A conductivity cell is filled with a 0.09 M solution of a weak acid. The measured resistance of the solution is 100 and the cell constant is 0.36 cm⁻¹ . If the limiting molar conductivity of the weak acid is 400 S cm^2 mol⁻¹ , its acid dissociation constant K_a is x 10⁻⁵ . The value of x is ___________.

Step-by-step solution

First, calculate the specific conductance ( ) of the solution using the cell constant ( G^* ) and resistance ( R ): = G^* R = 0.36 100 = 3.6 10⁻³ S cm⁻¹ Next, find the molar conductivity ( _m ): _m = 1000 C = 1000 3.6 10⁻³ 0.09 = 3.6 0.09 = 40 S cm^2 mol⁻¹ Calculate the degree of dissociation ( ): = _m _m^ = 40 400 = 0.1 Finally, determine the acid dissociation constant ( K_a ): K_a = C ^2 1- = 0.09 (0.1)^2 1 - 0.1 = 0.0009 0.9 = 0.001 K_a = 100 10⁻⁵ Thus, the value of x is 100 . Answer: 100