A conductivity cell of cell constant 0.75 cm^-1 filled with a 0.020 M solution of a weak monobasic acid HA shows a resistance of 2500 Omega. The limiting molar conductivity of HA is 375 S cm^2 mol^-1. Using K_a = (calpha^2)/(1-alpha), the pK_a of the acid (two decimals) is

JEE Advanced levelNumerical answerDifficulty 3 of 3ec204

A conductivity cell of cell constant $0.75 c m^{- 1}$ filled with a $0.020 M$ solution of a weak monobasic acid HA shows a resistance of $2500 Ω$ . The limiting molar conductivity of HA is $375 S c m^{2} mo l^{- 1}$ . Using $K_{a} = \frac{c α ^{2}}{1 - α}$ , the $p K_{a}$ of the acid (two decimals) is

Numerical answer type, tolerance 0.02.

κ = 0.75/2500 = 3.0 \times 1 0^{- 4} S c m^{- 1}

and

Λ_{m} = 1000 \times 3.0 \times 1 0^{- 4} /0.020 = 15.0 S c m^{2} mo l^{- 1}

α = 15.0/375 = 0.040

K_{a} = \frac{0.020 \times ( 0.040 ) ^{2}}{1 - 0.040} = \frac{3.2 \times 1 0 ^{- 5}}{0.960} = 3.33 \times 1 0^{- 5}

p K_{a} = - lo g (3.33 \times 1 0^{- 5}) = 4.48