Calculate the ionic product of water at 25°C from the following data: Conductivity of water = 5.5 × — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Apply Kohlrausch's Law of Independent Migration of Ions According to Kohlrausch's Law, the limiting molar conductance of a weak electrolyte at infinite dilution is equal to the sum of the limiting ionic conductances of its constituent cations and anions. For water ($\ce{H2O}$), which dissociates into hydrogen ions ($\ce{H^+}$) and hydroxide ions ($\ce{OH^-}$): $$\Lambda_m^\circ(\ce{H2O}) = \lambda^\circ(\ce{H^+}) + \lambda^\circ(\ce{OH^-})$$ Given data: * $\lambda^\circ_{\ce{H^+}} = 0.035\text{ mho m}^2\text{ mol}^{-1}$ * $\lambda^\circ_{\ce{OH^-}} = 0.020\text{ mho m}^2\text{ mol}^{-1}$ Substituting the given values into Kohlrausch's equation: $$\Lambda_m^\circ(\ce{H2O}) = 0.035\text{ mho m}^2\text{ mol}^{-1} + 0.020\text{ mho m}^2\text{ mol}^{-1}$$ $$\Lambda_m^\circ(\ce{H2O}) = 0.055\text{ mho m}^2\text{ mol}^{-1}$$ Step 2 - Relate Specific Conductivity ($\kappa$) to Limiting Molar Conductance ($\Lambda_m^\circ$) Since pure water is an extremely weak electrolyte, its concentration of ions is exceedingly low. At such immense dilution, we can safely assume that the molar conductance ($\Lambda_m$) of water is virtually equal to its limiting molar conductance ($\Lambda_m^\circ$). The relationship between specific conductivity ($\kappa$), molar conductance ($\Lambda_m^\circ$), and ionic concentration ($C$) in SI units ($\text{mol m}^{-3}$) is: $$\Lambda_m^\circ = \frac{\kappa}{C} \implies C = \frac{\kappa}{\Lambda_m^\circ}$$ Step 3 - Calculate the Concentration ($C$) of Ions in Molarity ($\text{M}$) Given conductivity of pure water: $$\kappa = 5.5 \times 10^{-6}\text{ mho m}^{-1}$$ Substitute the values of $\kappa$ and $\Lambda_m^\circ$ into the concentration formula: $$C = \frac{5.5 \times 10^{-6}\text{ mho m}^{-1}}{0.055\text{ mho m}^2\text{ mol}^{-1}}$$ $$C = 1.0 \times 10^{-4}\text{ mol m}^{-3}$$ To convert the concentration from SI units ($\text{mol m}^{-3}$) to molarity ($\text{mol L}^{-1}$ or $\text{M}$), we use the volume conversion factor $1\text{ m}^3 = 1000\text{ L}$: $$C = \frac{1.0 \times 10^{-4}\text{ mol}}{1\text{ m}^3} \times \frac{1\text{ m}^3}{1000\text{ L}}$$ $$C = 1.0 \times 10^{-7}\text{ mol L}^{-1} = 1.0 \times 10^{-7}\text{ M}$$ Step 4 - Calculate the Ionic Product of Water ($K_w$) Water autoionizes according to the following equilibrium: $$\ce{H2O(l) <=> H^+(aq) + OH^-(aq)}$$ In pure water, due to electrical neutrality, the concentration of hydrogen ions ($\ce{H^+}$) must be exactly equal to the concentration of hydroxide ions ($\ce{OH^-}$): $$[\ce{H^+}] = [\ce{OH^-}] = C = 1.0 \times 10^{-7}\text{ M}$$ The ionic product of water ($K_w$) is defined as the product of the molar concentrations of hydrogen and hydroxide ions: $$K_w = [\ce{H^+}] \times [\ce{OH^-}]$$ Substituting the calculated concentrations: $$K_w = \left(1.0 \times 10^{-7}\text{ M}\right) \times \left(1.0 \times 10^{-7}\text{ M}\right)$$ $$K_w = \mathbf{1.0 \times 10^{-14}\text{ M}^2}$$ Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This value ($2 \times 10^{-14}\text{ M}^2$) represents a calculation error where the coefficients of the ionic product are doubled. * **Option (B) is incorrect:** This value ($1 \times 10^{-7}\text{ M}^2$) represents the concentration of a single ion ($[\ce{H^+}]$ or $[\ce{OH^-}]$), not their product. * **Option (C) is incorrect:** This value ($1 \times 10^{-8}\text{ M}^2$) does not match any proper physical calculation of the system. * **Option (D) is correct:** As mathematically demonstrated, the calculated ionic product of water at $25^\circ\text{C}$ is exactly $1.0 \times 10^{-14}\text{ M}^2$. $$\text{Correct Option: } \boxed{\text{D}}$$