Electrochemistry and Ionic Equilibria Questions and Answers

Q: At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:SaltMolar conductivity (S cm2 mol−1) at concentration (M)0.010.04NaNO3111101NaCl117107AgNO3125116The conductivity of a saturated aqueous solution of AgCl is 1.40×10−6 S cm−1 at 300 K. If the solubility of AgCl in water at 300 K is X mol L−1, then log10(X−1) is(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

The solubility is \(10^{-5}\) mol L−1, so \(\log_{10}(X^{-1})\) is 5.For strong electrolytes Kohlrausch's law gives \(\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}\). With \(\sqrt{0.01} = 0.1\) and \(\sqrt{0.04} = 0.2\), each pair of values gives \(A\) and the limiting value:\[ \begin{aligned} \mathrm{NaNO_{3}}:&\ A = 100,\ \Lambda^{\circ} = 121 \\ \mathrm{NaCl}:&\ A = 100,\ \Lambda^{\circ} = 127 \\ \mathrm{AgNO_{3}}:&\ A = 90,\ \Lambda^{\circ} = 134 \end{aligned} \]By Kohlrausch's law of independent migration of ions, the nitrate and sodium contributions cancel:\[ \begin{aligned} \Lambda^{\circ}_{\mathrm{AgCl}} &= \Lambda^{\circ}_{\mathrm{AgNO_3}} + \Lambda^{\circ}_{\mathrm{NaCl}} \\ &\quad - \Lambda^{\circ}_{\mathrm{NaNO_3}} \\ &= 134 + 127 - 121 = 140 \end{aligned} \]in S cm2 mol−1.\[ \begin{aligned} X &= \frac{1000\,\kappa}{\Lambda^{\circ}} = \frac{1000 \times 1.40 \times 10^{-6}}{140} \\ &= 1.0 \times 10^{-5}\ \text{mol L}^{-1} \end{aligned} \]So \(\log_{10}(X^{-1}) = 5\). The factor 1000 converts cm3 to litres; misplacing it by a power of ten gives 4 or 6, and 3 is further off still.

Q: The concentration of hydrogen ions in a sample of soft drink is 2×10-4mol lit-1 . Its pH value is (log 2 = 0.3010)

The pH is 3.699.\[ \begin{aligned} pH &= -\log(2 \times 10^{-4}) \\ &= 4 - \log 2 = 4 - 0.301 = 3.699 \end{aligned} \]A soft drink is acidic, so a pH below 7 makes sense. 4.301 (not offered) would come from adding \(\log 2\); 3.301 and 5.301 put the 0.301 on the wrong side or use the wrong power of ten.

Q: A monobasic acid HA has pH of 3 in 0.1M solution at 298 K. What is the pKa of the acid at 298 K?

The pKa is 5.pH 3 means \([\mathrm{H^{+}}] = [\mathrm{A^{-}}] = 10^{-3}\) M. The acid is only 1% ionised, so \([\mathrm{HA}] \approx 0.1\) M:\[ \begin{aligned} K_{a} &= \frac{(10^{-3})^{2}}{0.1} = 10^{-5} \\ pK_{a} &= -\log K_{a} = 5 \end{aligned} \]pKa = 3 confuses it with the pH, and 4 or 6 come from mishandling the powers of ten.

Q: The ion with the highest limiting molar conductance at 298 K is

The highest limiting molar conductance belongs to H+.H+ has a limiting molar conductivity of about 350 S cm2 mol−1 at 298 K, far above any metal ion, because it moves by the Grotthuss mechanism: protons hop from one water molecule to the next along hydrogen bonds instead of the whole hydrated ion moving through the solvent.Na+ and K+ are around 50 and 74, and even counted per mole of ions, Ca2+ and Mg2+ (about 119 and 106) are far below H+.

Q: What is the quantity of current required to deposit one mole of metallic magnesium from fused magnesium chloride? (1F=96500C)

The charge needed is \(1.93 \times 10^{5}\) C.At the cathode, \(\mathrm{Mg^{2+} + 2e^{-} \rightarrow Mg}\), so one mole of magnesium needs two moles of electrons, that is 2 faradays:\[ Q = 2 \times 96500 = 193000 = 1.93 \times 10^{5}\ \text{C} \]\(9.65 \times 10^{4}\) C is one faraday, enough only for a univalent metal. The other options are these values with the wrong power of ten.

Q: An aqueous solution of which of the following has the highest pH value?

The highest pH is that of 0.10 M \(\mathrm{NaOH}\).A higher pH means a more basic solution. NaOH is a strong base, fully dissociated, so 0.10 M NaOH gives \(\mathrm{[OH^{-}]} = 0.10\) M:\[ \begin{aligned} \mathrm{pOH} &= -\log 0.10 = 1 \\ \mathrm{pH} &= 14 - 1 = 13 \end{aligned} \]The more dilute 0.01 M NaOH gives pOH 2 and pH 12, so it is less basic. The three acid solutions all have pH below 7: 0.10 M HCl has pH 1, and 0.5 M HCl and 0.50 M \(\mathrm{H_{2}SO_{4}}\) are more concentrated acids with pH near or below 0.3.

Q: The resistance of 0.10 M KCl solution when measured with a conductivity cell at 298 K is 100 Ω. If the conductivity of 0.10 M KCl solution is 1.29 Sm-1 , what is the value of cell constant of the same solution at 298 K?

The cell constant is \(1.29\ \text{cm}^{-1}\).Conductivity \(\kappa\) is related to the measured resistance \(R\) through the cell constant \(G^{*} = l/A\): \(\kappa = G^{*}/R\), so \(G^{*} = \kappa R\).\[ \begin{aligned} G^{*} &= 1.29\ \text{S m}^{-1} \times 100\ \Omega \\ &= 129\ \text{m}^{-1} \\ &= 1.29\ \text{cm}^{-1} \end{aligned} \]The last step uses 1 m = 100 cm, so 129 per metre is 1.29 per centimetre. \(1.29\ \text{m}^{-1}\) is the conductivity's number without multiplying by the resistance, \(1.29 \times 10^{-2}\ \text{m}^{-1}\) comes from dividing by 100 instead of multiplying, and the two 0.129 values are slips of a power of ten.

Q: Which of the following solutions of salts are neutral?(i) KBr (ii) \(NaNO_{2}\) (iii) KF (iv) \(NH_{4}NO_{3}\) (v) NaCl

The neutral solutions are (i) and (v): KBr and NaCl.A salt of a strong acid and a strong base does not hydrolyse, so its solution is neutral. KBr (from KOH and HBr) and NaCl (from NaOH and HCl) are such salts.NaNO2 and KF come from weak acids (HNO2, HF), so their anions hydrolyse to give basic solutions. NH4NO3 comes from a weak base, so its cation hydrolyses to give an acidic solution.

Q: The molar conductivity \((\Lambda_{m})\) acetic acid is 78.1 S \(cm^{2}\,mol^{-1}\) . Its degree of dissociation (α ) is\((\Lambda_{m}^{0})\) for acetic acid= 390.5 S \(cm^{2}\,mol^{-1}\)

The degree of dissociation is 0.20.\[ \begin{aligned} \alpha &= \frac{\Lambda_{m}}{\Lambda_{m}^{0}} \\ &= \frac{78.1}{390.5} \\ &= 0.20 \end{aligned} \]A weak electrolyte's molar conductivity is lower than its limiting value because only a fraction \(\alpha\) is ionised. 0.02 and 0.04 slip a decimal, and 0.40 and 0.12 do not follow.So only one acetic acid molecule in five is ionised at this concentration, as expected for a weak acid.

11.

With reference to the variation of molar conductivity (Λ_m) with concentration for a strong electrolyte in an aqueous solution, the CORRECT statement is

A.The asymmetry effect contributes to decrease \(\Lambda_{m}\) whereas the electrophoretic effect contributes to increase \(\Lambda_{m}\)
B.The asymmetry effect contributes to increase \(\Lambda_{m}\) whereas the electrophoretic effect contributes to decrease \(\Lambda_{m}\)
C.Both asymmetry effect and electrophoretic effect contribute to decrease \(\Lambda_{m}\)
D.Both asymmetry effect and electrophoretic effect contribute to increase \(\Lambda_{m}\)

Exam Year(s)

IIT JAM - 2018,

Answer: C

Both the asymmetry effect and the electrophoretic effect contribute to decrease Λ_m.

In the Debye-Hückel-Onsager picture each ion is surrounded by an ionic atmosphere of opposite charge. When the ion moves in a field, the atmosphere lags behind and becomes asymmetric, pulling the ion back: the asymmetry (relaxation) effect. The atmosphere also moves the other way, carrying its solvent with it, so the ion travels against a counter-flow: the electrophoretic effect.

Both retard the ion, and both grow with concentration, which is why the molar conductivity of a strong electrolyte falls linearly with \(\sqrt{c}\). Neither effect can raise Λ_m, so options A, B and D are wrong.

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12.

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

Salt	Molar conductivity (S cm² mol⁻¹) at concentration (M)
Salt	0.01	0.04
NaNO₃	111	101
NaCl	117	107
AgNO₃	125	116

The conductivity of a saturated aqueous solution of AgCl is 1.40×10⁻⁶ S cm⁻¹ at 300 K. If the solubility of AgCl in water at 300 K is X mol L⁻¹, then log₁₀(X⁻¹) is

(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

Exam Year(s)

JEE Advanced - 2026,

Answer: C

The solubility is \(10^{-5}\) mol L⁻¹, so \(\log_{10}(X^{-1})\) is 5.

For strong electrolytes Kohlrausch's law gives \(\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}\). With \(\sqrt{0.01} = 0.1\) and \(\sqrt{0.04} = 0.2\), each pair of values gives \(A\) and the limiting value:

\[ \begin{aligned} \mathrm{NaNO_{3}}:&\ A = 100,\ \Lambda^{\circ} = 121 \\ \mathrm{NaCl}:&\ A = 100,\ \Lambda^{\circ} = 127 \\ \mathrm{AgNO_{3}}:&\ A = 90,\ \Lambda^{\circ} = 134 \end{aligned} \]

By Kohlrausch's law of independent migration of ions, the nitrate and sodium contributions cancel:

\[ \begin{aligned} \Lambda^{\circ}_{\mathrm{AgCl}} &= \Lambda^{\circ}_{\mathrm{AgNO_3}} + \Lambda^{\circ}_{\mathrm{NaCl}} \\ &\quad - \Lambda^{\circ}_{\mathrm{NaNO_3}} \\ &= 134 + 127 - 121 = 140 \end{aligned} \]

in S cm² mol⁻¹.

\[ \begin{aligned} X &= \frac{1000\,\kappa}{\Lambda^{\circ}} = \frac{1000 \times 1.40 \times 10^{-6}}{140} \\ &= 1.0 \times 10^{-5}\ \text{mol L}^{-1} \end{aligned} \]

So \(\log_{10}(X^{-1}) = 5\). The factor 1000 converts cm³ to litres; misplacing it by a power of ten gives 4 or 6, and 3 is further off still.

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13.

The concentration of hydrogen ions in a sample of soft drink is 2×10^-4mol lit^-1 . Its pH value is (log 2 = 0.3010)

A.4.369
B.3.699
C.2.369
D.5.301
E.3.301

Exam Year(s)

KEAM - 2025,

Answer: B

The pH is 3.699.

\[ \begin{aligned} pH &= -\log(2 \times 10^{-4}) \\ &= 4 - \log 2 = 4 - 0.301 = 3.699 \end{aligned} \]

A soft drink is acidic, so a pH below 7 makes sense. 4.301 (not offered) would come from adding \(\log 2\); 3.301 and 5.301 put the 0.301 on the wrong side or use the wrong power of ten.

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14.

A monobasic acid HA has pH of 3 in 0.1M solution at 298 K. What is the pKa of the acid at 298 K?

Exam Year(s)

KEAM - 2025,

Answer: C

The pK_a is 5.

pH 3 means \([\mathrm{H^{+}}] = [\mathrm{A^{-}}] = 10^{-3}\) M. The acid is only 1% ionised, so \([\mathrm{HA}] \approx 0.1\) M:

\[ \begin{aligned} K_{a} &= \frac{(10^{-3})^{2}}{0.1} = 10^{-5} \\ pK_{a} &= -\log K_{a} = 5 \end{aligned} \]

pK_a = 3 confuses it with the pH, and 4 or 6 come from mishandling the powers of ten.

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15.

The ion with the highest limiting molar conductance at 298 K is

A.\(\mathrm{H^{+}}\)
B.\(\mathrm{Na^{+}}\)
C.\(\mathrm{K^{+}}\)
D.\(\mathrm{Ca^{2+}}\)
E.\(\mathrm{Mg^{2+}}\)

Exam Year(s)

KEAM - 2025,

Answer: A

The highest limiting molar conductance belongs to H⁺.

H⁺ has a limiting molar conductivity of about 350 S cm² mol⁻¹ at 298 K, far above any metal ion, because it moves by the Grotthuss mechanism: protons hop from one water molecule to the next along hydrogen bonds instead of the whole hydrated ion moving through the solvent.

Na⁺ and K⁺ are around 50 and 74, and even counted per mole of ions, Ca²⁺ and Mg²⁺ (about 119 and 106) are far below H⁺.

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16.

What is the quantity of current required to deposit one mole of metallic magnesium from fused magnesium chloride? (1F=96500C)

A.\(1.93 \times 10^{4} C\)
B.\(1.93 \times 10^{3} C\)
C.\(9.65 \times 10^{3} C\)
D.\(9.65 \times 10^{4} C\)
E.\(1.93 \times 10^{5} C\)

Exam Year(s)

KEAM - 2025,

Answer: E

The charge needed is \(1.93 \times 10^{5}\) C.

At the cathode, \(\mathrm{Mg^{2+} + 2e^{-} \rightarrow Mg}\), so one mole of magnesium needs two moles of electrons, that is 2 faradays:

\[ Q = 2 \times 96500 = 193000 = 1.93 \times 10^{5}\ \text{C} \]

\(9.65 \times 10^{4}\) C is one faraday, enough only for a univalent metal. The other options are these values with the wrong power of ten.

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17.

An aqueous solution of which of the following has the highest pH value?

A.0.10 M \(\mathrm{HCl}\)
B.0.50 M \(\mathrm{H_{2}SO_{4}}\)
C.0.10 M \(\mathrm{NaOH}\)
D.0.5 \(\mathrm{MHCl}\)
E.0.01 \(\mathrm{MNaOH}\)

Exam Year(s)

KEAM - 2025,

Answer: C

The highest pH is that of 0.10 M \(\mathrm{NaOH}\).

A higher pH means a more basic solution. NaOH is a strong base, fully dissociated, so 0.10 M NaOH gives \(\mathrm{[OH^{-}]} = 0.10\) M:

\[ \begin{aligned} \mathrm{pOH} &= -\log 0.10 = 1 \\ \mathrm{pH} &= 14 - 1 = 13 \end{aligned} \]

The more dilute 0.01 M NaOH gives pOH 2 and pH 12, so it is less basic. The three acid solutions all have pH below 7: 0.10 M HCl has pH 1, and 0.5 M HCl and 0.50 M \(\mathrm{H_{2}SO_{4}}\) are more concentrated acids with pH near or below 0.3.

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18.

The resistance of 0.10 M KCl solution when measured with a conductivity cell at 298 K is 100 Ω. If the conductivity of 0.10 M KCl solution is 1.29 Sm^-1 , what is the value of cell constant of the same solution at 298 K?

A.\(1.29\ m^{-1}\)
B.\(1.29 \times 10^{-2} m^{-1}\)
C.\(0.129\ cm^{-1}\)
D.\(1.29 cm^{-1}\)
E.\(0.129\ m^{-1}\)

Exam Year(s)

KEAM - 2025,

Answer: D

The cell constant is \(1.29\ \text{cm}^{-1}\).

Conductivity \(\kappa\) is related to the measured resistance \(R\) through the cell constant \(G^{*} = l/A\): \(\kappa = G^{*}/R\), so \(G^{*} = \kappa R\).

\[ \begin{aligned} G^{*} &= 1.29\ \text{S m}^{-1} \times 100\ \Omega \\ &= 129\ \text{m}^{-1} \\ &= 1.29\ \text{cm}^{-1} \end{aligned} \]

The last step uses 1 m = 100 cm, so 129 per metre is 1.29 per centimetre. \(1.29\ \text{m}^{-1}\) is the conductivity's number without multiplying by the resistance, \(1.29 \times 10^{-2}\ \text{m}^{-1}\) comes from dividing by 100 instead of multiplying, and the two 0.129 values are slips of a power of ten.

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19.

Which of the following solutions of salts are neutral?
(i) KBr (ii) \(NaNO_{2}\) (iii) KF (iv) \(NH_{4}NO_{3}\) (v) NaCl

A.(ii) and (iii)
B.(i) and (iv)
C.(i) and (v)
D.(iv) and (v)
E.(iii) and (v)

Exam Year(s)

KEAM - 2025,

Answer: C

The neutral solutions are (i) and (v): KBr and NaCl.

A salt of a strong acid and a strong base does not hydrolyse, so its solution is neutral. KBr (from KOH and HBr) and NaCl (from NaOH and HCl) are such salts.

NaNO₂ and KF come from weak acids (HNO₂, HF), so their anions hydrolyse to give basic solutions. NH₄NO₃ comes from a weak base, so its cation hydrolyses to give an acidic solution.

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20.

The molar conductivity \((\Lambda_{m})\) acetic acid is 78.1 S \(cm^{2}\,mol^{-1}\) . Its degree of dissociation (α ) is
\((\Lambda_{m}^{0})\) for acetic acid= 390.5 S \(cm^{2}\,mol^{-1}\)

A.0.12
B.0.40
C.0.02
D.0.20
E.0.04

Exam Year(s)

KEAM - 2025,

Answer: D

The degree of dissociation is 0.20.

\[ \begin{aligned} \alpha &= \frac{\Lambda_{m}}{\Lambda_{m}^{0}} \\ &= \frac{78.1}{390.5} \\ &= 0.20 \end{aligned} \]

A weak electrolyte's molar conductivity is lower than its limiting value because only a fraction \(\alpha\) is ionised. 0.02 and 0.04 slip a decimal, and 0.40 and 0.12 do not follow.

So only one acetic acid molecule in five is ionised at this concentration, as expected for a weak acid.

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