The best conductor of electricity is a 0.1 M solution of — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Factors Governing Electrical Conductivity of Electrolytes The electrical conductivity ($\kappa$) of an aqueous electrolyte solution depends directly on two primary factors: 1. The **concentration of free ions** present in the solution. 2. The **ionic mobility** and electrical charge of those individual ions. For solutions with the same nominal molar concentration ($C = 0.1\text{ M}$), the concentration of free ions in solution is determined by the **degree of dissociation** ($\alpha$) of the solute and the number of ions ($y$) produced per formula unit of the electrolyte: $$\text{Concentration of Free Ions} \propto C \cdot \alpha \cdot y$$ Step 2 - Classify and Analyze the Dissociation of the Given Electrolytes Let us analyze each of the substances provided in the options: 1. **Sulphuric acid ($\ce{H2SO4}$):** Sulphuric acid is a strong diprotic mineral acid. In aqueous solution, it undergoes complete dissociation ($\alpha \approx 1$): $$\ce{H2SO4(aq) -> 2H^+(aq) + SO4^{2-}(aq)}$$ Each formula unit of $\ce{H2SO4}$ yields $3\text{ moles}$ of ions (namely, $2\text{ moles}$ of $\ce{H^+}$ and $1\text{ mole}$ of $\ce{SO4^{2-}}$). Therefore, for a $0.1\text{ M}$ solution of $\ce{H2SO4}$: $$\text{Total ionic concentration} \approx 3 \times 0.1\text{ M} = 0.3\text{ M}$$ 2. **Boric acid ($\ce{H3BO3}$):** Boric acid is an extremely weak, monobasic Lewis acid that does not release protons directly, but instead abstracts a hydroxide ion from water ($\alpha \ll 0.01$): $$\ce{H3BO3(aq) + H2O(l) <=> [B(OH)4]^-(aq) + H^+(aq)}$$ Because its dissociation constant is extremely small ($K_a \approx 5.8 \times 10^{-10}$), the concentration of free ions in a $0.1\text{ M}$ solution is negligible: $$\text{Total ionic concentration} \approx 2 \cdot C \cdot \alpha \approx 2 \times 0.1 \times \left(7.6 \times 10^{-5}\right) \approx 1.5 \times 10^{-5}\text{ M}$$ 3. **Acetic acid ($\ce{CH3COOH}$):** Acetic acid is a weak organic monobasic carboxylic acid that dissociates only partially in water ($\alpha \ll 1$): $$\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H^+(aq)}$$ With a dissociation constant of $K_a \approx 1.8 \times 10^{-5}$, the degree of dissociation ($\alpha$) at $0.1\text{ M}$ is: $$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}} \approx 0.0134 \quad (1.34\%)$$ Thus, the total ionic concentration is: $$\text{Total ionic concentration} = 2 \cdot C \cdot \alpha \approx 2 \times 0.1 \times 0.0134 \approx 0.0027\text{ M}$$ 4. **Propanoic acid ($\ce{CH3CH2COOH}$):** Propanoic acid is also a weak organic monobasic carboxylic acid. Due to the electron-donating inductive effect (+I effect) of the larger ethyl group, its dissociation constant is even lower than that of acetic acid ($K_a \approx 1.3 \times 10^{-5}$): $$\ce{CH3CH2COOH(aq) <=> CH3CH2COO^-(aq) + H^+(aq)}$$ $$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.3 \times 10^{-5}}{0.1}} \approx 0.0114 \quad (1.14\%)$$ $$\text{Total ionic concentration} = 2 \cdot C \cdot \alpha \approx 2 \times 0.1 \times 0.0114 \approx 0.0023\text{ M}$$ Step 3 - Compare Ionic Concentrations and Mobility Comparing the calculated concentration of ions in the $0.1\text{ M}$ solutions: $$\text{Ionic Concentration of } \ce{H2SO4} \ (0.3\text{ M}) \gg \ce{CH3COOH} \ (0.0027\text{ M}) \approx \ce{CH3CH2COOH} \ (0.0023\text{ M}) \gg \ce{H3BO3} \ (1.5 \times 10^{-5}\text{ M})$$ Additionally, $\ce{H2SO4}$ produces a high concentration of $\ce{H^+}$ ions, which possess exceptionally high ionic mobility in water due to the Grotthuss proton-hopping mechanism. Therefore, the $0.1\text{ M}$ sulphuric acid solution will have the highest electrical conductivity and is the best conductor of electricity among the given choices. Step 4 - Evaluate the Options * **Option (A) is incorrect:** Boric acid is a very weak Lewis acid that hardly ionizes in water, making it an extremely poor conductor. * **Option (B) is correct:** Sulphuric acid is a strong, completely dissociated electrolyte that releases three ions per formula unit, giving it the highest concentration of highly mobile ions. * **Option (C) is incorrect:** Acetic acid is a weak acid with a low degree of dissociation (around $1.3\%$), resulting in very few ions. * **Option (D) is incorrect:** Propanoic acid is a weak organic acid with an even lower ionization than acetic acid, making it a poor conductor. $$\text{Correct Option: } \boxed{\text{B}}$$