Home
Class 11
CHEMISTRY
Assertion (A): A solution contains 0.1M ...

Assertion (A): A solution contains `0.1M` each of `pB^(2+), Zn^(2+),Ni^(2+)`, ions. If `H_(2)S` is passed into this solution at `25^(@)C`.
`Pb^(2+), Ni^(2+), Zn^(2+)` will get precpitated simultanously.
Reason (R): `Pb^(2+)` and `Zn^(2+)` will get precipitated if the solution contains `0.1M HCI`.
`[K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]`

A

If both (A) and (R) are correc, and (R) is the correct explanation of (A).

B

If both (A) and (R) are correct but (R) is not the correct explanation of (A).

C

If (A) is correct, but (R) is incorrect.

D

If (A) is incorrect, but (R) is correct.

Text Solution

Verified by Experts

The correct Answer is:
A
Here `0.1M H_(2)S` is passed into the solution directly. Selectively precipitation of `Pb^(2+), Zn^(2+)` and `Ni^(2+)` is not possible, as the ionic product to their respective sulphides is more than their `K_(sp)` values and `S^(-2)` ion is not added slowely or drowise.
If `H_(2)S` is passed into the solution contianing only `H_(2)O`, then
`[S^(2-)] ~~K_(2) of H_(2)S = 10^(-14)M`.
But in `0.1M HCI` solution.
`K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/(H_(2)S)`
`10^(-7)xx 10^(-14) = ((0.1)[S^(2-)])/(0.1) :. [S^(2-)] = 10^(-20)M`.
`:. Q_(sp)` of `PbS` and `ZnS(0.1 xx 10^(-20)) gt K_(sp) of PbS` and `Zns` (Hence both precipitate) `Q_(sp) of NiS (10^(-21)) lt K_(sp)` of `NiS` (Hence do not precipitate).
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises Integer|10 Videos

  • IONIC EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises True/ False|10 Videos

  • IONIC EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Exercises Single Correct|121 Videos

  • HYDROGEN, WATER AND HYDROGEN PEROXIDE

    CENGAGE CHEMISTRY|Exercise Subjective Archive (Subjective)|3 Videos

  • ISOMERISM

    CENGAGE CHEMISTRY|Exercise Assertion-Reasoning Type|1 Videos

Similar Questions

Explore conceptually related problems

Calculate [Pb^(2+)] in a 1M HCl solution that is saturated in PbS. K_(a) (H_(2)S)=1.1 xx 10^(-21), K_(sp) (PbS)=8 xx 10^(-28)

Calculate the solubility of CoS in 0.1M H_(2)S and 0.15M H_(3)O^(oplus) (K_(sp) of CoS = 3 xx 10^(-26)) . (K_(1) xx K_(2) (H_(2)S) = 10^(-21))

Knowledge Check

  • The pH of the a solution containing 0.4 M HCO_(3)^(-) is : [K_(a_(1)) (H_(2)CO_(3)) = 4 xx 10^(-7), K_(a_(2)) (HCO_(3)^(-)) = 4 xx 10^(-11)]

    A
    `10.4`
    B
    `10.1`
    C
    `6.1`
    D
    `8.4`

  • What is the maximum molarity of Co^(+2) ions in 0.1M HC1 saturated with 0.1M H_(2)S. (K_(a) = 4 xx 10^(-21)) . Given: K_(sp) of CoS = 2xx10^(-21) .

    A
    `0.10M`
    B
    `1.00M`
    C
    `4.48 xx 10^(-11)M`
    D
    `0.50M`

  • In qualitative analysis, cations of group II as well as group IV precipitated in the form of sulphides. Due to low value of Ksp of group II sulphides, group reagent is H_(2)S in presence of dil. HCl and due to high value of Ksp of group IV sulphides, group reagent is H_(2)S in presence of NH_(4)OH and NH_(4)Cl . In 0.1M H_(2)S solution, Sn^(2+), Cd^(2+) and Ni^(2+) ions are present in equimolar concentration (0.1M) . Given : Ka_(1)(H_(2)S) = 10^(-7), Ka_(2)(H_(2)S) = 10^(-14), K_(sp)(SnS) = 8 xx 10^(-29) , K_(sp) (CdS) = 10^(-28), K_(sp)(NiS) = 3 xx 10^(-21) At what pH precipitate of NiS will form

    A
    `12.76`
    B
    `7`
    C
    `1.24`
    D
    `4`

    • Similar Questions

    Explore conceptually related problems

    Should FeS precipitate from a solution that is saturated in H_2S (0.1 M), 0.002 M in Fe^(2+) and at a pH=3.5? K_(sp) (FeS)=6.3 xx 10^(-18), K_(a) (H_(2)S)=1.1 xx 10^(-21) .

    It is given that 0.001 mol each of Cd^(2+) and Fe^(2+) ions are contained in 1.0L of 0.02M HC1 solution. This solutions is now saturated with H_(2)S gas at 25^(@)C . a. Determine whether or not each of these ions will be precipitated as sulphide? b. How much Cd^(2+) ions remains in the solution at equilibrium? K_(1)(H_(2)S) = 1.0 xx 10^(-7), K_(2) (H_(2)S) = 1.0 xx 10^(-14) : ltbRgt K_(sp) (CdS) = 8 xx 10^(-27): K_(sp) (FeS) = 3.7 xx 10^(-19) .

    Will MsS be precipitated from a decimolar solution of Mn^(2+) containing 0.01 M H^+ ions by passing H_(2)S? Neglect hydrolysis of S^(2-) ions. K_(sp) (MnS)=1.4 xx 10^(-15), K_(H_(2)S)=1.1 xx 10^(-21)

    Zn salt is mixed with (NH_(4))_(2)S of 0.021M . What amount of Zn^(2+) will remain uprecipitated in 12mL of the solution? K_(sp) of ZnS = 4.51 xx 10^(-24) .

    The [H^(+)] of 0.10 MH_(2)S solution is (given K_(1)=1.0 xx 10^(-7) , K_(2)=1.0 xx 10^(-14))

    Home
    Profile