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 chemistry substances in common use preparation, properties and uses of baking soda chemical from common salt compounds of carbon

The balanced equations for : "Carbon dioxide is passed through a concentrated aqueous solution of sodium chloride saturated with ammonia".

  1. $NaCl + NH _4OH + CO _2\rightarrow NH _4Cl +NaHCO _3$
  2. $2NaCl + NH _4OH + CO _2\rightarrow NH _4Cl +2NaHCO _3$
  3. $NaCl + NH _4OH + CO _2\rightarrow NH _4CO +NaHCl _3$
  4. $NaCl + NH _4OH + CO _2\rightarrow NH _3 +NaCl+H _2CO _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$NaCl + NH _4OH + CO _2\rightarrow NH _4Cl +NaHCO _3$ 

Sodium bicarbonate is formed. This reaction is used in the Solvay process for the manufacture of sodium carbonate.

Multiple choice conductivity and its types electrochemistry

The specific conductance of a $0.01\ M$ solution of $KCl$ is $0.0014\ ohm^{-1} cm^{-1}$ at $25^{\circ}C$. Its equivalent conductance is____________.

  1. $14$
  2. $140$
  3. $1.4$
  4. $0.14$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\kappa =0.0014\ S{ cm }^{ -1 }\ \Lambda _{ eq }=\cfrac { 1000\times \kappa  }{ C } \ C=0.01M\ { \Lambda  } _{ eq }=\cfrac{1000 \times 0.0014}{0.01}=140$

Multiple choice conductivity and its types electrochemistry

The resistance of a N/10 KCI solution is 245$\Omega $. Calculate the equivalent conductance of the solution if the electrodes in the cell are 4cm apart and each having an area of 7.0sq,cm.

  1. $23.32S{ cm }^{ 2 }{ eq }^{ -1 }$
  2. $23.23S{ cm }^{ 2 }{ eq }^{ -1 }$
  3. $2.332S{ cm }^{ 2 }{ eq }^{ -1 }$

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

Given resistance = 245$\Omega $

Formula for the specific conductance (K) = $ \dfrac{1}{R} $ $\times \dfrac{l}{a}$
= $ \dfrac{1}{245} $ $\times \dfrac{4}{7}$
 K = 2.33 $\times 10^-$$^3$ 
Formula for the euivalent conductance=  K $\times$ $\dfrac{1000}{C}$
= 2.33 $\times 10^-$$^3$ $\times 10000$
= 23.32S $cm^2$ eq $^-$$^1$

Multiple choice conductivity and its types electrochemistry

The equivalent conductance of a weak monobasic acid at infinite dilution is $100cm^3$ $eq^{-1}$ and that of its $0.01$M solution is $5cm^2$ $eq^{-1}$ at $25^o$C. The dissociation constant $K _a$ of the acid is:

  1. $2.5\times 10^{-4}$
  2. $5\times 10^{-4}$
  3. $1.25\times 10^{-5}$
  4. $2.5\times 10^{-5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The degree of dissociation, $\alpha=\cfrac {\text{Equivalent conductance at given dilution}}{\text{Equivalent conductance at infinite dilution}}$

                                                   $=\cfrac {5cm^2eq^{-1}}{100cm^2eq^{-1}}=\cfrac {1}{20}$
For weak monobasic acid,
$K _a=\cfrac {C \alpha^2}{1-\alpha}$
       $=C\alpha^2(\because 1>> \alpha)$
       $=0.01\times \left(\cfrac {1}{20}\right)^2= 2.5 \times 10^{-5}$

Multiple choice conductivity and its types electrochemistry

Equivalent conductance and molar conductance of $Fe _2(SO _4) _3$ are related?

  1. $\bigwedge _e=\bigwedge _m$
  2. $\bigwedge _{eq} = \dfrac{\bigwedge _m}{3}$
  3. $\bigwedge _{eq} = 3 \bigwedge _m$
  4. $\bigwedge _{eq} = \dfrac{\bigwedge _m}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For Fe2(SO4)3, the salt dissociates into 2 Fe^3+ and 3 SO4^2- ions. The total charge is 6. The relationship between equivalent conductance and molar conductance is Lambda_eq = Lambda_m / n, where n is the total charge (valency factor). Here n = 6.

Multiple choice conductivity and its types electrochemistry

Molar ionic conductance of ${Ca}^{+2}$ is $x$ $S{m}^{2}$ ${mole}^{-1}$. Equivalent conductance of calcium phosphate is ____ $S{m}^{2}g$ ${eq}^{-1}$

  1. $x+y$
  2. $(3x+2y)$
  3. $6(3x+2y)$
  4. $\cfrac{3x+2y}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calcium phosphate $\rightarrow { Ca } _{ 3 }{ \left( { PO } _{ 4 } \right)  } _{ 2 }$

ionic conductance of ${ Ca }^{ 2+ }=x$ ${ sm }^{ 2 }{ mol }^{ -1 }$
ionic conductance of ${ PO } _{ 4 }^{ 3- }=Y$ ${ sm }^{ 2 }{ mol }^{ - }$
${ Ca } _{ 3 }{ \left( { PO } _{ 4 } \right)  } _{ 2 }\rightarrow 3{ Ca }^{ 2+ }+2{ PO } _{ 4 }^{ 3- }$
$\therefore $  equivalent conductivity $=x+y$

Multiple choice conductivity and its types electrochemistry

The equivalent conductance of $CH _3COONa, \ HCl$ and $NaCl$ at infinite dilution are $91, 426$ and $126 \ S \ cm^3 \ eq^{-1}$ respectively at $25^oC$. The equivalent conductance of $1 \ M \ CH _3COOH$ solution is $19.55 \ S \ cm^2 \ eq^{-1}$. The pH of the solution is:

  1. $5.3$
  2. $4.3$
  3. $2.3$
  4. $1.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\lambda^o _{CH _3COONa}= \lambda^o _{CH _3COO^-}+ \lambda^o _{Na}=91 \ Scm^2eq^{-1}$.....(i)

$\lambda^o _{HCl}= \lambda^o _{H}+ \lambda^o _{Cl^-}= 426 \ Scm^2eq^{-1}$.....(ii)

$\lambda^o _{NaCl}= \lambda^o _{Na}+ \lambda^o _{Cl^-}=126 \ Scm^2eq^{-1}$.....(iii)

$\lambda^o _{CH _3COOH}= \lambda^o _{CH _3COO^-}+ \lambda^o _{H^+}= (i) + (ii) – (iii) = 391 \ Scm^2eq^{-1}$

$ \lambda _{CH _3COOH}= 19.5 \ Scm^2eq^{-1}$ (given)

Degree of dissociation $= \cfrac {\lambda _m}{\lambda^o _m}= \cfrac {19.55}{391}=0.05$

$CH _3COOH \longrightarrow CH _3COO^- + H^+$

$1(1-0.05)$                $0.05$             $0.05$

$[H^+]=0.05 \ M$

$pH= -log [H^+]=1.3$

Multiple choice conductivity and its types electrochemistry

Equivalent conductance of $BaCl _2, H _2SO _4$ and $HCl$ are $x _1, x _2$ and $x _3 S cm^2 equiv^{-1}$ at infinite dilution.If specific conductance of saturated $BaSO _4$ solution is of $y S cm^1$ then $K _{sp}$ of $BaSO _4$ is:

  1. $\frac{10^3y}{2(x _1 + x _2 - 2x _3)}$
  2. $\frac{10^6y^2}{4(x _1 + x _2 - 2x _3)^2}$
  3. $\frac{10^6y^2}{2(x _1 + x _2 - x _3)^2}$
  4. $\frac{x _1 + x _2 - 2x _3}{10^6y^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ \Lambda  } _{ eq }\quad (Ba{ SO } _{ 4 })={ \Lambda  } _{ eq }\left( Ba{ Cl } _{ 2 } \right) +{ \Lambda  } _{ eq }\left( { H } _{ 2 }{ SO } _{ 4 } \right) -2{ \Lambda  } _{ eq }\left( HCl \right) $

                            $=\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right) S{ cm }^{ 2 }{ eq }^{ -1 }$
$\because { \Lambda  } _{ eq }=K\times \cfrac { 1000 }{ N } \quad \Rightarrow \quad N=\left( \cfrac { y\times { 10 }^{ 3 } }{ { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } }  \right) \quad \left( \because { n } _{ f }=2 \right) $
$\therefore \quad M=\left{ \cfrac { y\times { 10 }^{ 3 } }{ 2\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }  \right} $
$\therefore \quad { K } _{ sp }={ \left( M \right)  }^{ 2 }=\left{ \cfrac { { y }^{ 2 }\times { 10 }^{ 6 } }{ 4{ \left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }^{ 2 } }  \right} $

Multiple choice conductivity and its types electrochemistry

For $HCl$ solution at ${25}^{o}C$ equivalent conductance at infinite dilution is $425 \ {ohm}^{-1}{cm}^{2}{equiv}^{-1}$. The specific conductance of a solution of $HCl$ is $3.825$ ${ohm}^{-1}{cm}^{-1}$. If the apparent degree of dissociation is $90$% the normality of the solution is :

  1. $0.90N$
  2. $1.0N$
  3. $10\ N$
  4. $1.2N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The degree of dissociation alpha = Lambda_m / Lambda_m_infinity. Given alpha = 0.9, Lambda_m_infinity = 425, and specific conductance kappa = 3.825. Lambda_m = kappa * 1000 / M. Solving for M (molarity/normality for HCl) gives 10 N.

Multiple choice conductivity and its types electrochemistry

The equivalent conductivity of $0.1 N \ CHNCH _{3}COOH$ at $25^{0}C$ is 80 and at infinite dilution it is 400, the degree of dissociation of $CH _{3}COOH$ is :

  1. 1

  2. 0.2

  3. 0.1

  4. 0.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:-

$\wedge _{eq}(CH _3COOH)= 80 S cm^{-2} eq^{-1}$

$\wedge^{\infty} _{eq}(CH _3COOH)$ at infinite dilution= $400S cm^2 eq^{-1}$

$\alpha \longrightarrow$ Degree of dissociation

$\alpha= \cfrac {\wedge^m _{eq}}{\wedge^{\infty} _{eq}}=\cfrac {80}{400}= 0.2$

$\alpha= 0.2$

Multiple choice conductivity and its types electrochemistry

In infinite dilusions, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ are $127$ and $76 ohm^{-1} \, cm^{-1} \, eqvt^{-1}$. The equivalent conductivity of $BaCl _2$ at indefinite dilution is?

  1. $101.5$
  2. $203.5$
  3. $139.5$
  4. $279.5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equivalent conductance of BaCl2 at infinite dilution, 


λ∞ of BaCl2=1/2 λ∞ of Ba2+ + λ∞ of Cl−
       

 =127/2+76 

=139.

Option C is correct answer

Multiple choice conductivity and its types electrochemistry

Equivalent conductivity of $BaCl _2,H _2SO _4$ and HCI, are $x _1,x _2$ and $x _3scm^{-1}eq^{-1}$ at infinite dilution. If conductivity of saturated $BaSO _4$ solution is x $Scm^{-1}$, then $K _{sp}$ of $BaSO _4$ is

  1. $\dfrac {500x} {(x _1+x _2-2x _3)}$
  2. $\dfrac {10^6x^2} {(x _1+x _2-2x _3)^3}$
  3. $\dfrac {2.5\times10^5 x^2} {x _1-2x _2-x _3)^2}$
  4. $\dfrac {0.25 x^2} {x _1 + x _2-x _3)^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice conductivity and its types electrochemistry

Molar conductance of $C{a^{2 + }}$ and $C{l^ - }$ are  $120\,c{m^2}\,$ $mo{l^{ - 1}}$ and $77\,S\,c{m^2}\,mo{l^{ - 1}}$ respectively. What is the equivalent conductance of $CaC{l _2}$ ?

  1. $98.5\,Sc{m^2}e{q^{ - 1}}$
  2. $137\,Sc{m^2}e{q^{ - 1}}$
  3. $197\,Sc{m^2}e{q^{ - 1}}$
  4. $247\,Sc{m^2}e{q^{ - 1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equivalent conductance is calculated by dividing the molar conductance by the valence factor of the electrolyte. For calcium chloride, the valence factor is 2, and applying Kohlrausch law using the given ionic conductances gives 120 plus two times 77, all divided by 2, which equals 137.

Multiple choice conductivity and its types electrochemistry

At infinite dilution equivalent conductance of ${B^{ + 2}}$ & CI ions are 127  &  76$oh{m^{ - 1}}$ $c{m^{ - 1}}$` $e{q^{ - 1}}$ respectively. Equivalent conductance  of $BaC{I _2}$ at infinite diluition is : 

  1. 139.5

  2. 101.5

  3. 203

  4. 279
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice conductivity and its types electrochemistry

Equivalent constant of standard $BaSO _{4}$ is $400ohm^{-1}\ cm^{2}$ equiv$^{-1}$ and specific conduction is $8\times 10^{-5}\ ohm^{-1}\ cn^{-1}$. Hence $K _{SP}$ of $BaSO _{4}$ is

  1. $4\times 10^{-8}M^{2}$
  2. $1\times 10^{-8}M^{2}$
  3. $2\times 10^{-4}M^{2}$
  4. $1\times 10^{-4}M^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer