Question

2. a. Calculate the pH of a 0.6 M solution of HN₃ with K_a = 1.9

Answer 1

Kb = Kw (1.0 * 10^-14 ) / Ka (1.9 * 10^-5)

= 5.26 * 10^-10

NH3 + H2O ----------> NH₄⁺  + OH^-

0.6 - - I

-x +x +x C

(0.6-x) -x +x +x E

Ka = 5.26 * 10^-10   = x²

(0.6-x)

Ignoring x in thew denominator as "x" represents moles of weak base which hardly dissociate

x = 5.617 * 10^-5 = moles of OH^- ions

so pOH = - log ( 5.617 * 10^-5 )

= 4.25

pH = 14- 4.25

= 9.7

b.

% dissociated = x / 0.6

= (5.617 * 10^-5 / 0.6) * 100

= 0.0094 %