The specific conductance of AgCl solution in water was determined to be 1.8 × 10^-6 Ω^-1 cm^-1 at 29 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Limiting Molar Conductivity of \ce{AgCl} According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity (molar conductivity at infinite dilution) of an electrolyte is the sum of the individual limiting molar conductivities of its constituent cations and anions: $$\Lambda^\circ_m(\ce{AgCl}) = \lambda^\circ_{\ce{Ag^+}} + \lambda^\circ_{\ce{Cl^-}}$$ Given values: * Limiting molar conductivity of silver ions ($\lambda^\circ_{\ce{Ag^+}}$) = $67.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ * Limiting molar conductivity of chloride ions ($\lambda^\circ_{\ce{Cl^-}}$) = $82.1\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ Substituting these values into the equation: $$\Lambda^\circ_m(\ce{AgCl}) = 67.9\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} + 82.1\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\Lambda^\circ_m(\ce{AgCl}) = 150.0\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Step 2 - Establish the Relationship Between Limiting Molar Conductivity and Solubility Silver chloride ($\ce{AgCl}$) is a sparingly soluble salt. When dissolved in water, it dissociates as: $$\ce{AgCl(s) <=> Ag^+(aq) + Cl^-(aq)}$$ Because the solubility ($S$) of $\ce{AgCl}$ in water is extremely low, the resulting saturated solution is exceptionally dilute (approaching infinite dilution). Consequently, its molar conductivity ($\Lambda_m$) at this saturation concentration can be approximated to its limiting molar conductivity ($\Lambda^\circ_m$): $$\Lambda_m \approx \Lambda^\circ_m$$ The standard formula relating molar conductivity ($\Lambda_m$), specific conductance ($\kappa$), and concentration ($C$) in $\text{mol L}^{-1}$ is: $$\Lambda_m = \frac{\kappa \times 1000}{C}$$ Since the concentration of a saturated sparingly soluble salt solution is equal to its solubility ($S$), we can substitute $C = S$ and $\Lambda_m = \Lambda^\circ_m$: $$\Lambda^\circ_m = \frac{\kappa \times 1000}{S}$$ Rearranging this formula to solve for solubility ($S$): $$S = \frac{\kappa \times 1000}{\Lambda^\circ_m}$$ Step 3 - Substitute Values and Calculate Solubility Given specific conductance (conductivity) of the solution: * $\kappa = 1.8 \times 10^{-6}\ \Omega^{-1}\text{ cm}^{-1}$ Substituting $\kappa$ and $\Lambda^\circ_m$ with their respective units: $$S = \frac{(1.8 \times 10^{-6}\ \Omega^{-1}\text{ cm}^{-1}) \times 1000\text{ cm}^3\text{ L}^{-1}}{150.0\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ $$S = \frac{1.8 \times 10^{-3}\text{ mol L}^{-1}}{150.0}$$ $$S = 1.2 \times 10^{-5}\text{ M}$$ Thus, the solubility of silver chloride in water at $298\text{ K}$ is $\mathbf{1.2 \times 10^{-5}\text{ M}}$. Step 4 - Explain Each Option * **Option (A) is incorrect:** This value ($1.2 \times 10^{-8}\text{ M}$) is three orders of magnitude lower than the correct value, which occurs if one fails to multiply the conductivity ($\kappa$) by the unit conversion factor of $1000$. * **Option (B) is incorrect:** This value ($1.44 \times 10^{-10}\text{ M}$) represents the solubility product ($K_{\text{sp}}$) of $\ce{AgCl}$ ($K_{\text{sp}} = S^2 = (1.2 \times 10^{-5})^2 = 1.44 \times 10^{-10}$) rather than its solubility ($S$). * **Option (C) is correct:** As calculated, the correct solubility of $\ce{AgCl}$ in water is exactly $1.2 \times 10^{-5}\text{ M}$. * **Option (D) is incorrect:** This value ($1.44 \times 10^{-16}\text{ M}$) is mathematically incorrect and represents a major power-of-ten calculation mistake. $$\text{Correct Option: } \boxed{\text{C}}$$