Question

Calculate the pH in 0.050 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5×10−5.

Answer 1

Ans :- pH = 8.44

Explanation :-

Given ,

K_a of benzoic acid i.e. C₆H₅CO₂H = 6.5 x 10^-5

So,

K_b of C₆H₅CO₂^-= 1.0 x 10^-14 / K_a = 1.0 x 10^-14 / 6.5 x 10^-5 = 1.538 x 10^-10

ICE table of C₆H₅CO₂^- is :

...........................C₆H₅CO₂^- (aq).........+...........H₂O (l) <----------------> C₆H₅CO₂H (aq)..........+.......OH^- (aq)

Initial (I)................0.050 M........................................................................0.0 M.................................0.0 M

Change (C)...........-y..................................................................................+y.........................................+y

Equilibrium (E).....(0.050-y) M.....................................................................y M.......................................y M

Expression of K_b is :

K_b = [C₆H₅CO₂H ].[OH^-] / [C₆H₅CO₂^- ]

1.538 x 10^-10 = y² / (0.050-y)

Let y <<<0.050, So neglect y as compare to 0.050 , we have

1.538 x 10^-10 = y²/0.050

y² = 7.69 x 10^-12

y = 2.77 x 10^-6

So, [OH^-] = y = 2.77 x 10^-6 M

We know,

pOH = -log [OH^-]

pOH = - log 2.77 x 10^-6

pOH = 5.56

Also, pH = 14 - pOH

So,

pH = 14 - 5.56

pH = 8.44

Hence, pH of sodium benzoate = 8.44