na_(2)CO_(3)rarrCaCO_(3)+2NaCl`
`NaSO_(4)+Na_(2)CO_(3)rarrMgCO_(3)+Na_(2)SO_(4)`
`:.` Required `Na_(2)CO_(3)` for 100g of water `=((0.00012)/(120)+(0.000111)/(111))` mole
`=2xx10^(-6)` mole
`:.` Required `Na_(2)CO_(3)` for 1000 litre water `=(2xx10^(-6))/(100)xx(2)/(100)` mole (`:.` d=1g/mL)
`=(20)/(1000)` mole = 20 m mole