A0.20 m solution of a weak acid has a ph of 5.40. what is the ka for the acid?

The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <> H3O+  +   A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation  (the acid form of water)

How do you calculate Ka??  Ka is the acid equilibrium constant.

( [H3O+]  . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

 HA + H2O <> H3O+  +   A-

initial- 0.2 M             I don't have H3O+, either A-

reaction - an specific amount reacted  (X)

in equilibrium (0,2 - X)  <>  X   +  X

And now, how's the formula for Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, s small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11