JEE Challenger
More from Electrochemistry
ChemHardAdv 2017 • Paper 1

Limiting Molar Conductivity of Monobasic Acid Solution

The measured conductance of a 0.0015 M0.0015\text{ M} aqueous solution of a weak monobasic acid was determined using a conductivity cell equipped with platinized Pt electrodes. The electrode separation distance is 120 cm120\text{ cm} with a cross-sectional area of 1 cm21\text{ cm}^2. The observed conductance of this solution is 5×10−7 S5 \times 10^{-7}\text{ S}. Given that the pH of the solution equals 44, the limiting molar conductivity (Λm0\Lambda_m^0) of this weak monobasic acid in aqueous solution is Z×102 S cm2 mol−1Z \times 10^2\text{ S cm}^2\text{ mol}^{-1}. Calculate the value of ZZ.

Official Numerical Answer6

Topics & Concepts

ElectrochemistryElectrochemical CellAcid-Base Equilibrium

Step-by-Step Solution

To find the value of ZZ, we can break down the problem into the following logical steps:

Step 1: Calculate the cell constant (G∗G^*) The cell constant is defined as the ratio of the electrode separation distance (ll) to the cross-sectional area (AA): G∗=lA=120 cm1 cm2=120 cm−1G^* = \frac{l}{A} = \frac{120\text{ cm}}{1\text{ cm}^2} = 120\text{ cm}^{-1}

Step 2: Calculate the conductivity (κ\kappa) of the solution Conductivity is the product of observed conductance (GG) and the cell constant (G∗G^*): κ=G×G∗=(5×10−7 S)×(120 cm−1)=6×10−5 S cm−1\kappa = G \times G^* = (5 \times 10^{-7}\text{ S}) \times (120\text{ cm}^{-1}) = 6 \times 10^{-5}\text{ S cm}^{-1}

Step 3: Calculate the molar conductivity (Λm\Lambda_m) Given the concentration of the monobasic acid solution C=0.0015 M=1.5×10−3 mol L−1C = 0.0015\text{ M} = 1.5 \times 10^{-3}\text{ mol L}^{-1}: Λm=κ×1000C=6×10−5 S cm−1×1000 cm3 L−11.5×10−3 mol L−1=40 S cm2 mol−1\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-5}\text{ S cm}^{-1} \times 1000\text{ cm}^3\text{ L}^{-1}}{1.5 \times 10^{-3}\text{ mol L}^{-1}} = 40\text{ S cm}^2\text{ mol}^{-1}

Step 4: Determine the degree of dissociation (α\alpha) The pH\text{pH} of the solution is given as 44, which means: [H+]=10−pH=10−4 M[\text{H}^+] = 10^{-\text{pH}} = 10^{-4}\text{ M}

For a weak monobasic acid HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-, the hydrogen ion concentration is related to the degree of dissociation (α\alpha) by: [H+]=Cα[\text{H}^+] = C\alpha 10−4=(1.5×10−3)α  ⟹  α=10−41.5×10−3=11510^{-4} = (1.5 \times 10^{-3})\alpha \implies \alpha = \frac{10^{-4}}{1.5 \times 10^{-3}} = \frac{1}{15}

Step 5: Calculate the limiting molar conductivity (Λm0\Lambda_m^0) Using Arrhenius' relation for degree of dissociation: α=ΛmΛm0\alpha = \frac{\Lambda_m}{\Lambda_m^0} Λm0=Λmα=40115=600 S cm2 mol−1\Lambda_m^0 = \frac{\Lambda_m}{\alpha} = \frac{40}{\frac{1}{15}} = 600\text{ S cm}^2\text{ mol}^{-1}

Equating this to the given expression Z×102 S cm2 mol−1Z \times 10^2\text{ S cm}^2\text{ mol}^{-1}: Z×102=600  ⟹  Z=6Z \times 10^2 = 600 \implies Z = 6

PreviousNext