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RH2 (ion exchange resin) can replace Ca^...

`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 :

A

10 ppm

B

40 ppm

C

100 ppm

D

20 ppm

Text Solution

Verified by Experts

The correct Answer is:
D
`[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)
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Knowledge Check

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

    A
    20
    B
    10
    C
    40
    D
    100

  • RH_(2) ( ion exchange resin) can replace Ca^(2+) d in hard water as. RH_(2)+Ca^(2+)rarrRCa+2H^(+) 1 "litre" of hard water passing through RH_(2) has pH2 . Hence hardness in pp m "of" Ca^(2+) is:

    A
    `200`
    B
    `100`
    C
    `50`
    D
    `125`

  • Ca^(2+) ion is isoelectronic with

    A
    `Mg^(2+)`
    B
    `Na^(+)`
    C
    `Ar `
    D
    `Kr`.

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    A sample of water has its hardness due to only CaSO_(4) . When this water is passed through on anion exchange resin, SO_(4)^(2-) ions are replaced by OH^(-) . A 25.0 mL sample of water so treated requires 21.58 mL of 10^(-3) MH_(2)SO_(4) for its titration. What is the hardness of water expressed in terms of CaCO_(3) in ppm? Assume density of water 1.0g//mL .

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