Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 130 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If Q- happened to be equal to Z-, then the difference of the pKa values of the two weak acids (pKa(HQ) - pKa(HZ)) is (Nearest integer). [Given: degree of dissociation ( ) 1 for both weak acids, : limiting molar conductivity of ions]
Answer
2
Solution
m(HQ) = H+ + Q- = H+ + Z- = m(HZ)
Since = m m and the m values are equal,
HQ HZ = m(HQ) m(HZ) = 130
For 1: Ka = c 2
⇒ Ka(HQ)Ka(HZ) = 0.180.02 ( 130 )2 = 9900 = 1100
∴ pKa(HQ) - pKa(HZ) = - Ka(HQ)Ka(HZ) = - 10-2 = 2
Answer: 2