`RH_2` (ion exchange resin) can replace `Ca^(2+)` ions in hard water as:
`RH_(2) + Ca^(2+) to "Rca" + 2H^(+)`
If 1 L of hard water after passing through `RH_2` has pH = 3, then hardness in parts per million of `Ca^(2+)` is :

Options

A. 10 ppm
B. 40 ppm
C. 100 ppm
D. 20 ppm

Answer

D

Explanation

`[H^(+)] = 10^(-3) M[therefore pH = 3]=- log [H^(+)]`
`=10^(-3)` moles per litre
`[Ca^(2+)] = 5 xx 10^(-4)` moles per litre (Since, 1 mole of `Ca^(2+)` gives 2 moles of `H^+`)
`= 5 xx 10^(-1)` moles of `Ca^(2+)` per `10^(3)` L
`=5 xx 10^(-1) xx 40g Ca^(2+)` per `10^(6)` g of `H_(2)O`
`Ca^(2+) = 20` ppm = (H)