Question

If an acid is dissolved in water to make a 0.600 M solution, and the resulting pOH is 12.6, calculate the K_a for the acid.

Answer 1

pH + pOH = 14

Resulting pOH is 12.6

So, PH = 14 – 12.6 = 1.4

Write the dissociation equation for the acid-

HA ⇌ H+ + A¯

Now write the equilibrium expression-

Ka = ( [H+] [A¯] ) / [HA]

Use the pH to calculate the [H+].

Therefore [H+] = 10¯pH

[H+] = 10^-1.4

       = 3.981 x 10^-2 M

From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [A¯].

Therefore,

[A¯] = 3.981 x 10^-2 M

HA          ⇌         H+      +               A¯

I 0.600M             0                          0

C - 0.03981         + 0.03981           +0.03981

E 0.600 -0.03981 + 0.03981         +0.03981

Ka = ( [H+] [A¯] ) / [HA]

Ka = [3.981 x 10¯2 M ] [3.981 x 10¯2 M]/ 0.600 – 0.03981

1.58489 x 10^-3/0.56019

Ka = 2.8292 x 10^-3

Ka = 2.8 x 10^-3