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 p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Aqueous solution of $AIC{I _3}$ on heating to dryness will give 

  1. $AIC{I _3}$
  2. $A{l _2}{O _3}$
  3. $Al(OH)C{L _2}\,$
  4. $A{L _2}C{l _6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heating an aqueous solution of AlCl3 causes hydrolysis where the aluminum ion reacts with water to form aluminum hydroxide, which then dehydrates to aluminum oxide (Al2O3) upon heating to dryness.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

$AlCl _3$ achieves stability by forming a dimer. In trivalent state the compound is hydrolysed in water. $AlCl _3$ in acidified aqueous solution forms :

  1. $Al(OH) _3+HCl$
  2. $[Al(H _2O) _6]^{3+}+3Cl^-$
  3. $AlCl _3.2H _2O$
  4. $Al _2O _3+HCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Answer:- (B) $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$

The electron deficiency of Aluminium atom in $Al{Cl} _3$ is compensated by formation of co-ordinate bond between lone pair of Chlorine atom of another $Al{Cl} _3$ molecule and the empty unhybridised p orbital of Aluminium atom, thus forms dimer and in acidified aqueous solution forms $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$.

$Al{Cl} _3 + 6H _2O \; \longrightarrow \; [Al({H _2O)} _6]^{+3} + 3{Cl}^{-}$
Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Aluminium chloride exists as dime , $Al _2Cl _6$ in solid state as well as in solution of non-polar solvents like benzene.When dissolved in water, it gives :

  1. $Al^{3+} +2Cl$
  2. $Al _2O _3+6HCl$
  3. $[Al(OH) _6]^3+3HCl$
  4. $[Al(H _2O) _6]^{3+}+3Cl$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Similar to question 430733, the dissolution of AlCl3 in water results in the formation of the octahedral hexaaquaaluminum(III) complex and chloride ions.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

The solubility of anhydrous $AlCl _{3}$ and hydrous $AlCl _{3}$ in diethyl ether are $S _{1}$ and $S _{2}$ respectively. Then -

  1. $S _{1}=S _{2}$
  2. $S _{1}> S _{2}$
  3. $S _{1}< S _{2}$
  4. $S _{1}< S _{2}$ but not $S _{1}=S _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Anhydrous AlCl3 will dissolve the diethyl ether because anhydrous ​AlCl3 is a good lewis acid. It is electron deficient. The lone pairs of diethyl ether will be donated to the anhydrous ​AlCl3 due to which it is soluble. On the other hand, hydrous ​AlCl3 have the water molecules which make it poor lewis acid.


So, $S _{1}> S _{2}$.

Multiple choice chemistry basic analytical techniques partition coefficients solvent extraction separation methods

Mixture of ${\text{Al}}{\left( {{\text{OH}}} \right) _3}\,\,{\text{and}}\,\,{\text{Fe}}{\left( {{\text{OH}}} \right) _3}$ can be separated by

  1. ${\text{HCl}}\,$
  2. ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}\,$
  3. ${\text{NaOH}}$
  4. ${\text{HN}}{{\text{O}} _{\text{3}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Aluminum hydroxide is amphoteric and dissolves in strong acids like HCl, whereas ferric hydroxide is basic and also dissolves in strong acids. However, in the context of qualitative analysis, HCl is the standard reagent used to dissolve these hydroxides to separate them from other insoluble residues or to bring them into solution for further testing.

Multiple choice chemistry basic analytical techniques filtration and crystallisation other methods of separation separation techniques

In fractional crystallization the proportion of components in the precipitate will depend on their solubility product.

  1. True

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

Fractional crystallization is the method of refining substances based on the differences in their solubility. If a mixture of two or more substances in a solution is allowed to crystallise, the least soluble substance will crystallise first. The proportion of the components in the precipitate will depend on their solubility product.

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

The constituents of the electrolyte for the extraction of aluminium is ____________.

  1. 40 parts cryolite + 40 parts fluorspar + 20 parts pure $Al _2O _3$
  2. 60 parts cryolite + 30 parts fluorspar + 10 parts pure $Al _2O _3$
  3. 30 parts cryolite + 30 parts fluorspar + 40 parts pure $Al _2O _3$
  4. 60 parts cryolite + 20 parts fluorspar + 20 parts pure $Al _2O _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryoliteand fluorspar is added, the mixture melts at $900^oC$ and $Al _2O _3$ becomes a good conductorof electricity.
Electrolyte: 60 parts cryolite + 20 parts fluorspar + 20 parts pure  $Al _2O _3$

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 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

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

$\alpha= 0.2$