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
The solubility of $Hg{I} _{2}$ in water decreases in presence of $KI$.

State whether the given statement is true or false.


  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Hg{I} _{2}$ forms soluble complex with $KI$
$2KI+Hg{I} _{2}\longrightarrow {K} _{2}Hg{I} _{4}$

So solubility of $Hg{I} _{2}$ in water increases in presence of $KI$.
Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

Assertion: AgCl will not dissolve in a concentrated solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assertion: AgCl will not dissolve in a concentrated NaCl solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.
This is an example of a common ion effect. The chloride ions are common ions.
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

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

Assertion:
Solubility of $AgCl$ in water decreases if $NaCl$ is added to it.
Reason:
$NaCl$ is soluble freely in water but $AgCl$ is sparingly soluble.

  1. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion

  3. Assertion is correct but Reason is not correct

  4. Assertion is not correct but Reason is correct

  5. Both Assertion and Reason are not correct
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$NaCl$ is a strong electrolyte. It completely dissociates to form ions. Hence, it is highly soluble in water.
$AgCl$ dissociates to little extent. Hence, it is sparingly soluble in water.
When $NaCl$ is added to a solution of $AgCl$, due to common ion effect (chloride ion is the common ion), the dissociation of $AgCl$ is suppressed. Also as the concentration of chloride ion (from $NaCl$) increases, the ionic product of $AgCl$ exceeds its solubility product and precipitation occurs.

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

$\displaystyle { BaCl } _{ 2 }$ dissociates in water to give one $\displaystyle { Ba }^{ 2+ }$ ion and two $\displaystyle { Cl }^{ - }$ ions. If concentrated $\displaystyle HCl$ is added to this solution :

  1. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ increases
  2. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ remains constant
  3. $\displaystyle \left[ { OH }^{ - } \right] $ increases
  4. The number of moles of undissociated $\displaystyle { BaCl } _{ 2 }$ increases
  5. $\displaystyle \left[ { H }^{ + } \right] $ decreases
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.
$BaCl _2\rightarrow Ba^{2+} + 2Cl^{-}$
$HCl \rightarrow H^{+} + Cl^{-}$
As $Cl^{-} $ is the common ion so, the $ Ba^{2+}$and $ 2Cl^{-}$ combines ( associate ) to give the undissociated $BaCl _2$. 
Hence , there is increase in concentration of undissociated $BaCl _2$.

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 following reaction occurs in a beaker: $\displaystyle { Ag }^{ + }\left( aq \right) +{ Cl }^{ - }\left( aq \right) \rightarrow AgCl\left( s \right) $. If a solution of sodium chloride were added to this beaker,

  1. The solubility of the sodium chloride would decrease

  2. The reaction would shift to the left

  3. The concentration of silver ions in solution would increase

  4. The solubility of the silver chloride would decrease

  5. The equilibrium would not shift at all
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Ag^+ + Cl^- \rightarrow AgCl$
$NaCl \rightarrow Na^+ + Cl^-$
As $Cl^- $ is the common ion in both the solution ( i.e. $NaCl $ and $AgCl$ ) , hence $AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

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

A solution contains both $Mg^{2+}$(aq) and $Sr^{2+}$(aq) at the same concentration.
The solution is divided into two equal portions. Aqueous sodium hydroxide is added dropwise to one portion. Dilute sulfuric acid is added dropwise to the other portion.
Which row is correct?

precipitate seen first when NaOH(aq) is added precipitate seen first when $H _2SO _4$(aq) is added
A magnesium hydroxide magnesium sulfate
B magnesium hydroxide strontium sulfate
C strontium hydroxide magnesium sulfate
D strontium hydroxide strontium sulfate


  1. A

  2. B

  3. C

  4. D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{On reacting with NaOH(aq) the Mg+2 gives precipitation reaction.}$


$\text{Sr2+ forms strontium sulfate by reaction with sulphuric acid.}$

$\text{So option B is correct.}$

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

In KI solution, $ I _2 $ readily dissolves and forms :

  1. $ I^{-}$
  2. $ KI _2 $
  3. $ KI _2^- $
  4. $ KI _3 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In KI solution, $I _2$ readily dissolves and forms $KI _3$ which contains polyhalide ion $I _3^-$


$KI + I _2  \rightarrow KI _3$

Due to this, iodine is partly soluble in water but highly soluble in the $KI$ solution.

So, the correct option is $D$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Conductivity ($\kappa $) of $0.01M$ $NaCl$ solution is $0.000145S$ ${cm}^{-1}$. What happens to the conductivity if extra $100mL$ of ${H} _{2}O$ be added to the above solution?

  1. Decreases

  2. Increases

  3. Remains unchanged

  4. First increase and then decrease
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Conductivity (kappa) is defined as the conductance of a solution contained between two electrodes of unit area and unit distance apart. Adding water (dilution) decreases the number of ions per unit volume, thus decreasing the conductivity.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Electrolytic conductivity of $0.3 M$ solution of $KCI$ at $298 K $is $3.72$ x ${10}^{-2}Scm^{-1}$.Calculate its molar conductivity ($S cm^{2} mol ^{-1}$) 

  1. $124$
  2. $30.56$
  3. $192$
  4. $185$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$k=3.72\times 10^{-2}\ Scm^{-1}$
$c=0.3M$

Molar conductivity($m$) is given by,

$m=\dfrac{k\times 1000}{c}$

$\Rightarrow m=\dfrac{3.72\times 10^{-2}\times 1000}{0.3}$

$\Rightarrow m=124\ S\ cm^2mol^{-1}$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Which of the following solutions contains the greatest number of ions assuming that each solute is fully ionized?

  1. $1$ cm$^{-3}$ of $0.2M$ ${Na} _{2}{SO} _{4}$
  2. $1$ cm$^{-3}$ of $0.2M$ $Ca{Cl} _{2}$
  3. $1$ cm$^{-3}$ of $0.2M$ $NaCl$
  4. $1$ cm$^{-3}$ of $0.2M$ ${Cr} _{2}{({SO} _{4})} _{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

No. of ions in 11 molecule of $Na _2SO _4=3$ 

No. of ions in 11 molecule of $CaCl _2=3$
No. of ions in 11 molecule of $NaCl=2$
No. of ions in 11 molecule of $Cr _2(SO _4) _3=5$
As the molarity and volume is same for all the given substances, number of ions will be greatest for the compound having greatest number of ions in one molecule. Therefore, $Cr _2(SO _4) _3$ will have the greatest number of ions.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Which among the following solutions is NOT used in determination of the cell constant?

  1. ${ 10 }^{ -2 }MKCl$
  2. ${ 10 }^{ -1 }MKCl$
  3. $1 M KCl$
  4. Saturated $KCl$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Standard KCl solutions (like 0.1M or 0.01M) are used for cell constant determination because their conductivity is accurately known. Saturated KCl is not used because its concentration and conductivity can be unstable or difficult to reproduce precisely.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Ionic conductance of ${ H }^{ + }$ and ${ SO } _{ 4 }^{ 2- }$ are $x$ and $yS\ { cm }^{ 2 }\ { mol }^{ -1 }$. Hence, equivalent conductivity of ${ H } _{ 2 }{ SO } _{ 4 }$ is:

  1. $2x+\cfrac { y }{ 2 } $
  2. $x+\cfrac { y }{ 2 } $
  3. $\cfrac { x }{ 2 } +y$
  4. $\cfrac { x }{ 2 } +\cfrac { y }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equivalent conductance can be defined as the conductance produce by an equivalent of ion present in the solution.

For ${ H }^{ + }$, no.of ${ e }^{ - }$ change $=1$
For ${ SO } _{ 4 }^{ 2- }$, no.of ${ e }^{ - }$ change $=2$
Thus equivalent conductance of ${ H } _{ 2 }{ SO } _{ 4 }$
${ H } _{ 2 }{ SO } _{ 4 }$ $\rightleftharpoons 2{ H }^{ + }+{ SO } _{ 4 }^{ 2- }$
Eg Conductance $=\cfrac { 2[Ionic\quad conductance\quad of\quad { H }^{ + }] }{ nf } +\cfrac { Ionic\quad Conductance\quad of\quad { SO } _{ 4 }^{ 2- } }{ nf } $
Eg. Conductance $=2\times \cfrac { x }{ 1 } s{ cm }^{ 2 }+\cfrac { y }{ 2 } s{ cm }^{ 2 }$
Eg. Conductance $=2x+\cfrac { y }{ 2 }$