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Ionic Equilibrium

NEET Chemistry · 99 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
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}}
Q72
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
Q73
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
Q74
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

Q75
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

Q76
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

Q77
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

Q78
Which of the following are Lewis acids?
A BCl3 and AlCl3
B PH3 and BCl3
C AlCl3 and SiCl4
D PH3 and SiCl4
Correct Answer
Option A
Solution

The compound which have the ability to accepted at least are lone pair electron.

Structure of BCl3 is Here B is electron deficient atom, so it can accepted lone pair.

So it is a lewis acid.

Structure of AlCl3 Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair.

So it is also lewis acid.

Here in PH3 there is vacant 3d orbital but it Can't take.

Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases.

So PH3 is not lewis acid.

Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital.

This can be shown by following reaction.

So, here option (A) and (C) both are correct.

But as SiCl4 is not as strong lewis acid as BCl3 and AlCl3, So we can say option (A) is more correct.

Q79
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :
A 3 ×\times 10–1
B 1 ×\times 10–3
C 1 ×\times 10–5
D 1 ×\times 10–7
Correct Answer
Option C
Solution
H+=Cα;α=[H+]C{H^ + } = C\alpha ;\alpha = {{\left[ {{H^ + }} \right]} \over C}

or

   α=10−30.1=10−2\,\,\,\alpha = {{{{10}^{ - 3}}} \over {0.1}} = {10^{ - 2}}
Ka=Cα2=0.1×10−2×10−2Ka = C{\alpha ^2} = 0.1 \times {10^{ - 2}} \times {10^{ - 2}}
=10−5= {10^{ - 5}}
Q80
1 M NaCL and 1 M HCL are present in an aqueous solution. The solution is
A not a buffer solution with pH < 7
B not a buffer solution with pH > 7
C a buffer solution with pH < 7
D a buffer solution with pH > 7
Correct Answer
Option A
Solution

NOTE : A buffer is a solution of weak acid and its salt with strong base and vice versa.

HCLHCL

is strong acid and

NaCLNaCL

is its salt with strong base.

pHpH

is less than

77

due to

HCLHCL

.

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