Electrochemistry and Ionic Equilibria Questions and Answers

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.

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.

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.

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.

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⁺.

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.

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.

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.

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.

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.