[Four-digit Integer] We have taken a saturated solution of AgBr. of AgBr is 12 × 10^-14. If 10^-7 mo — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the Dissociation Reactions and Express Equilibrium Concentrations When silver nitrate ($\ce{AgNO3}$), a strong electrolyte, is added to water, it dissociates completely into its constituent ions: $$\ce{AgNO3(aq) -> Ag^+(aq) + NO3^-(aq)}$$ Since $10^{-7}\text{ moles}$ of $\ce{AgNO3}$ is dissolved in $1\text{ L}$ of solution: * Initial concentration of added $\ce{Ag^+}$ $= 10^{-7}\text{ M}$ * Concentration of $\ce{NO3^-}$ $= 10^{-7}\text{ M}$ Silver bromide ($\ce{AgBr}$) is a sparingly soluble salt that establishes a dynamic solubility equilibrium: $$\ce{AgBr(s) <=> Ag^+(aq) + Br^-(aq)}$$ Let $S$ be the molar solubility of $\ce{AgBr}$ in this solution. Due to the presence of the common ion $\ce{Ag^+}$ from $\ce{AgNO3}$, the total equilibrium concentrations of the ions in the solution are: * $$[\ce{Ag^+}] = (S + 10^{-7})\text{ M}$$ * $$[\ce{Br^-}] = S\text{ M}$$ * $$[\ce{NO3^-}] = 10^{-7}\text{ M}$$ Step 2 - Apply the Solubility Product Constant ($K_{sp}$) and Solve for Solubility ($S$) The solubility product expression for $\ce{AgBr}$ is: $$K_{sp} = [\ce{Ag^+}][\ce{Br^-}]$$ Substitute the given value $K_{sp} = 12 \times 10^{-14}$ and the equilibrium concentration expressions: $$12 \times 10^{-14} = (S + 10^{-7}) \times S$$ $$S^2 + 10^{-7}S - 12 \times 10^{-14} = 0$$ This is a quadratic equation in terms of $S$. We solve it using the quadratic formula: $$S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$S = \frac{-10^{-7} \pm \sqrt{(10^{-7})^2 - 4(1)(-12 \times 10^{-14})}}{2}$$ $$S = \frac{-10^{-7} \pm \sqrt{10^{-14} + 48 \times 10^{-14}}}{2}$$ $$S = \frac{-10^{-7} \pm \sqrt{49 \times 10^{-14}}}{2}$$ $$S = \frac{-10^{-7} \pm 7 \times 10^{-7}}{2}$$ Since solubility must be a positive quantity: $$S = \frac{6 \times 10^{-7}}{2} = 3 \times 10^{-7}\text{ M}$$ Step 3 - Calculate the Equilibrium Molar Concentrations of All Ionic Species Using the calculated value of $S = 3 \times 10^{-7}\text{ M}$, we find the equilibrium concentration of each ion in the solution: * $$[\ce{Br^-}] = S = 3 \times 10^{-7}\text{ M}$$ * $$[\ce{Ag^+}] = S + 10^{-7} = 3 \times 10^{-7} + 10^{-7} = 4 \times 10^{-7}\text{ M}$$ * $$[\ce{NO3^-}] = 10^{-7}\text{ M}$$ Step 4 - Convert Concentration Units from Molarity to SI Units ($\text{mol m}^{-3}$) The limiting ionic conductivities ($\lambda^\circ$) are given in SI units ($\text{S m}^2\text{ mol}^{-1}$). To ensure unit consistency, we must convert the concentrations from $\text{mol L}^{-1}$ (Molarity) to $\text{mol m}^{-3}$: $$1\text{ m}^3 = 1000\text{ L} \implies 1\text{ mol L}^{-1} = 10^3\text{ mol m}^{-3}$$ Multiplying each concentration by $10^3$: * $$[\ce{Br^-}] = 3 \times 10^{-7}\text{ mol L}^{-1} \times 10^3\text{ L m}^{-3} = 3 \times 10^{-4}\text{ mol m}^{-3}$$ * $$[\ce{Ag^+}] = 4 \times 10^{-7}\text{ mol L}^{-1} \times 10^3\text{ L m}^{-3} = 4 \times 10^{-4}\text{ mol m}^{-3}$$ * $$[\ce{NO3^-}] = 10^{-7}\text{ mol L}^{-1} \times 10^3\text{ L m}^{-3} = 10^{-4}\text{ mol m}^{-3}$$ Step 5 - Calculate the Specific Conductance ($\kappa$) and Express in the Required Units According to Kohlrausch's law of independent migration of ions, the total specific conductance ($\kappa$) of the solution is the sum of the individual conductances of all ions present: $$\kappa = \lambda^\circ(\ce{Ag^+})[\ce{Ag^+}] + \lambda^\circ(\ce{Br^-})[\ce{Br^-}] + \lambda^\circ(\ce{NO3^-})[\ce{NO3^-}]$$ Substitute the given values of $\lambda^\circ$ and our converted concentrations: $$\kappa = \left(6 \times 10^{-3} \times 4 \times 10^{-4}\right) + \left(8 \times 10^{-3} \times 3 \times 10^{-4}\right) + \left(7 \times 10^{-3} \times 10^{-4}\right)$$ $$\kappa = \left(24 \times 10^{-7}\right) + \left(24 \times 10^{-7}\right) + \left(7 \times 10^{-7}\right)\text{ S m}^{-1}$$ $$\kappa = (24 + 24 + 7) \times 10^{-7}\text{ S m}^{-1}$$ $$\kappa = 55 \times 10^{-7}\text{ S m}^{-1}$$ The question asks for the value of conductivity in terms of $10^{-7}\text{ S m}^{-1}$ units, which is $55$. Formatting as a four-digit integer: $$\boxed{0055}$$