Chemistry

Electrochemistry and Solutions

206 Questions

Work through advanced chemistry questions focusing on electrochemistry and properties of solutions. Key themes include electrolytic conductivity, colloidal solutions, and solubility products. These concepts are frequently tested in JEE advanced, NEET, and university level chemistry competitive examinations.

Colloidal solutionsMolar conductivitySolubility productElectrolytic conductivityHydrolysis of saltsEquivalent conductivity

Electrochemistry and Solutions Questions

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases
State whether the given statement is true or false:

$AgCl$ is less soluble in aqueous sodium chloride solution than in pure water.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$NaCl \rightarrow Na^+ + Cl^-$

Sodium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of AgCl. Hence, the solubility of AgCl decreases.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility of $Ca{ F } _{ 2 }  \left( { K } _{ sp } = 3.4 \times { 10 }^{ -11 } \right)$ in $0.1  M$ solution of $NaF$ would be

  1. $3.4\times { 10 }^{ -12 }M$
  2. $3.4\times { 10 }^{ -10 }M$
  3. $3.4\times { 10 }^{ -9 }M$
  4. $3.4\times { 10 }^{ -13 }M$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$CaF _2 \rightarrow Ca^{+2} + 2F^- $
                    $S$        $0.1$
$CaF _2: { K } _{ sp } = 3.4 \times { 10 }^{ -11 }   = [Ca^{+2}][F^-]^2$
$S = \displaystyle \frac {3.4 \times { 10 }^{ -11 }}{(0.1)^2} = 3.4 \times { 10 }^{ -9 }$
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

When solid $KCl$ is added to a saturated solution of $AgCI$ in $H _2O$

  1. Nothing happens
  2. Solubility of $AgCl$ decreases
  3. Solubility of $AgCl$ increases
  4. Solubility product of $AgCl$ increases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$KCl \rightarrow K^+ + Cl^-$

Potassium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of $AgCl$. Hence, the solubility of $AgCl$ decreases.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility of silver chloride ___________ in the presence of sodium chloride because of __________.

  1. increases; common ion effect

  2. increases; aldol condensation

  3. decreases; common ion effect

  4. decreases; aldol condensation
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The solubility of silver chloride decreases in the presence of sodium chloride because of common ion effect.

$\displaystyle AgCl \rightleftharpoons  Ag^+ + Cl^-$

$\displaystyle NaCl \rightarrow Na^+ + Cl^-$

Sodium chloride is a strong electrolyte and completely dissociates to provide chloride ions that are common ions. The chloride ions shift the equilibrium for dissociation of $AgCl$ towards left.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

$Ca{SO} _{4}$ is somewhat soluble in water.
$I$. When ${H} _{2}{SO} _{4}$ is added to a solution of $Ca{SO} _{4}$, the solubility of the $Ca{SO} _{4}$ will be increased.
$II$. The addition of ${H} _{2}{SO} _{4}$ will lower the $pH$ of the solution.

  1. Statement $I$ is true, Statement $II$ is true
  2. Statement $I$ is true, Statement $II$ is false
  3. Statement $I$ is false, Statement $II$ is true
  4. Statement $I$ is false, Statement $II$ is false
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When $H _2SO _4$ is added to a $CaSO _4$ solution, due to common ion effect, the solubility of $CaSO _4$ decreases. Further, as the acid concentration increases, the $pH$ of the solution lowers due to addition of $H _2SO _4$.

Hence, statement $I$ is false, statement $II$ is true.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility of ${ A } _{ 2 }{ X } _{ 5 }$ is $x\ mol\ { dm }^{ -3 }$. Its solubility product is:

  1. $36{ x }^{ 6 }$
  2. $64\times { 10 }^{ 4 }{ x }^{ 7 }$
  3. $126{ x }^{ 7 }$
  4. $1.25\times { 10 }^{ 4 }{ x }^{ 7 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the solubility be $S.$
${ As } _{ 2 }X _{ 3 }\leftrightharpoons 2As^{ 5+ }+5X^{ 2- }\\ \quad \quad \quad   \quad 2S\quad \quad \quad 5S$
Solubility product is 
$K _{ sp }=[As^{ 5+ }]^{ 2 }\times [S^{ 2- }]^{ 5 }\\$ 
Let $[{ As } _{ 2 }S _{ 5 }]=S,[As^{ 5+ }]=2S, [S^{ 2- }]=5S\\ Ksp=(2S)^{ 2 }\times (5S)^{ 5 }=4S^{ 2 }\times 3125S^{ 5 }=12500S^{ 7 }$
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility product constant $Ksp$ of $Mg(OH) _{2}$ is $9.0\times 10^{-12}$. If a solution is $0.010\ M$ with respect to $Mg^{2+}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH) _{2}$?

  1. $1.5\times 10^{-7}M$
  2. $3.0\times 10^{-7}M$
  3. $1.5\times 10^{-5}M$
  4. $3.0\times 10^{-5}M$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Ksp(Mg(OH) _2)=9.0\times 10^{-12}$

$(Mg(OH) _2 \leftrightharpoons Mg^{2+}+2[OH]^-$
$Ksp=[Mg^{2+}][OH^-]^2$
$9\times 10^{-12}=(10^{-12})(OH^-)^2$
$[OH^-]^2=3^2\times (10^{-5})^2$
$[OH^-]=3.0\times 10^{-5}M$
Maximum Hydroxide-ion concentration.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The $K _{sp}$ for AgCl is $2.8\times 10^{-10}$ at a given temperature. The solubility of AgCl in 0.01 molar HCl solution at this temperature will be :

  1. $2.8\times 10^{-12}mol L^{-1}$
  2. $2.8\times 10^{-8}mol L^{-1}$
  3. $5.6\times 10^{-8}mol L^{-1}$
  4. $2.8\times 10^{-4}mol L^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The chloride ion concentration in 0.01M HCl will be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
$K _{sp}=[Ag^+][Cl^-]$
Substitute values in the above expression.
$2.8 \times 10^{-10}=[Ag^+] \times 0.01$
Hence, $[Ag^+]= \frac {2.8 \times 10^{-10}} {0.01}=2.8 \times 10^{-8}$mol/L.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

Solubility of AgCl will be minimum in ___________.

  1. 0.01 M $Na _{2} SO _{4}$
  2. 0.01 M $Ca Cl _{2}$

  3. Pure water
  4. 0.001 M $Ag NO _{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The solubility of insoluble substances can be decreased by the presence of a common ion. 

Present in silver chloride are silver ions $(Ag^+)$ and chloride ions $(Cl^-)$. $AgCl$ is not soluble in water. 

Silver nitrate (which is soluble) has silver ion in common with silver chloride. But the concentration of the common ion is low (0.001M)

Calcium chloride (also soluble) has chloride ion in common with silver chloride. In $CaCl _2$ has two moles of common ion (0.02M) that decreases solubility very rapidly.  

In $Na _2SO _4$ there is no common ion effect apply therefore is soluble in it. 
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

Solid $Ba{({NO} _{3})} _{2}$ is gradually dissoved in a $1\times {10}^{-4}M$ ${Na} _{2}{CO} _{3}$ solution. At what minimum conc. of ${Ba}^{-2}$ will a precipitate of $Ba{CO} _{3}$ begin to form? (${K} _{sp}$ for $Ba{CO} _{3}=5.1\times {10}^{-9}$)

  1. $4.1\times {10}^{-5}M$
  2. $8.1\times {10}^{-7}M$
  3. $5.1\times {10}^{-5}M$
  4. $8.1\times {10}^{-8}M$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

If $S _0, S _1, S _2$ and $S _3$ are the solubilities in water of $AgCl$, $0.01 \,M \,CaCl _2, 0.01 \,M \,NaCl$ and $0.5 \,M \,AgNO _3$ solutions, respectively, then which of the following is true?

  1. $S _0 > S _2 > S _1 > S _3$
  2. $S _0 = S _2 = S _1 > S _3$
  3. $S _3 > S _1 > S _2 > S _0$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The solubility of $AgCl$ or its ion formation will depend inversely on the concentration

$S _0=H _2O$
$S _1=0.01\ M\ CaCl _2$
$S _2=0.01\ M\ NaCl$
$S _3=0.05\ M\ AgNO _3$
$H _2O$ is dilute solution and least concentrated hence have maximum solubility of $AgCl$ in it. Out of $NaCl$ and $CaCl _2$, the solubility of $NaCl$ is high due to loess number of ions produced from $NaCl$
as compare to $CaCl _2$. More the ion produced lesser is the solubility of coming salt.
So, the Correct order is $S _0 > S _2 > S _1 > S _3$
We know that concentration of common ion $\alpha \dfrac{1}{solubility}$. The order of solubility of $AgCl : S _1 > S _3 > S _2 > S _4$

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

$Ag _3 PO _4$ would be least soluble at 25$^o$C in

  1. 0.1 M $AgNO _3$
  2. 0.1 M $HNO _3$

  3. pure water
  4. 0.1 M $Na _3PO _4$

  5. solubility in (a), (b), (c) or (d) is not different
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Ag _3 PO _4$ is a weak electrolyte and $AgNO _3$ is a strong electrolyte containing common ion $(NO _3^-ion)$. Thus common ion effect is observed and the solubility of $Ag _3 PO _4$ is suppressed.
Hence, $Ag _3 PO _4$ is least soluble in 0.1M $AgNO _3$.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The solubility of $AgI$ in $NaI$ solution is less than that in pure water because

  1. $AgI$ forms complex with $NaI$

  2. Of common ion effect
  3. Solubility product of $AgI$ is less than that of $NaI$

  4. The temperature of the solution decreases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The common ion presence in the solution decrease the solubility of a given sparingly soluble compound. So, solubility of AgI in NaI solution is less.

Multiple choice chemistry chemical equilibrium and acids-bases dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The ionic strength of $C{H} _{3}CO{O}^{-}$ ion in $0.1\ M$ $C{H} _{3}COOH$ solution having ${K} _{a}= 1.8\times {10}^{-5}$ is

  1. $0.1$
  2. $0.05$
  3. $6.7\times {10}^{-4}$
  4. $1.34\times {10}^{-3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$C{H} _{3}COOH\rightleftharpoons C{H} _{3}CO{O}^{-} + {H}^{+}$


$[C{H} _{3}CO{O}^{-}]= C\alpha =C\sqrt {\frac{K _a}{C}}=\sqrt {K _aC} =\sqrt {1.8\times {10}^{-5}\times 0.1} =1.34\times {10}^{-3}$

Ionic strength of $C{H} _{3}CO{O}^{-}=\dfrac{1}{2}n\ [CH _3COO^-]{Z}^{2}=\dfrac{1}{2}\times 1\times 1.34\times {10}^{-3}\times (-1)^2=6.7\times {10}^{-4}$

where, n = no. of ions and Z= charge on the ion.

Option C is correct.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

Simultaneous solubility of $AgCNS\ (a)$ and $AgBr\ (b)$ in a solution of water will be

${K} _{{sp} _{(AgBr)}}=5\times {10}^{-13}$ and ${K} _{{sp} _{(AgCNS)}}={10}^{-12}$ 

  1. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$
  2. $a=4.08\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$
  3. $a=8.16\times {10}^{-7}mol$ ${litre}^{-1}$; $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Suppose solubility of $AgCNS$ and $AgBr$ in a solution are $a$ and $b$ respectively.

$AgCNS(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + }(aq) } +\underset { a }{ { CNS }^{ - }(aq) } $

$AgBr(s)\rightleftharpoons \underset { a+b }{ { Ag }^{ + } } +\underset { b }{ { Br }^{ - } } $

$\therefore$ $[{Ag}^{2+}]=(a+b); [CN{S}^{-}]=a$ and $[{Br}^{-}]=b$

For $AgCNS:{ K } _{ { sp } _{ AgCNS }\quad  }=\left[ { Ag }^{ + } \right] \left[ { CNS }^{ - } \right] $

or $1\times {10}^{-12}=(a+b)(a)......(i)$

For $AgBr: { K } _{ { sp } _{ AgBr }\quad  }=\left[ { Ag }^{ + } \right] \left[ { Br }^{ - } \right] $

or $5\times {10}^{-13}=(a+b)(b)........(ii)$

By Eqs $(i)$ and $(ii)$, we get

$\cfrac{a}{b}=\cfrac{{10}^{-12}}{5\times {10}^{-13}}=2$  ($a=2b$)

$\therefore $ By Eq $(i)$

$(2b+b)(2b)=1\times {10}^{-12}$

$\therefore$ $6{b}^{2}=1\times {10}^{-12}$

or $b=4.08\times {10}^{-7}$ $mol$ ${litre}^{-1}$

Similarly by Eq $(ii)$

$[a+(a/2)]a=1\times {10}^{-12}$

$a=8.16\times {10}^{-7}$ $mol$ ${litre}^{-1}$