Question

HA is a weak acid. Its ionization constant, Ka, is 2.4 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.046 M.

Answer 1

*NaA is a salt of strong base and weak acid which dissociate completely, A^- ,the conjugate of weak acid undergo hydrolysis (ANIONIC HYDROLYSIS) to make the solution ALKALINE (pH > 7 at 25°C).

* Let 'h' is the degree of hydrolysis, K_H is the hydrolysis constant. Then

NaA ---------> Na⁺ + A^-

  A^- + H₂O $\rightleftharpoons$    HA + OH^-

0.046(1- h) ------ 0.046h 0.046h

K_H = [HA][OH^-]/[A^-]    [ K_H = K_w/K_a

=> (0.046h)²/[0.046(1-h)]        [= 10^-14/(2.4 ×10^-13)

  => 0.046h²/(1-h) = 0.0417     [   = 0.0417  

=> h²/(1-h) =0.0417/0.046 = 0.906

=> h = 0.6011

[OH^-] = 0.046h = 0.046 × 0.6011

= 0.0276M

pOH = -log(0.0276) = 1.56

pH = 14 - 1.56 = 12.44