Question

\( \Lambda_{m}^{\infty}( \) weak mono basic HA acid \( )=390.7 S cm ^{2} mol ^{-1} \) \( \Lambda_{m} \) of \( HA \) at \( 0.01 M \) is \( 3.907 S cm ^{2} mol ^{-1} \) Find \( pH \) of \( 0.01 MHA \) ? (1) 3 (2) 4 (3) 5 (4) 6

Answer 1

Correct option is (2) 4

Degree of dissociation of HA (α) = \(\frac{∧m}{∧^{\infty} m}\)

 = \(\frac{3.907 scm^2mol^{-1}}{390.7 scm^2mol^{-1}}\)

 = 10^-2

α = 0.01

\(\therefore\) [H⁺] = Cα

 = 0.01 x 0.01 = 10^-4M

\(\therefore\) P^H = -log[H⁺] = -log10^-4

P^H = 4