Question

Calculate the pH at 25°C of a 0.31M solution of potassium acetate KCH3CO2 . Note that acetic acid HCH3CO2 is a weak acid with a pKa of 4.76 . Round your answer to 1 decimal place.

Answer 1

Ans :- pH = 9.24

Explanation :-

ICE table is :

...........................CH₃CO₂^-(aq).............. +.......... H₂O ----------> HCH₃CO₂ (aq)............. +............ OH^- (aq)

Initial (I)................0.31 M................................................................0.0 M........................................0.0 M

CHange (C).............-y........................................................................+y..............................................+y

Equilibrium (E)......(0.31-y) M...............................................................y M..........................................y M

Pk_a = 4.76

So,

-logK_a  = 4.76

K_a= 10^-4.76  = 1.8 x 10^-5  

and , k_b= K_w/K_a = 1x10^-14/1.8 x 10^-5   = 5.6 x 10^-10

K_b   = [HCH₃CO₂][OH^-]/[CH₃CO₂^-]

5.6 x 10^-10 = y² /(0.31 - y)

As y <<<0.31, So neglect y as compare to 0.31 M.

y² = (5.6 x 10^-10)(0.31)

y = 1.74 x 10^-5

So, [OH^-] = y = 1.74 x 10^-5 M

pOH   = -log[OH^-]

= - log 1.74 x 10^-5 M

= 4.76

So,

pH = 14 - pOH

= 14 - 4.76

= 9.24

Hence, pH = 9.24