Practice Start Test

Ionic Equilibrium

JEE Chemistry · 82 questions · Page 6 of 9 · Click an option or "Show Solution" to reveal answer

Q51
If Ksp of Ag2CO3 is 8 × \times 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is -
A 8 × \times 10–12 M
B 8 × \times 10–10 M
C 8 × \times 10–13 M
D 8 × \times 10–11 M
Correct Answer
Option B
Solution

∴\therefore 8 ×\times 10−-12 = (2s + 0.1)2 s ⇒\Rightarrow s ×\times 10−-2 = 8 ×\times 10−-12 ⇒\Rightarrow s = 8 ×\times 10−-10

Q52
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 ×\times 10−-16 and Ka(HCN) = 6.2 ×\times 10−-10]
A 1.9 ×\times 10−-5
B 1.6 ×\times 10−-6
C 2.2 ×\times 10−-16
D 0.625 ×\times 10−-6
Correct Answer
Option A
Solution

Let solubility is x

Ksp×1Ka=[Ag+][CN−]×[HCN][H+][CN−]{K_{sp}} \times {1 \over {{K_a}}} = [A{g^ + }][C{N^ - }] \times {{[HCN]} \over {[{H^ + }][C{N^ - }]}}
2.2×10−16×16.2×10−10=[S][S]10−32.2 \times {10^{ - 16}} \times {1 \over {6.2 \times {{10}^{ - 10}}}} = {{[S][S]} \over {{{10}^{ - 3}}}}
S2=2.26.2×10−9{S^2} = {{2.2} \over {6.2}} \times {10^{ - 9}}
S2=3.55×10−10{S^2} = 3.55 \times {10^{ - 10}}
S=3.55×10−10S = \sqrt {3.55 \times {{10}^{ - 10}}}
S=1.88×10−5=1.9×10−5S = 1.88 \times {10^{ - 5}} = 1.9 \times {10^{ - 5}}
Q53
Consider the following statements (a) The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 mL of 0.1 M NaOH will be approximately 1.3 (b) Ionic product of water is temperature dependent. (c) A monobasic acid with Ka = 10–5 has pH = 5. The degree of dissociation of this acid is 50 %. (d) The Le Chatelier's principle is not applicable to common-ion effect. The correct statements are :
A (a) and (b)
B (a), (b) and (c)
C (a), (b) and (d)
D (b) and (c)
Correct Answer
Option B
Solution

(a) Equivalance of strong acid = 0.1 ×\times 2 ×\times 400 = 80 Equivalance of strong base = 0.1 ×\times 400 = 40 ∴\therefore [H+] of mixture =

80−40800{{80 - 40} \over {800}}

=

120{1 \over {20}}

∴\therefore pH =

−log⁡[H+]- \log \left[ {{H^ + }} \right]

=

−log⁡(120)- \log \left( {{1 \over {20}}} \right)

= 1.3 (b) Ionic product of water increases with increase of temperature because ionisation of water is endothermic. (c) ka =

10−5×cαc(1−α){{{{10}^{ - 5}} \times c\alpha } \over {c\left( {1 - \alpha } \right)}}

⇒\Rightarrow 10-5 =

10−5×α(1−α){{{{10}^{ - 5}} \times \alpha } \over {\left( {1 - \alpha } \right)}}

⇒\Rightarrow

α(1−α){\alpha \over {\left( {1 - \alpha } \right)}}

= 1 ⇒\Rightarrow α\alpha = 0.5 ∴\therefore Degree of dissociation(α\alpha) = 50% (d) The Le Chatelier's principle is always applicable to common-ion effect.

Q54
The solubility in water of a sparingly soluble salt AB2 is 1.0 ×\times 10-5 mol L-1. Its solubility product number will be :
A 4 ×\times 10-10
B 1 ×\times 10-15
C 1 ×\times 10-10
D 4 ×\times 10-15
Correct Answer
Option D
Solution
AB2⇌ A+2+2B−A{B_2}\rightleftharpoons\,{A^{ + 2}} + 2{B^ - }
[A]=1.0×10−5,  \left[ A \right] = 1.0 \times {10^{ - 5}},\,\,
[B]=[2.0×10−5],\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right],
Ksp=[B]2[A]{K_{sp}} = {\left[ B \right]^2}\left[ A \right]
=[2×10−5]2[1.0×10−5]= {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right]
=4×10−15= 4 \times {10^{ - 15}}
Q55
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is :
A 7.0
B 4.5
C 2.5
D 9.5
Correct Answer
Option D
Solution

For acidic buffer

pH=pKa+log⁡[saltacid]pH = p{K_a} + \log \left[ {{{salt} \over {acid}}} \right]

or

   pH=pKa+log⁡[A−][HA]\,\,\,pH = p{K_a} + \log {{\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}

Given

pKa=4.5  p{K_a} = 4.5\,\,

and acid is

50%50\%

ionised.

[HA]=[A−]   \left[ {HA} \right] = \left[ {{A^ - }} \right]\,\,\,

(when acid is

50%50\%

ionised) ∴\therefore

   pH=pKa+log⁡ 1\,\,\,pH = p{K_a} + \log \,1

∴\therefore

   pH=pKa=4.5\,\,\,pH = p{K_a} = 4.5
pOH=14−pHpOH = 14 - pH
=14−4.5= 14 - 4.5
=9.5= 9.5
Q56
At 25°C, the solubility product of Mg(OH)2 is 1.0 ×\times 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?
A 9
B 10
C 11
D 8
Correct Answer
Option B
Solution
Mg(OH)2 ⇌ Mg+++2OH−Mg{\left( {OH} \right)_2}\,\rightleftharpoons\,M{g^{ + + }} + 2O{H^ - }
Ksp=[Mg++][OH−]2{K_{sp}} = \left[ {M{g^{ + + }}} \right]{\left[ {O{H^ - }} \right]^2}
1.0×10−11=10−3×[OH−]21.0 \times {10^{ - 11}} = {10^{ - 3}} \times {\left[ {O{H^ - }} \right]^2}
[OH−]=10−1110−3=10−4\left[ {O{H^ - }} \right] = \sqrt {{{{{10}^{ - 11}}} \over {{{10}^{ - 3}}}}} = {10^{ - 4}}

∴\therefore

   pOH=4\,\,\,pOH = 4

∴\therefore

   pH+pOH=14\,\,\,pH + pOH = 14

∴\therefore

   pH=10\,\,\,pH = 10
Q57
An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 ×\times 10–7 and that of S2- from HS– ions is 1.2 ×\times 10–13 then the concentration of S2- ions in aqueous solution is :
A 5 ×\times 10–19
B 5 ×\times 10–8
C 3 ×\times 10–20
D 6 ×\times 10–21
Correct Answer
Option C
Solution

HCl →\to H+ + Cl−- H+ concentration is = 0.2 M.

H2S ⇌\rightleftharpoons H+ + HS−-; K1 = 1.0 ×\times 10−-7 HS−- ⇌\rightleftharpoons H+ + S2−-; K2 = 1.2 ×\times 10−-13 H2S ⇌\rightleftharpoons S2−- + 2H+ K = K1 ×\times K2 = 1.0 ×\times 10−-7 ×\times 1.2 ×\times 1.0−-13 = 1.2 ×\times 10−-20 as K1 and K2 both are very low for this reaction so dissociation of H2S and HS−- will be very low so, the produced H+ from this reaction will also be very low.

So, we can say the concentration of H+ will be almost same as H+ in HCl.

∴   \therefore\,\,\,

[ H+ ] = 0.2 M. From the reaction, H2S

⇌ \rightleftharpoons\,

2H+ + S2−- We get [ H+ ]2 [ S2−-] = K ×\times [ H2 S ]

⇒   \Rightarrow \,\,\,

[ S2−- ] =

1.2×10−20×0.1(0.2)2{{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}
⇒   \Rightarrow \,\,\,

[ S2−- ] = 3 ×\times 10−-20 M

Q58
20 mL of 0.1 M H2SO4 solution is added to 30 mL of of 0.2 M NH4OH solution. The pH of the resultant mixture is : [pkb of NH4OH = 4.7].
A 5.2
B 9.0
C 5.0
D 9.4
Correct Answer
Option B
Solution

H2SO4 + 2NH4OH →\to (NH4)2SO4 + H2O Initially, H2SO4 present = 20 ×\times 0.1 ×\times 2 = 4 miliequivalent NH4OH present = 30 ×\times 0.2 = 6 miliequivalent Here H2SO4 is the limiting reagent, So, finally.

H2SO4 present = 0 and NH4OH present = (6 −- 4) = 2 and (NH4)2SO4 produced = 4 miliequivalent.

As in the solution there is (NH4)2 SO4 present so it a basic buffer.

∴\therefore POH = PKb + log

[Salt][base]{{\left[ {Salt} \right]} \over {\left[ {base} \right]}}

= 4.7 + log

42{4 \over 2}

= 4.7 + log2 = 4.7 + 0.3 = 5 ∴\therefore PH = 14 −- POH = 14 −- 5 = 9

Q59
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively, are : (Molar mass of Ca (OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1 , respectively; Ksp of Ca(OH)2 is 5.5 × 10–6 )
A 13.6g, 0.28 mol L−-1
B 13.6g, 0.14 mol L−-1
C 1.9g, 0.28 mol L−-1
D 1.9g, 0.14 mol L−-1
Correct Answer
Option C
Solution

Ca(OH)2 + Na2SO4 →\to CaSO4 + 2NaOH 100 m mol 14 m mol −- −- −- −- −- −- 14 m mol 28 m mol wCasO4 = 14 ×\times 10−-3 ×\times 136 = 1.9 gm [OH−-] =

28100{{28} \over {100}}

= 0.28 M

Q60
20 mL20 \mathrm{~mL} of 0.1 M NH4OH0.1\, \mathrm{M} \,\mathrm{NH}_{4} \mathrm{OH} is mixed with 40 mL40 \mathrm{~mL} of 0.05MHCl0.05 \mathrm{M} \mathrm{HCl}. The pH\mathrm{pH} of the mixture is nearest to : (Given : Kb(NH4OH)=1×10−5,log⁡2=0.30,log⁡3=0.48,log⁡5=0.69,log⁡7=0.84,log⁡11=1.04)\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11= 1.04)
A 3.2
B 4.2
C 5.2
D 6.2
Correct Answer
Option C
Solution

[NH4+]=2mmole60ml=130M\left[\mathrm{NH}_4^{+}\right]=\dfrac{2 \mathrm{mmole}}{60 \mathrm{ml}}=\dfrac{1}{30} \mathrm{M} pH=pKw−pKb−log⁡C2=14−5+1.482=5.24\mathrm{pH}=\dfrac{\mathrm{pK}_{\mathrm{w}}-\mathrm{pK}_{\mathrm{b}}-\log \mathrm{C}}{2}=\dfrac{14-5+1.48}{2}=5.24

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