The electrical resistance of a 0.05 ~M solution of a weak acid HA is 2000 ~ . The conductivity cell used has electrodes of surface area 2.0 ~cm ² separated by a distance of 1.6 ~cm . The limiting molar conductivities of NaA , HCl , and NaCl are 90 , 426 , and 116 ~S ~cm ² ~mol ⁻¹ , respectively. If the acid dissociation constant K_ a of the weak acid is x 10⁻⁵ , the value of x is _ _ _ _ . (Assume the degree of disso

Step-by-step solution

Given: Concentration C = 0.05 ~M Resistance R = 2000 ~ Distance between electrodes l = 1.6 ~cm Area of electrodes A = 2.0 ~cm ² First, calculate the cell constant G^ * : G^ * = l A = 1.6 2.0 = 0.8 ~cm ⁻¹ Calculate the specific conductivity : = G^ * R = 0.8 2000 = 4 10⁻⁴ ~S ~cm ⁻¹ Calculate the molar conductivity _ m : _ m = 1000 C = 4 10⁻⁴ 1000 0.05 = 8 ~S ~cm ² ~mol ⁻¹ Using Kohlrausch's law, find the limiting molar conductivity of HA : ^ _ m ( HA ) = ^ _ m ( NaA ) + ^ _ m ( HCl ) - ^ _ m ( NaCl ) ^ _ m ( HA ) = 9