Home
Class 12
CHEMISTRY
What [H^(+)] must be maintained in a sat...

What `[H^(+)]` must be maintained in a saturated `H_(2)S(0.1)M` to precipitate `CdS` but not `ZnS` if `[Cd^(2+)]=[Zn^(2+)]=0.1` initially? `K_(sp)` of ZnS = `1 X 10^(-21)` `K_a` of `H_2S` is `1.1 X 10^(-21)`

Text Solution

AI Generated Solution

To solve the problem of determining the concentration of \([H^+]\) that must be maintained in a saturated \(H_2S\) solution to precipitate \(CdS\) but not \(ZnS\), we will follow these steps: ### Step 1: Understand the Conditions for Precipitation To prevent the precipitation of \(ZnS\), the ionic product of \(Zn^{2+}\) and \(S^{2-}\) must be less than the solubility product (\(K_{sp}\)) of \(ZnS\). ### Step 2: Write the Expression for Ionic Product The ionic product for \(ZnS\) can be expressed as: \[ ...
Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Board Level Exercise|27 Videos
  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Exercise -1|28 Videos
  • IONIC EQUILIBRIUM

    RESONANCE ENGLISH|Exercise partIII one or more than one options correct type|10 Videos
  • HYDROCARBON

    RESONANCE ENGLISH|Exercise ORGANIC CHEMISTRY(Hydrocarbon)|50 Videos
  • IUPAC NOMENCLATURE & STRUCTURAL ISOMERISM

    RESONANCE ENGLISH|Exercise Advanced Level Problems Part-5|2 Videos

Similar Questions

Explore conceptually related problems

What (H_(3)O^(+)) must be maintained in a saturated H_(2)S solution to precipitate Pb^(2+) , but not Zn^(2+) from a solution in which each ion is present at a concetration of 0.01 M ? (K_(SP) for H_(2)S= 1.1xx10^(-22), K_(SP) for ZnS=1.0xx10^(-21))

A lead salts is dissolved in HC1 which si 94% ionised. It is found to have 0.1M Pb^(2+) and 0.28M H^(o+) ions. The solution is satured with H_(2)S(g) . Calculate the amount of Pb^(2+) ions that remains unprecipitated. K_(sp) of PbS = 4 xx 10^(-29) , K_(sp) of H_(2)S = 1.1 xx 10^(-22)

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))

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) .

A solution containing both Zn^(2+) and Mn^(2+) ions at a concentration of 0.01M is saturated with H_(2)S . What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of Zn^(2+) ions remaining in the solution ? Given K_(sp) of ZnS is 10^(-22) and K_(sp) of MnS is 5.6 xx 10^(-16), K_(1) xx K_(2) of H_(2)S = 1.10 xx 10^(-21) .

A solution constains Zn^(2+) ions and Cu^(2+) ions each of 0.02M . If the solution is made 1M in H^(o+) , and H_(2)S is passed untill the solution is satured, should a precipitate be formed? Given: K_(sp) ZnS = 10^(-22) , K_(sp) Cus = 8 xx 10^(-37) . In satured solution, K_(sp) (H_(2)S) = 10^(-22)

In equalitative analysis, cations of graph II as well as group IV both are precipitated in the form of sulphides. Due to low value of K_(sp) of group II sulphides, group reagent is H_(2)S in the presence of dil. HC1 , and due to high value of K_(sp) of group IV sulphides, group reagent is H_(2)S in the presence of NH_(4)OH and NH_(4)C1 . In a solution containing 0.1M each of Sn^(2+), Cd^(2+) , and Ni^(2+) ions, H_(2)S gas is passed. K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 10^(-28), K_(sp) of NiS - 3 xx 10^(-21) K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14) If 0.1M HC1 is mixed in the solution containing only 0.1 M Cd^(2+) ions and saturated with H_(2)S , then [Cd^(2+)] remaining in the solution after CdS stopes to precipitate is:

In equalitative analysis, cations of graph II as well as group IV both are precipitated in the form of sulphides. Due to low value of K_(sp) of group II sulphides, group reagent is H_(2)S in the presence of dil. HC1 , and due to high value of K_(sp) of group IV sulphides, group reagent is H_(2)S in the presence of NH_(4)OH and NH_(4)C1 . In a solution containing 0.1M each of Sn^(2+), Cd^(2+) , and Ni^(2+) ions, H_(2)S gas is passed. K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 15 10^(-28), K_(sp) of NiS - 3 xx 10^(-21), K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14) At what value of pH, NiS will start to precipitate?

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) .

In equalitative analysis, cations of graph II as well as group IV both are precipitated in the form of sulphides. Due to low value of K_(sp) of group II sulphides, group reagent is H_(2)S in the presence of dil. HC1 , and due to high value of K_(sp) of group IV sulphides, group reagent is H_(2)S in the presence of NH_(4)OH and NH_(4)C1 . In a solution containing 0.1M each of Sn^(2+), Cd^(2+) , and Ni^(2+) ions, H_(2)S gas is passed. K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 15 10^(-28), K_(sp) of NiS - 3 xx 10^(-21), K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14) Which of the following sulphides is more soluble in pure water?