Chemistry

Electrochemistry and Solutions

205 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 conductivity and its types electrochemistry

Which of the following is correct regarding current carrying ions in the solution of $C _{2}H _{5}COOH$ upon dilution?

  1. The number of ions in $1 cm^{3}$, as well as in total volume increases.
  2. The number of ions in $1 cm^{3}$ decreases, whereas that in the total volume remains constant.
  3. The number of ions in $1 cm^{3}$ decreases, but that in the total volume increases.
  4. The number of ions in $1 cm^{3}$, as well as in total volume decreases.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For weak electrolytes, as the concentration decreases, the percentage dissociation increases and the total number of ions increases. But as the volume increases, number of ions per unit volume decreases.


Hence, option C is correct.

Multiple choice conductivity and its types electrochemistry

The resistance of 0.1 N solution of a salt is found to be $2.5\times10^{3}$. The equivalent conductance of the solution is: (cell constant=1.15 $cm^{-1}$)

  1. 3.6

  2. 4.6

  3. 5.6

  4. 6.6
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relationship between the specific conductance, resistance and the cell constant is $ \kappa =\cfrac { 1 }{ R } \times \cfrac { l }{ a }$.
Substitute $ R=2.5\times { 10 }^{ 3 }\quad ohm $ and $ \cfrac { l }{ a } =1.15\quad {cm }^{ -1 } $.
Hence $ \kappa =\cfrac { 1 }{ 2.5\times { 10 }^{ 3 }\quad ohm } \times1.15\quad { cm }^{ -1 }=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
The relationship between the equivalent conductance and specifc conductance is $  { \Lambda  } _{ eq }=\cfrac { \kappa \times 1000 }{ M }  $.
Substitute $ M=0.1\quad N $ and $ \kappa=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
Hence $ { \Lambda  } _{ eq }=\cfrac { \cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \times 1000 }{ 0.1 } \quad =\quad 4.6\quad \quad { ohm }^{ -1 }\quad { cm }^{ 2 }\quad { equiv }^{ -1 } $.

Multiple choice conductivity and its types electrochemistry

At 291 K, the equivalent conductivities at infinite dilution of $NH _4Cl$, $NaOH $ and $NaCl $ are $129.8$, $217.5$ and $108.9$ S $cm^2 eq^{-1}$ respectively. The equivalent conductivity at infinite dilution of $NH _4OH$ is:

  1. $208.4 S cm^2 eq^{-1}$
  2. $238.4 S cm^2 eq^{-1}$
  3. $283.4 S cm^2 eq^{-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equivalent conductance of $NH _4OH $ = Equivalent conductance of $NH _4Cl $+$NaOH$-$NaCl$

                                                                $= 129.8+218.4-108.9$

Equivalent conductance of $NH _4OH = 239.3$

Therefore, option B is the correct answer.
Multiple choice conductivity and its types electrochemistry

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is:

  1. $LiCl > NaCl > KCl$
  2. $KCl > NaCl > LiCl$
  3. $NaCl > KCl > LiCl$
  4. $LiCl > KCl > NaCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is $KCl > NaCl > LiCl$.

Ionic mobility depends upon size of the ion. The ionic size in case of hydrated cation, is $K^+(aq) < Na^+(aq) < Li^+(aq)$.

As the size of the hydrated ion increases, the equivalent conductivity at infinite dilution decreases.

Multiple choice conductivity and its types electrochemistry

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1\ Ohm^{-1} cm^{2}$. If $NaF$ is replaced by $KF$ what is the value of equivalent conductance?

  1. $90.1\ Ohm^{-1} cm^{2}$
  2. $11.2\ Ohm^{-1} cm^{2}$
  3. $0$
  4. $222.4\ Ohm^{-1} cm^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.

For this case, Both NaF and KF are strong electrolyte and also furnish 2 ions in the solution.
Hence, Both will have same equivalent conductance.
So, option A is correct.

Multiple choice conductivity and its types electrochemistry

The resistance of $N/10$ solution is found to be $2.5\times 10^{3}ohm$. The equivalent conductance of the solution is (cell constant $= 1.25\ cm^{-1})$.

  1. $2.5\ ohm^{-1} cm^{2} equiv^{-1}$
  2. $5\ ohm^{-1} cm^{2} equiv^{-1}$
  3. $2.5\ ohm^{-1} cm^{-2} equiv^{-1}$
  4. $5\ ohm^{-1} cm^{-2} equiv^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given Data : 1)  Resistance = $R$ = $2.5 \times 10^3$

                     2) Cell constant = $k$ = $\cfrac{l}{a}$= $1.15$$cm^2$
                     3)  Normality =$N$= $0.1 N$
To Find : Equivalent conductance = $\Lambda$$ _e$$ _q$

The relation between $K$ ,$R$ and $ k$ is,
$K$ = $\cfrac{1}{R} \times \cfrac{l}{a}$
where $K$ is specific conductance.
$\therefore$ Substituting the given values we get,

$K$ = $\cfrac{1}{2.5×10^3}$ × $1.15cm^-$$^1$

     = $\cfrac{1.15}{2.5×10^3}$ $ohm$$^-$$^1$$cm$$^-$$^1$

The relation between $\Lambda$$ _e$$ _q$ and $K$ is,

$\Lambda$$ _e$$ _q$ = $\cfrac{K×1000}{N}$

Substituting the value of $M$ and $K$ we get,

$\Lambda _{eq}= \cfrac{1.15}{2.5\times10^3\times0.1} \times 1000$

       = $4.6$$ohm^-$$^1$$cm$$^2$$equi$$^{-1}$      [Note:$ \text {Normality= no. of equiv.} /cm^3$]

Here appproximation is taken,

      $\approx$ $5$ $ohm$$^-$$^1$$cm$$^2$$equi$$^-$$^1$

Hence the correct option is 'B'.

Multiple choice conductivity and its types electrochemistry

The equivalent conductances of $NaCl$ at concentration $c$ and at infinite dilution are $\lambda _{c}$ and $\lambda _{\infty}$ respectively. The correct relationship between $\lambda _{c}$ and $\lambda _{\infty}$ is given as: (where the constant $b$ is positive).

  1. $\lambda _{c} = \lambda _{\infty} - b\sqrt {c}$
  2. $\lambda _{c} = \lambda _{\infty} + b\sqrt {c}$
  3. $\lambda _{c} = \lambda _{\infty} + bc$
  4. $\lambda _{c} = \lambda _{\infty} - bc$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to debye huckel onsager equation we have,

equivalent conductance of NaCl is given by
$\lambda _C$ = $\lambda _{\infty}$ - $b C^{1/2}$
where b is positive.

Multiple choice conductivity and its types electrochemistry

The equivalent conductances at infinite dilution of $HCl$ and $NaCl$ are $426.15$ and $126.15\ mho\ cm^{2}g\ eq^{-1}$ respectively. If can be said that the mobility of:

  1. $H^{+}$ ions is much more than that of $Cl^{-}$ ions
  2. $Cl^{-}$ ions is much more than that of $H^{+}$ ions
  3. $H^{+}$ ions is much more than that of $Na^{+}$ ions
  4. $Na^{+}$ ions is much more than that of $H^{+}$ ions
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\rightarrow$ The equivalent conductance of $HCl$ is more than $NaCl$ , it is because of difference in speed of ions.
$\rightarrow\,HCl$ has conductance at infinite dilution more than three times as high as that of $NaCl$.
$\rightarrow$ As Chlorine ion is common in both of them, it seems that speed of $H^{+}$ ion is much more than that of $Na^{+}$ ions.

Multiple choice conductivity and its types electrochemistry

The resistance of $0.01\ N$ solution at $25^{\circ}$ is $200\ ohm$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

  1. $200\ ohm^{-1}cm^{2} eq^{-1}$.
  2. $300\ ohm^{-1}cm^{2} eq^{-1}$.
  3. $400\ ohm^{-1}cm^{2} eq^{-1}$.
  4. $500\ ohm^{-1}cm^{2} eq^{-1}$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

since, we have

$conductance * cell constant$ =  specific conductance
cell consyant = 1
conductance = specific conductance= $\dfrac{1}{200}$

equivalent conductance= $\dfrac{K*1000}{N}$ $Scm^{2}eq^{-1}$

equivalent conductance= $\dfrac{1*1000}{200*0.01}$

equivalent conductance= $500$ $ Scm^{2}eq^{-1}$
 

Multiple choice conductivity and its types electrochemistry

The equivalent conductivity of $N/10$ solution of acetic acid at $25^o$C is $14.3$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$. What will be the degree of dissociation of acetic acid?
$(\Lambda _{\infty CH _3COOH}=390.71$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$).

  1. $3.66\%$
  2. $3.9\%$
  3. $2.12\%$
  4. $0.008\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equivalent conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.
Given,
$\Lambda^{\infty} _{m(CH _3COOH)}=390.71 ohm^{-1} cm^{2} eq^{-1}$
$\Lambda _{m(CH _3COOH)}=14.3 ohm^{-1} cm^{2} eq^{-1}$
Degree of dissociation is given by,
$\alpha=\dfrac{\Lambda _m}{\Lambda^{0} _m}=\dfrac{14.3}{390.71}=0.0366 \implies$3.66%


Multiple choice conductivity and its types electrochemistry

The ionic conductivity of $B{a^{2 + }}$ and $C{l^ - }$ at infinite dilution are 127 and 76 respectively. The equivalent conductivity of $BaC{l _2}$ at infinite dilution (in $oh{m^ - }\,c{m^2}\,e{q^{ - 1}}$) would be:

  1. 279

  2. 280

  3. 139.5

  4. 102
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given data,


$\lambda (Ba^{2+}) = 127$


$\lambda(Cl) = 76$

$\wedge _m^{\infty} = \lambda (Ba^{2+}) + 2\lambda(Cl)$

$= 127 + 2\times 76$

$\wedge _m^{\infty} = 279$

We know that-

$\wedge _{eq}^{\infty} = \dfrac{\lambda _m^{\infty}(Ba^{2+})}{2} + \lambda _m^{\infty}(Cl^-)$ 

$= \dfrac{127}{2} + 76$

$\wedge _{eq}^{\infty} = 139.5\ {ohm}^{-1}{cm}^2$

Hence, option C is correct.

Multiple choice conductivity and its types electrochemistry

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:

  1. LiCl $>$ NaCl $>$ KCl
  2. KCl $>$ NaCl $>$ LiCl
  3. NaCl $>$ KCl $>$ LiCl
  4. LiCl $>$ KCl $>$ NaCl
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, the anion are same, comparing the size of cation.

$Li^+<Na^+<K^+$
but due to hydration of cation, the radii order is inverted.
$Li^+>Na^+>K^+$
$Conductance \propto\cfrac{1}{radii}$
$\therefore$ The order is $KCl>NaCl>LiCl$

Multiple choice conductivity and its types electrochemistry

Equivalent conductance at infinite dilution for weak electrolyte HF:

  1. can be determined by measurement of equivalent conductance at infinite dilution for dilute solution of $HCL, \; HBr$ and $HI$
  2. can be determined by measurement of equivalent conductance at infinite dilution for very dilute $HF$ solutions

  3. can best be determined from measurements on dilute

  4. can not be calculated
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For weak electrolytes, equivalent conductance at infinite dilution can be determined by their very dilute solutions. Hence, for $HF$. equivalent conductance at infinite dilution can be determined by measurements on very dilute $HF$ solutions.
Multiple choice conductivity and its types electrochemistry

The conductivity of a saturated solution of $Ba{ SO } _{ 4 }$ is $306\times { 10 }^{ -6 }{ ohm }^{ -1 }{ cm }^{ -1 }$ and its equivalent conductance is $1.53 \ { ohm }^{ -1 }{ cm }^{ 2 }{ equiv }^{ -1 }$. 


The ${ K } _{ sp }$ for ${ BaSO } _{ 4 }$ will be :

  1. $4\times { 10 }^{ -12 }$
  2. $2.5\times { 10 }^{ -9 }$
  3. $2.5\times { 10 }^{ -13 }$
  4. $4\times { 10 }^{ -6 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given : $\wedge eq=1.53 \Omega^{-1}eq^{-1}$

$\wedge = 3.06\times 10^{-6}\Omega^{-1}$

Solubilicty, $s=\dfrac{\wedge\times 1000}{\wedge eq}=\dfrac{3.06\times 10^{-6}\times 10^{3}}{1.53}=2\times 10^{-3}M$

$K _{\wedge p}=[Ba^{2+}][SO _{4}^{2-}]=S^{2}$

$K _{\wedge p}=(2\times 10^{-3})^{2}=4\times 10^{-6}M^{2}$

Therefore, the correct option is D.
Multiple choice conductivity and its types electrochemistry

If the specific resistance of a solution of concentration C g equivalent/litre is R, then its equivalent conductance is:

  1. $\dfrac{100R}{C}$
  2. $\dfrac{RC}{1000}$
  3. $\dfrac{1000}{RC}$
  4. $\dfrac{C}{1000R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Specific resistance for $C _g eq/lt=R$

Dont know the meaning of conductance it must be conductance
Conductance of solution $=k=\dfrac{1}{k}$
Equivalent conductance $=\dfrac{k\propto 1000}{c}$
                                         $=\dfrac{1000}{RC}$