The solubility of PbF2 in water at 25ºC is ∼ 10–3 M. What is its solubility in 0.05 M NaF solution? Assume the latter to be fully ionised.
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Solution
Solubility of PbF2≈10-3 M
∴Ksp = 4S3 = 4 × 10-9In 0.05 M NaF we have 0.05 M of F–ion contributed by NaF. If the solubility of PbF2 in this solution is S M,then
total [F–] = [2S+ 0.05] M.
∴S[2S + 0.05]2= 4×10-9 Assuming 2S << 0.05, S × 25 × 10-4 = 4 × 10-9
∴S = 0.16 × 102-5M⇒1.6 ⇒ 10-6 M
We observe that our approximation that 2S << 0.05 is justified.
The equilibrium constant for a reaction, N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g) is 4× 10-4 at 2000 K. In the presence of catalyst, the equilibrium is attained 10 times faster. The equilibrium constant in presence of catalyst at 2000 K is :
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Solution
Catalyst only changes time to achieve equilibrium, it does not affect the value of equilibrium constant.
A weak acid, HA is found to be 10% ionized in 0.01 M aqueous solution.Calculate the pH of a solution which is 0.1 M in HA and 0.05 M in Na A.
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Solution
α= 0.1,
\(K_{a}=\frac{\alpha ^{2}C}{1-\alpha }=\frac{(0.1)^{2}\times (0.01)}{(1-0.1)}= 1.11\times 10^{-4}\)Now pKa= –log 1.11×10-4=3.9542
\(pH=pK_{a}+log\frac{[salt]}{[acid]}\)
= 3.9542 +log\(\left [ \frac{0.05}{0.10} \right ]= 3.653\)
The reaction quotient (Q) for the reaction N2(g)+3H2(g)\(\rightleftharpoons\)2NH3(g)is given by \(Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\). The reaction will proceed fromright to left if
where Kc is the equilibrium constant
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Solution
For reaction to proceed from right to left
Q>Kc i.e the reaction will be fast in backward direction i.e rb > rf.
According to Le-chatelier’s principle, adding heat to a solid\(\rightleftharpoons\)liquid equilibrium will cause the
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Solution
Solid\(\rightleftharpoons\)Liquid
It is an endothermic process. So when temperature is raised, more liquid is formed. Hence adding heat will shift the equilbrium in the forward direction.
Which reaction is not affected by change in pressure?
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Solution
Reactions in which number of moles of gaseous reactants and gaseous products is equal (i.e. Δ n = 0)are not affected by change in pressure (Le Chatelier principle)
H2(g)+I2(g)\(\rightleftharpoons\)2HI(g)
On adding 0.1 M solution each of [Ag+], [Ba2+], [Ca2+] in a Na2SO4 solution, species first precipitated is [KspBaSO4 = 10-11, KspCaSO4= 10-6, KspAg2SO4 = 10-5]
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Solution
The species having minimum value of Ksp will get precipitated first of all because ionic product will exceed the solubility product of such a species.The Ksp value is minimum for BaSO4(10-11),so,BaSO4 will get precipitated first of all.
For the gaseous reaction
C2H2+H2\(\rightleftharpoons\) C2H6, ΔH= –130 kJ mol-1 carried in a closed vessel, the equilibrium concentration of C2H6 can definitely be increased by :
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Solution
C2H2+H2\(\rightleftharpoons\) C2H6, ΔH= –130 kJ mol-1
The reaction is exothermic and involves decrease involume, so forward reaction will be favoured by decreasing temperature and increasing pressure(Le Chatelier principle).
In which of the following reactions, the concentration of the product is higher than the concentration of reactant at equilibrium?
(K= equilibrium constant)
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Solution
\(K=\frac{[Product]}{[Reactant]}\)
Hence, [Product]=K[Reactant]
∴[Product] > Reactant, when K > 1
In a gaseous reversible reaction
N2+O2\(\rightleftharpoons\)2NO+heat
If pressure is increased then the equilibrium constant would be :
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Solution
N2(g)+O2(g)\(\rightleftharpoons\)2NO(g)+Heat
No. of moles of reactants = No. of moles of products.The reaction in which number of moles of reactants are equal to number of moles of products are not affected by change in pressure (Le-chatelier's principle).