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In equalitative analysis, cations of gra...

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?

A

CdS

B

NiS

C

SnS

D

all have equal solubility

Text Solution

Verified by Experts

The correct Answer is:
B
since `k_(sp)` of Nis is greater than the remaining compounds, then `NiS is more soluble
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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?

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) 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) If H_(2)S is passed into the above mixture in the presence of HC1 , which ion will be precipitated first?

In qualitative analysis ,in order to detect second group basic radical, H_(2)S gas is passed in the presence of dil. HCl to

H_(2)S in the presence of HCl precipitates II group but not IV group because:

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