H2S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
A
presence of HCl decreases the sulphide ion concentration
B
solubility product of group II sulphides is more than that of group IV sulphates
C
presence of HCl increases the sulphide ion concentration
D
sulphides of group IV cations are unstable in HCl.
Explanation
H2S ⇌ H+ + HS–
HCl ⇌ H+ + Cl–
In presence of HCl this ionization of H2S is
suppressed due to the presence of extra H+ ions
from HCl and produces
less amount of sulphide ions due to common
ion effect, thus HCl decreases the solubility
of H2S which is sufficient to precipitate IInd
group radicals.
2
AIPMT 2004
MCQ (Single Correct Answer)
The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In$$-$$) forms of the indicator by the expression
The solubility product of AgI at 25oC is 1.0 $$ \times $$ 10$$-$$16 mol2 L$$-$$2. The solubility of AgI in 10$$-$$4 N solution of KI at 25oC is approximately (in mol L$$-$$1
A
1.0 $$ \times $$ 10$$-$$16
B
1.0 $$ \times $$ 10$$-$$12
C
1.0 $$ \times $$ 10$$-$$10
D
1.0 $$ \times $$ 10$$-$$8
Explanation
AgI
⇌
Ag+
+
I-
s
s
s
Ksp = s2
$$ \Rightarrow $$ 1.0 × 10–16 = s2
$$ \Rightarrow $$ s = 1.0 × 10–8 mol L–1
$$ \therefore $$ [Ag+] = 1.0 × 10–8 mol L–1
Also, in 10–4 N KI solution,
[I–1]
= (10–4 + 1.0 × 10–8) mol L–1
$$ \Rightarrow $$ [I–1]
= (10–4) mol L–1
[As 1.0 × 10–8 mol L–1 << 1.0 × 10–4 mol L–1]
$$ \therefore $$ Ksp of AgI = [Ag+][I–]
= (1.0 × 10–8)(10–4)
= 1.0 × 10–12 mol L–1
Questions Asked from Equilibrium
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