A 0.01 ~M solution of a weak monoprotic acid is placed in a conductivity cell with a cell constant of 0.4 ~cm ⁻¹ . The electrical resistance of the solution is measured to be 10000 ~ . If the limiting molar conductivity of the weak acid is 400 ~S ~cm ² ~mol ⁻¹ , the pH of the solution is _ _ _ _ .

Step-by-step solution

Given: Concentration C = 0.01 ~M Cell constant G^ * = 0.4 ~cm ⁻¹ Resistance R = 10000 ~ Limiting molar conductivity ^ _ m = 400 ~S ~cm ² ~mol ⁻¹ First, calculate the specific conductivity : = G^ * R = 0.4 10000 = 4 10⁻⁵ ~S ~cm ⁻¹ Next, calculate the molar conductivity _ m : _ m = 1000 C = 4 10⁻⁵ 1000 0.01 = 4 ~S ~cm ² ~mol ⁻¹ Now, find the degree of dissociation : = _ m ^ _ m = 4 400 = 0.01 Calculate the hydrogen ion concentration [H⁺] : [H⁺] = C = 0.01 0.01 = 10⁻⁴ ~M Finally, determine the pH : pH = - ₁₀[H⁺] = - ₁