50 ml of water sample, containing temporary hardness only, required 0.1 ml of M/50 HCl for complete neutralisation. Calculate the temporary hardness of water in ppm.

Answer

2ppm

Explanation

`M_1V_1=M_2V_2`
`50xxM_1=1/50xx0.1`
`M_1=(HCO_2^(-))=1/(25xx10^3)`
mmoles of `HCO_3=1/(25xx10^3)xx50=1/500`
mmoles of `Ca^(2+)=1/1000`
mmoles of `CaCO_3=1/1000`
weight of `CaCO_3=1/1000xx10^(-3) xx 100=10^(-4)`
hardness=[weight of `CaCO_3`/weight of water]x`10^6=10^(-4)/50xx10^6=2` ppm.