Ionic Equilibrium — Page 2 | JEE Chemistry

Q11

Which of the following salts is the most basic in aqueous solution?

A Pb(CH3COO)2

B Al(CN)3

C CH3COOK

D FeCl3

Correct Answer

Option C

Solution

Note : When a salt is made with strong base and weak acid then that salt will be basic nature in aqueous solution. (a) Pb(CH3COO)2 is a salt of weak base Pb(OH)2 and weak acid CH3COOH. (b) Al(CN)3 is a salt of weak base Al(OH)3 and weak acid HCN. (c) CH3COOK is a salt of weak acid CH3COOH and strong base KOH. (d) FeCl3 is a salt of weak base Fe(OH)3 and strong acid HCl.

Q12

An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1

\times

10–10. What is the original concentration of Ba2+?

A 1.0

\times

10–10 M

B 5

\times

10–9 M

C 2

\times

10–9 M

D 1.1

\times

10–9 M

Correct Answer

Option D

Solution

Let initially concentration of Ba+2 = x m. After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.

\therefore\,\,\,

Initial volume of Ba+2 solution = (500 $-$ 50) ml = 450 ml As at the begining of precipitation, ionic product = solubility product.

\Rightarrow \,\,\,

[Ba2+] [SO

{_4^{ - 2}}

] = Ksp of BaSO4

\Rightarrow \,\,\,

[Ba2+]

\left( {{{50 \times 1} \over {500}}} \right)

= 1 $\times$ 10 $-$ 10

\Rightarrow \,\,\,

[Ba2+] = 10 $-$ 9 M. So, the concentration of Ba+2 in final solution is 10 $-$ 9 M.

\therefore\,\,\,

concentration of Ba+2 in original solution, M1 V1 = M2 V2

\Rightarrow \,\,\,

x $\times$ 450 = 10 $-$ 9 $\times$ 500

\Rightarrow \,\,\,

x = 1.1 $\times$ 10 $-$ 9 M

Q13

The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is :

A 2.49 × 10–10 M

B 1.84 × 10–9 M

C 6.23 × 10–11 M

D

{{2.49} \over {1.84}} \times {10^{ - 9}}M

Correct Answer

Option A

Solution

Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH $-$ (aq) S 2S $\therefore$ Ksp[Cd(OH)2] = 4(S)3 = 4(1.84 × 10–5)3 [Given S = 1.84 × 10–5] For buffer solution of pH = 12 : As pH = 12 so pOH = 2 $\Rightarrow$ [OH $-$ ] = 10-2 Let the solubilty = S1 .tg .tg Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH $-$ (aq) S1 2S1 + 10-2 $\therefore$ Ksp[Cd(OH)2] = S1(10-2)2 = 4(1.84 × 10–5)3 $\Rightarrow$ S1 = 2.49 × 10–10 M

Q14

The decreasing order of electrical conductivity of the following aqueous solutions is : 0.1 M Formic acid (A), 0.1 M Acetic acid (B), 0.1 M Benzoic acid (C)

A A > B > C

B C > B > A

C A > C > B

D C > A > B

Correct Answer

Option C

Solution

0.1 M Formic acid(HCOOH) (A), 0.1 M Acetic acid(CH3COOH) (B), 0.1 M Benzoic acid(C6H5COOH) (C) Conductivity (K) depends on no of ions present in unit volume of solution.

When no of ions increases conductivity also increases.

Also no of ions depends on degree of dissociation( $\alpha$ ).

We know, $\alpha$ =

\sqrt {{{{K_a}} \over C}}

For all solutions C = 0.1 M So, $\alpha$ depends on only K

a

here. K

a

of HCOOH = 10-4 K

a

of CH3COOH = 1.8 $\times$ 10-5 and K

a

of C6H5COOH = 6 $\times$ 10-5 $\therefore$ $\alpha$ (HCOOH) > $\alpha$ (C6H5COOH) > $\alpha$ (CH3COOH) Therefore conductivity order = A > C > B

Q15

If solublity product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then which of the following relation between S and Ksp is correct ?

A

S = {\left( {{{{K_{sp}}} \over {929}}} \right)^{1/9}}

B

S = {\left( {{{{K_{sp}}} \over {6912}}} \right)^{1/7}}

C

S = {\left( {{{{K_{sp}}} \over {144}}} \right)^{1/6}}

D

S = {\left( {{{{K_{sp}}} \over {216}}} \right)^{1/7}}

Correct Answer

Option B

Solution

Zr3(PO4)4 ⇌ 3Zr+4 + 4PO4-3 S 3S 4S Ksp = [Zr+4]3 [PO4-3]4 = [3S]3 [4S]4 = 27S3 $\times$ 256S4 = 6912S7 $\therefore$

S = {\left( {{{{K_{sp}}} \over {6912}}} \right)^{1/7}}

Q16

The pH of rain water, is approximately :

A 5.6

B 7.5

C 7.0

D 6.5

Correct Answer

Option A

Solution

pH of rain water is approximate 5.6

Q17

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.

Q18

For the following Assertion and Reason, the correct option is : Assertion : The pH of water increases with increase in temperature. Reason : The dissociation of water into H+ and OH– is an exothermic reaction.

A Both assertion and reason are false.

B Both assertion and reason are true, but the reason is not the correct explanation for the assertion.

C Both assertion and reason are true, and the reason is the correct explanation for the assertion.

D Assertion is not true, but reason is true.

Correct Answer

Option A

Solution

H2O ⇌ H+ + OH- As

\Delta

H

>

0 for this reaction. So Dissociation of H2O is endothermic. pH =

- \ln \left[ {{H^ + }} \right]

kw =

\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]

\ln \left\{ {{{{k_{{T_2}}}} \over {{k_{{T_1}}}}}} \right\} = {{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}

When temperature increases then

{{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\} > 0

, as reaction is endothermic. $\therefore$ kT2

>

kT1 $\Rightarrow$

{\left[ {{H^ + }} \right]_{{T_2}}}

>

{\left[ {{H^ + }} \right]_{{T_1}}}

$\therefore$ [pH]T2

<

[pH]T1

Q19

Which of the following compound CANNOT act as a Lewis base?

A SF4

B NF3

C ClF3

D PCl5

Correct Answer

Option D

Solution

NF3 has no vacant orbital neither in nitrogen nor in fluorine so it cannot accept the electron & hence cannot acts as lewis acid and but for PCl5 P has no L.P & hence it cannot acts as base but ClF3 (3 B.P + 2 L.P) & SF4 (4 B.P + 1 L.P)

Q20

Given below are two statements : One is labelled as Assertion A and the other labelled as reason R Assertion A : During the boiling of water having temporary hardness, Mg(HCO3)2 is converted to MgCO3. Reason R : The solubility product of Mg(OH)2 is greater than that of MgCO3. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are true but R is not the correct explanation of A

B A is true but R is false

C Both A and R are true and R is the correct explanation of A

D A is false but R is true

Correct Answer

Option D

Solution

A : For temporary hardness, Mg(HCO3)2

\overset{{heating}}\longrightarrow

Mg(OH)2 + 2CO2 Assertion is false.

MgCO3 has high solubility product than Mg(OH)2.

R : MgCO3 is more water soluble than Mg(OH)2.

KSP(Mg(OH)2) = 1.8 × 10–11 KSP(MgCO3) = 3.5 × 10–8