Equal volumes of 0.015 M - CH3COOH and 0.015 M - solutions are mixed together. What would be the mol — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Chemical Equation and Analyze Stoichiometry When acetic acid ($\ce{CH3COOH}$, a weak acid) is mixed with sodium hydroxide ($\ce{NaOH}$, a strong base), they undergo a quantitative acid-base neutralization reaction to form sodium acetate ($\ce{CH3COONa}$) and water ($\ce{H2O}$): $$\ce{CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)}$$ Since both reactant solutions have the identical initial concentration of $0.015\text{ M}$ and are mixed in equal volumes, they react completely in a $1:1$ stoichiometric ratio. There are no limiting or excess reactants remaining, and the only solute contributing significantly to the electrolytic conductivity of the final mixture is the salt, $\ce{CH3COONa}$. Step 2 - Calculate the Concentration ($C$) of the Salt in the Mixture Let the volume of each solution mixed be $V\text{ L}$. * The initial number of moles of both $\ce{CH3COOH}$ and $\ce{NaOH}$ is: $$\text{Moles of reactants} = 0.015\text{ M} \times V\text{ L} = 0.015V\text{ mol}$$ * According to the $1:1$ stoichiometry of the neutralization, the moles of $\ce{CH3COONa}$ produced is: $$\text{Moles of }\ce{CH3COONa} = 0.015V\text{ mol}$$ * Upon mixing, the final total volume of the mixture doubles to $2V\text{ L}$. Thus, the final molar concentration ($C$) of $\ce{CH3COONa}$ in the mixture is exactly halved: $$C = \frac{\text{Moles of }\ce{CH3COONa}}{\text{Total Volume}}$$ $$C = \frac{0.015V\text{ mol}}{2V\text{ L}} = \frac{0.015}{2}\text{ M} = 0.0075\text{ M}$$ Step 3 - State the Formula for Molar Conductivity ($\Lambda_m$) The relationship between molar conductivity ($\Lambda_m$), specific conductivity ($\kappa$), and molar concentration ($C$) of an electrolyte is given by: $$\Lambda_m = \frac{\kappa \times 1000}{C}$$ Where: * $\Lambda_m$ is the molar conductivity in $\text{S cm}^2\text{ mol}^{-1}$ * $\kappa$ is the conductivity in $\text{S cm}^{-1}$ * $C$ is the concentration of the electrolyte in $\text{mol L}^{-1}$ Step 4 - Substitute the Values and Calculate the Molar Conductivity Given values: * Conductivity of the mixture ($\kappa$) = $6.3 \times 10^{-4}\text{ S cm}^{-1}$ * Concentration ($C$) = $0.0075\text{ M}$ Substitute these values into the molar conductivity equation: $$\Lambda_m = \frac{(6.3 \times 10^{-4}\text{ S cm}^{-1}) \times 1000\text{ cm}^3\text{ L}^{-1}}{0.0075\text{ mol L}^{-1}}$$ $$\Lambda_m = \frac{0.63\text{ S cm}^{-1}\text{ cm}^3\text{ L}^{-1}}{0.0075\text{ mol L}^{-1}}$$ $$\Lambda_m = \mathbf{84\text{ S cm}^2\text{ mol}^{-1}}$$ Step 5 - Explain Each Option * **Option (A) is incorrect:** This value ($0.84\text{ S cm}^2\text{ mol}^{-1}$) is a power-of-ten error (underestimated by a factor of 100). * **Option (B) is incorrect:** This value ($8.4\text{ S cm}^2\text{ mol}^{-1}$) is a decimal-placement error (underestimated by a factor of 10). * **Option (C) is correct:** As mathematically calculated, the molar conductivity of the resulting mixture is exactly $84\text{ S cm}^2\text{ mol}^{-1}$. * **Option (D) is incorrect:** This value ($42\text{ S cm}^2\text{ mol}^{-1}$) is obtained if one fails to halve the concentration when the volume doubles during mixing (i.e., using $0.015\text{ M}$ instead of $0.0075\text{ M}$). $$\text{Correct Option: } \boxed{\text{C}}$$