Practice Start Test

Ionic Equilibrium

NEET Chemistry · 99 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
Conjugate acid of NH 2 −- is
A NH 4 OH
B NH 4 +
C NH 2 −-
D NH 3
Correct Answer
Option D
Solution

NH 3 after losing a proton (H + ) gives NH 2 – .

So, NH 3 is the conjugate acid of NH 2 – .

Note : Conjugate acid-base pair differ only by a proton.

Q62
Which statement is wrong about pH and H + ?
A pH of neutral water is not zero.
B Adding 1 N solution of CH 3 COOH and 1 N solution of NaOH, pH will be seven.
C [H + ] of dilute and hot H 2 SO 4 is more than concentrated and cold H 2 SO 4
D Mixing solution of CH 3 COOH and HCl. pH will be less than 7.
Correct Answer
Option B
Solution

CH 3 COOH is weak acid while NaOH is strong base, so one equivalent of NaOH can not be neutralized with one equivalent of CH 3 COOH.

Hence the solution of one equivalent of each does not have pH value as 7.

Its pH will be towards basic side as NaOH is a strong base hence conc. of OH – will be more than the conc. of H + .

Q63
The strength of an aqueous NaOH solution is most accurately determined by titrating : (Note : consider that an appropriate indicator is used)
A Aq. NaOH in a pipette and aqueous oxalic acid in a burette
B Aq. NaOH in a burette and aqueous oxalic acid in a conical flask
C Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask
D Aq. NaOH in a burette and concentrated H2SO4 in a conical flask
Correct Answer
Option B
Solution

Aq. NaOH in a burette and aqueous oxalic acid in a conical flask.

Q64
An acidic buffer is obtained on mixing :
A 100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa
B 100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl
C 100 mL of 0.1 M CH3COOH and 100 mL of 0.1 M NaOH
D 100 mL of 0.1 M CH3COOH and 200 mL of 0.1 M NaOH
Correct Answer
Option A
Solution

So finally we get mixture of CH3COOH + CH3COONa that will work like acidic buffer solution.

Q65
For the following Assertion and Reason, the correct option is Assertion (A): When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid. Reason (R): The equilibrium constant of Cu2+(aq) + S2–(aq) ⇌ CuS(s) is high because the solubility product is low.
A (A) is false and (R) is true.
B Both (A) and (R) are true but (R) is not the explanation for (A).
C Both (A) and (R) are true and (R) is the explanation for (A).
D Both (A) and (R) are false.
Correct Answer
Option C
Solution

Ksp value of CuS is very low 10–36 (3.6 × 10–36) due to low Ksp value Cu+2 ion gets precipitated very quickly even with very low concentration of S–2 ion.

Q66
Solubility product of silver bromide is 5.0 ×\times 10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :
A 1.2 ×\times 10–10 g
B 1.2 ×\times 10–9 g
C 6.2 ×\times 10–5 g
D 5.0 ×\times 10–8 g
Correct Answer
Option B
Solution
AgBr ⇌ Ag++Br−AgBr\,\rightleftharpoons\,A{g^ + } + B{r^ - }
Ksp=[Ag+][Br−]{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]

For precipitation to occur Ionic product

>>

Solubility product

[Br−]=Ksp[Ag+]=5×10−130.05=10−11\left[ {B{r^ - }} \right] = {{{K_{sp}}} \over {\left[ {A{g^ + }} \right]}} = {{5 \times {{10}^{ - 13}}} \over {0.05}} = {10^{ - 11}}

i.e., precipitation just starts when

10−11 {10^{ - 11}}\,

moles of

KBrKBr

is added to

1ℓ AgNO3 1\ell \,AgN{O_3}\,

solution ∴\therefore

   \,\,\,

Number of moles of

   Br−  \,\,\,B{r^ - }\,\,

needed from

KBr=10−11KBr = {10^{ - 11}}

∴\therefore

   \,\,\,

Mass of

KBr=10−11×120KBr = {10^{ - 11}} \times 120
=1.2×10−9g= 1.2 \times {10^{ - 9}}g
Q67
The Ksp for bismuth sulphide (Bi2S3) is 1.08 ×\times 10−-73. The solubility of Bi2S3 in mol L−-1 at 298 K is :
A 1.0 ×\times 10−-15
B 2.7 ×\times 10−-12
C 3.2 ×\times 10−-10
D 4.2 ×\times 10−-8
Correct Answer
Option A
Solution

Ksp = (2s)2(3s)3 = 108s5 108s5 = 108 × 10–75 s = 1.0 × 10–15 mol/L

Q68
Class XII students were asked to prepare one litre of buffer solution of pH 8.26\mathrm{pH} \,8.26 by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in 0.2 M0.2\, \mathrm{M} ammonia solution to make one litre of the buffer is : (Given: pKb(NH3)=4.74\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=4.74 Molar mass of NH3=17 g mol−1\mathrm{NH}_{3}=17 \mathrm{~g} \mathrm{~mol}^{-1} Molar mass of NH4Cl=53.5 g mol−1\mathrm{NH}_{4} \mathrm{Cl}=53.5 \mathrm{~g} \mathrm{~mol}^{-1} )
A 53.5 g
B 72.3 g
C 107.0 g
D 126.0 g
Correct Answer
Option C
Solution

For basic Buffer,

pOH=pKb+log⁡[ salt ][ Base ]\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text{ salt }]}{[\text{ Base }]}
pOH=14−8.26=5.74\mathrm{pOH}=14-8.26=5.74
5.74=4.74+log⁡[NH4Cl]0.25.74=4.74+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{0.2}
[NH4Cl]=2M\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=2 \mathrm{M}

Moles of

NH4Cl=2×1=2\mathrm{NH}_{4} \mathrm{Cl}=2 \times 1=2

moles Weight of

NH4Cl=2×53.5=107 g\mathrm{NH}_{4} \mathrm{Cl}=2 \times 53.5=107 \mathrm{~g}
Q69
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}}
Q70
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.

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