The limiting molar conductivities of NaI, NaNO and AgNO are 12.7, 12.0 and 13.3 S m mol , respective — Electrochemistry Chemistry Question
Question
The limiting molar conductivities of NaI, NaNO and AgNO are 12.7, 12.0 and 13.3 S m mol , respectively (all at 25 C). The limiting molar conductivity of AgI at this temperature is ______ S m mol . 3 3 2 –1 o 2 –1
💡 Solution & Explanation
**Step 1: Identify the key concept** Use Kohlrausch's Law of Independent Migration of Ions: The limiting molar conductivity of a salt equals the sum of the limiting molar conductivities of its constituent ions. For any salt: Λ°(salt) = λ°(cation) + λ°(anion) **Step 2: Write equations for the given salts** - NaI: Λ°(NaI) = λ°(Na⁺) + λ°(I⁻) = 12.7 - NaNO₃: Λ°(NaNO₃) = λ°(Na⁺) + λ°(NO₃⁻) = 12.0 - AgNO₃: Λ°(AgNO₃) = λ°(Ag⁺) + λ°(NO₃⁻) = 13.3 **Step 3: Find λ°(Ag⁺) and λ°(I⁻)** From equations (2) and (3): Λ°(AgNO₃) - Λ°(NaNO₃) = λ°(Ag⁺) - λ°(Na⁺) 13.3 - 12.0 = 1.3 Therefore: λ°(Ag⁺) - λ°(Na⁺) = 1.3 **Step 4: Calculate λ°(I⁻)** From equation (1): λ°(I⁻) = 12.7 - λ°(Na⁺) **Step 5: Calculate Λ°(AgI)** Λ°(AgI) = λ°(Ag⁺) + λ°(I⁻) = [λ°(Na⁺) + 1.3] + [12.7 - λ°(Na⁺)] = 12.7 + 1.3 = 14.0 Therefore, the answer is 14.