Question

Calculate the pH of an aqueous solution of 0.225 M salicylic acid, which has a Ka =1.0 x 10^-3

An explanation will be appreciated, a simple answer won't really help. Thank you!

Answer 1

Given :

Molarity of salicylic acid = 0.225 M , ka = 1.0E-3

Lets assume HA is for salicylic acid

Its ka shows that the acid is weak

ICE :

    Reaction :

               HA (aq) + H2O (l) --- > H3O+ (aq) + A- (aq)

I            0.225                                  0              0

C          -x                                 +x                    +x

E          (0.225-x)                     x                      x

ka = [H3O+][A-]/[HA]

1. E-3 = x² / (0.225-x)

Lets simplify the equation

1.0E-3 (0.225-x) = x²

2.25E-4 – 1.0 E-3 x = x²

We got quadratic equation

x² + 1.0 E-3 x – 2.25 E-4 = 0

after solving this equation we get roots of x

x = 0.0145

x= [H3O+]= 0.0145 M

pH = -log [H3O+] = -log ( 0.0145 )

= 1.83

pH = 1.83