A 200 mL sample of hard water requires 33.0 " mL of " 0.01 M `H_2SO_4` for complete neutralisation.
200 " mL of " the same sample was boiled with 15.0 " mL of " 0.1 M NaOH solution, filtered and made up to 200 mL again. This sample now requires 53.6 " mL of " 0.01 M `H_2SO_4`. Calculate Mg hardness.

Explanation

33.0 " mL of " `0.01xx2NH_2SO_4`
`-=(333.0xx0.01xx2xx50)/(1000)`
`-=0.033gCaCO_3(EwCaCO_3=50)`
`-=(0.033xx10^(6))/(200mL)-=165ppm of CaCO_3` Eq
Thus temporary hardness `-=165ppm`
`mEqNaOH=15xx0.1xx1=1.5`
`m" Eq of "H_2SO_4=53.6xx0.01xx2=1.072`
`m" Eq of "NaOH` (excess) that reacted with water
`=1.5-1.072=0.428mEq`
`0.428mEq NaOH-=(0.428xx50xx10^(6))/(1000xx200)ppmCaCO_3`
`-=107ppm of CaCO_3`
Thus Mg hardness `-=107ppm`