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Ionic Equilibrium

JEE Chemistry · 82 questions · Page 8 of 9 · Click an option or "Show Solution" to reveal answer

Q71
50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :
A 3.75
B 4.75
C 8.25
D 9.25
Correct Answer
Option D
Solution

NH3 + HCl →\to NH4Cl moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed) moles of NH3 = 0.2 M × 50 × 10–3 L = 0.01 moles HCl excess NH3 = 0.01 – 0.005 = 0.005 moles From reaction, 1 mole ammonia = 1 mole NH4Cl ∴\therefore 0.005 NH3 = 0.005 NH4Cl Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL [NH3] = [NH4Cl] =

0.00575×10−3{{0.005} \over {75 \times {{10}^{ - 3}}}}

= 0.066 M pOH = pKb + log

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

= pKb + log

[NH4Cl][NH3]{{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}

= 4.75 + log

[0.066][0.066]{{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}

⇒\Rightarrow pOH = 4.75 ∴\therefore pH = 14 – 4.75 = 9.25

Q72
The Ksp for the following dissociation is 1.6 × 10–5 PbCl2(s)⇋Pb(aq)2++2Cl(aq)−PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
A Q > Ksp
B Not enough data provided
C Q < Ksp
D Q = Ksp
Correct Answer
Option A
Solution

[Pb2+] =

300×0.134400{{300 \times 0.134} \over {400}}

= 1.005 × 10–1 M [Cl-] =

100×0.4400{{100 \times 0.4} \over {400}}

= 10–1 M

PbCl2(s)⇋Pb(aq)2++2Cl(aq)−PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ -

Q = [Pb2+] × [Cl–]2 = 0.1005 × (0.1)2 = 1.005 × 10–3 Given Ksp = 1.6 × 10–5 ∴\therefore Q > Ksp

Q73
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24 :
A 3 × 10–22
B 3 × 10–19
C 12 × 10–21
D 12 × 10–22
Correct Answer
Option A
Solution
Ksp={K_{sp}} =

[Al3+][OH-]3 ⇒\Rightarrow 2.4 × 10–24 = s(0.2)3 ⇒\Rightarrow s =

2.4×10−248×10−3{{2.4 \times {{10}^{ - 24}}} \over {8 \times {{10}^{ - 3}}}}

⇒\Rightarrow s = 3 × 10–22

Q74
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid. Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid. In the light of the above statements, choose the correct answer from the options given below
A Both A\mathrm{A} and R\mathrm{R} are true and R\mathrm{R} is the correct explanation of A\mathrm{A}
B Both A\mathrm{A} and R\mathrm{R} are true but R\mathrm{R} is NOT the correct explanation of A\mathrm{A}
C A\mathrm{A} is true but R\mathrm{R} is false
D A\mathrm{A} is false but R\mathrm{R} is true
Correct Answer
Option A
Solution

2KMnO4+16HCl→2MnCl2+2KCl+8H2O+Cl22 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{MnCl}_2+2 \mathrm{KCl}+8 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 HCl\mathrm{HCl} is not used in the process of titration because it reacts with the (KMnO4)\left(\mathrm{KMnO}_{4}\right) that is used in the process and gets oxidized.

Q75
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH3CH2COO−][CH3CH2COOH]{{[C{H_3}C{H_2}CO{O^ - }]} \over {[C{H_3}C{H_2}COOH]}} required to make buffer is ___________. Given : Ka(CH3CH2COOH)=1.3×10−5{K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}
A 0.03
B 0.13
C 0.23
D 0.33
Correct Answer
Option B
Solution

CH3CH2COOH⇌CH3CH2COO−+H+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+} From Henderson equation pH=pKa+log⁡[CH3CH2COO−][CH3CH2COOH]\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]} 4=−log⁡1.3×10−5+log⁡[CH3CH2COO−][CH3CH2COOH]4=-\log 1.3 \times 10^{-5}+\log \dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]} −log⁡10−4=−log⁡1.3×10−5+log⁡[CH3CH2COO−][CH3CH2COOH]-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]} −log⁡10−4=−log⁡1.3×10−5[CH3CH2COOH][CH3CH2COO−]-\log 10^{-4}=-\log 1.3 \times 10^{-5} \dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]} 10−4=1.3×10−5[CH3CH2COOH][CH3CH2COO−]10^{-4}=1.3 \times 10^{-5} \dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]} [CH3CH2COO−][CH3CH2COOH]=0.13\dfrac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13

Q76
The solubility of Ca(OH)2 in water is : [Given : The solubility product of Ca(OH)2 in water = 5.5 ×\times 10−-6]
A 1.11 ×\times 10−-6
B 1.11 ×\times 10−-2
C 1.77 ×\times 10−-6
D 1.77 ×\times 10−-2
Correct Answer
Option B
Solution

Let, solubility of Ca(OH)2 in pure water = S mol/L

Ca(OH)2Ca{(OH)_2}

⇌\rightleftharpoons

Ca2+S mol/L+2OH−2×S (mol/L)\mathop {C{a^{2 + }}}\limits_{S\,mol/L} + \mathop {2O{H^ - }}\limits_{2 \times S\,(mol/L)}

Ksp = [Ca2+] [OH−-]2 = S ×\times (2S)2 = 4 S3 (mol/L) The expression of Ksp can also be written as, Ksp = xx . yy .

Sx + y = 11 .

22 .

S1 + 2 = 4 S3 [∵\because For Ca(OH)2 : x = 1, y = 2] x and y are the coefficients of cations and anions respectively

S=(Ksp4)1/3=(5.5×10−64)1/3S = {\left( {{{{K_{sp}}} \over 4}} \right)^{1/3}} = {\left( {{{5.5 \times {{10}^{ - 6}}} \over 4}} \right)^{1/3}}
=1.11×10−2= 1.11 \times {10^{ - 2}}

ml/L

Q77
For a sparingly soluble salt AB2\mathrm{AB}_2, the equilibrium concentrations of A2+\mathrm{A}^{2+} ions and B−B^{-} ions are 1.2×10−4M1.2 \times 10^{-4} \mathrm{M} and 0.24×10−3M0.24 \times 10^{-3} \mathrm{M}, respectively. The solubility product of AB2\mathrm{AB}_2 is :
A 0.069×10−120.069 \times 10^{-12}
B 0.276×10−120.276 \times 10^{-12}
C 6.91×10−126.91 \times 10^{-12}
D 27.65×10−1227.65 \times 10^{-12}
Correct Answer
Option C
Solution

For a sparingly soluble salt AB2\mathrm{AB}_2, the dissolution in water can be represented by the following equilibrium equation: AB2⇌A2++2B− \mathrm{AB}_2 \rightleftharpoons \mathrm{A}^{2+} + 2\mathrm{B}^{-} When AB2\mathrm{AB}_2 dissolves in water, it generates one A2+\mathrm{A}^{2+} ion and two B−\mathrm{B}^{-} ions.

The solubility product constant, KspK_{sp}, for this reaction can be expressed as: Ksp=[A2+][B−]2 K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^2 Given in the problem, the equilibrium concentrations are: [A2+]=1.2×10−4M [\mathrm{A}^{2+}] = 1.2 \times 10^{-4} \mathrm{M} [B−]=0.24×10−3M=2.4×10−4M [\mathrm{B}^{-}] = 0.24 \times 10^{-3} \mathrm{M} = 2.4 \times 10^{-4} \mathrm{M} Substituting these values into the KspK_{sp} expression: Ksp=(1.2×10−4)×(2.4×10−4)2 K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2 Calculating KspK_{sp}: Ksp=1.2×10−4×(5.76×10−8) K_{sp} = 1.2 \times 10^{-4} \times (5.76 \times 10^{-8}) Ksp=6.912×10−12 K_{sp} = 6.912 \times 10^{-12} Therefore, the correct option is: Option C: 6.91×10−126.91 \times 10^{-12}

Q78
A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - pKa\mathrm{pK}_{\mathrm{a}})?
A log⁡(1−xx)\log \left(\dfrac{1-x}{x}\right)
B 00
C log⁡(1+2x)\log (1+2 \mathrm{x})
D log⁡(x1−x)\log \left(\dfrac{x}{1-x}\right)
Correct Answer
Option D
Solution
HA⇌H⊕+AΘt=0at=ta(1−x)axax Ka=(ax)(x)1−x;[H+]=ax−log⁡(Ka)=−log⁡(ax)−log⁡(x1−x)pKa=pH−log⁡(x1−x)pH−pKa=log⁡(x1−x)\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}^{\oplus}+\mathrm{A}^{\Theta} \\ & \mathrm{t}=0 \quad \mathrm{a} \\ & \mathrm{t}=\mathrm{t} \quad \mathrm{a}(1-\mathrm{x}) \quad \mathrm{ax} \quad \mathrm{ax} \\ & \mathrm{~K}_{\mathrm{a}}=(\mathrm{ax}) \frac{(\mathrm{x})}{1-\mathrm{x}} ;\left[\mathrm{H}^{+}\right]=\mathrm{ax} \\ & -\log \left(\mathrm{K}_{\mathrm{a}}\right)=-\log (\mathrm{ax})-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\ & \mathrm{pKa}=\mathrm{pH}-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\ & \mathrm{pH}-\mathrm{pKa}=\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \end{aligned}
Q79
Given below are two statements : Statement (I) : Aqueous solution of ammonium carbonate is basic. Statement (II) : Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on KaK_a and KbK_b value of acid and the base forming it. In the light of the above statements, choose the most appropriate answer from the options given below :
A Both Statement I and Statement II are correct
B Statement I is correct but Statement II is incorrect
C Both Statement I and Statement II are incorrect
D Statement I is incorrect but Statement II is correct
Correct Answer
Option A
Solution

The correct answer is Option A: Both Statement I and Statement II are correct.

Explanation: Statement I explains that an aqueous solution of ammonium carbonate is basic.

This is because ammonium carbonate (

(NH4)2CO3\text{(NH}_4)_2\text{CO}_3

) dissociates in water to form ammonium ions (

NH4+\text{NH}_4^+

) and carbonate ions (

CO32−\text{CO}_3^{2-}

). The ammonium ion (

NH4+\text{NH}_4^+

) is a weak acid and can donate a proton to form ammonia (

NH3\text{NH}_3

) and a hydronium ion (

H3O+\text{H}_3\text{O}^+

), but this happens to a minimal extent due to its weak acidic nature. On the other hand, the carbonate ion (

CO32−\text{CO}_3^{2-}

) can accept a hydrogen ion (

H+\text{H}^+

) from water to form bicarbonate (

HCO3−\text{HCO}_3^-

) and hydroxide ions (

OH−{OH}^-

), which makes the solution basic.

The reaction of carbonate ions removing hydrogen ions from water is more pronounced than the reaction of ammonium ions donating hydrogen ions to water, hence the overall solution becomes basic.

Statement II is about the general principle that determines the acidic or basic nature of a salt solution formed by the salt of a weak acid and a weak base.

The acidic or basic nature of such a salt solution depends on the acid dissociation constant (

KaK_a

) of the acid and the base dissociation constant (

KbK_b

) of the base. If

Ka>KbK_a > K_b

, the solution tends to be acidic because the weak acid is stronger (more willing to donate protons) than the weak base is at accepting protons.

Conversely, if

Kb>KaK_b > K_a

, the solution tends to be basic because the weak base is stronger (more willing to accept protons) than the weak acid is at donating protons.

In the case of ammonium carbonate, carbonic acid (

H2CO3H_2CO_3

) is a weak acid, and ammonia (

NH3NH_3

) is a weak base. The basic nature of ammonium carbonate solution suggests that the

KbK_b

of ammonia exceeds the

KaK_a

of carbonic acid for the reactions in an aqueous solution, making the solution basic.

Therefore, both statements I and II are correct, thereby making Option A the right choice.

Q80
Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10−-5, the ratio of salt to acid concentration in the buffer solution will be :
A 4 : 5
B 1 : 10
C 10 : 1
D 5 : 4
Correct Answer
Option C
Solution

HA ⇌\rightleftharpoons H+ + A−- Ka =

[H+][A−][HA]{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}

= 10−-5 pH = pKa + log

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

⇒\Rightarrow

   \,\,\,

6 = −- log [10−-5] + log

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

⇒\Rightarrow

   \,\,\,

6 = 5 + log

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

⇒\Rightarrow log

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

= 1 ⇒\Rightarrow

   \,\,\,
[Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}

= 10

∴   \therefore\,\,\,

Salt : Acid = 10 : 1

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