Question 1: The Ksp of Ag_{2}CrO_{4 }is 1.1 ×10^{–12} at 298K. The solubility (in mol/L) of Ag_{2}CrO_{4} in a 0.1M AgNO_{3} solution is (IITJEE 2013)
(a) 1.1×10^{11}
(b) 1.1×10^{10}
(c) 1.1×10^{9}
(d) 1.1×10^{12}
Answer: b
Solution:
Ag^{2+} is a common ion here. Hence, the contribution of Ag^{+} from AgNO_{3} would be ignored, and the solubility of Ag_{2}CrO_{4 }will depend on CrO_{4}^{2} only.
Let the concentration of CrO_{4}^{2} in solution = x molL^{1}
K_{sp} = 1.1 ´10^{–12 }= [Ag^{+}][CrO_{4}^{2}]
or
1.1´10^{–12} = [0.1]^{2}[x]
or
X = 1.1×10^{10}
Hence, the correct option is B.
Question 2: Solubility product constant (K_{SP}) of salts of type MX, MX_{2} and M_{3}X at temperature “T” are 4.0×10^{8}, 3.2×10^{14} and 2.7×10^{15}, respectively. Solubilities (mol dm^{3}) of the salts at temperature ‘T’ are (IIT JEE 2008)
(a) MX> MX_{2} > M_{3}X
(b) M_{3}X> MX_{2}> MX
(c) MX_{2}>M_{3}X> MX
(d) MX >M_{3}X> MX_{2}
Answer: d
Solution:
For salt MX:
MX → M^{+} +X^{}
Ksp = [M^{+}][X^{}]
Here [M^{+}] = [X^{}]
So, Ksp =[M^{+}]^{2}
or
[M^{+}] =[X^{}] =√Ksp = √(4.0×10^{8}) = 2×10^{4}
For salt MX_{2}:
MX_{2} → M^{+} +2X^{}
Ksp = [M^{+}][X^{}]^{2}
Here 2[M^{+}] = [X^{}]
So, Ksp =4[M^{+}]^{3}
or
[M^{+}] =?Ksp/4 = ? [(3.2×10^{14} )/4] = 2×10^{5}
For salt M_{3}X,
M_{3}X → 3M^{+} +X^{}
Ksp = [M^{+}]^{3}[X^{}]
Here [M^{+}] = 3[X^{}]
So, Ksp =27[X^{}]4
or
[X^{}]4 = Ksp/27 =(2.7×10^{15})/27
or
[X^{}] = 10^{4}
So, order of the solubilies is
MX >M_{3}X> MX_{2}
Hence, the correct option is D.
Question 3: In the following equilibrium, N_{2}O_{4}(g)↔ NO_{2}(g) When 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that
ΔG^{0}_{f} ( N_{2}O_{4}) = 100 KJ
ΔG^{0}_{f} ( NO_{2}) = 50 KJ
(i) Find ΔG of the reaction
(ii) The direction of the reaction in which the equilibrium shifts. (IIT IEE 2004)
Solution:
(i) Given that, P total = P (N_{2}O_{4}) = P (NO_{2}) = 20 bar
But at equilibrium, P (N_{2}O_{4}) = P (NO_{2}) ,
So, P (N_{2}O_{4}) = P (NO_{2}) = 10 bar
Now
For the equilibrium reaction,
N_{2}O_{4}(g)↔ 2NO_{2}(g)
Q = [P (NO_{2})]^{2 }/[P (N_{2}O_{4})] = 100/10 =
Now,
ΔG^{0 }= 2 ΔG^{0}_{f} ( NO_{2})  ΔG^{0}_{f} ( N_{2}O_{4}) = 2×50 100 = 0
So,
ΔG = ΔG^{0} – 2.303RT log Q_{p} = 0 – 2.303×8.314×298×log 10
= 5706 J = 5.706 kJ
(ii) Negative value of ΔG indicates that reaction is spontaneous one and it will shift in forward direction.
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