Calculating Ka of an acid

Question

Calculate Ka of an acid, HA, whose 0.150 M solution has a pH of 4.35?

Answer 1

Ka describes the ratio of products to reactants when an acid dissociates. When the HA molecule splits into A- and H+ (that joins to water molecules to form H3O+), the equation looks like this:

HA + H2O --> A- + H3O+

where HA is a reactant (water is never included in the equation, so it's ignored) and A- & H3O+ are the products.

The Ka equation would look like this: Ka = ([A-]*[H3O+]) / [HA]

where brackets "[ ]" are used to show "concentration of." We need to find the concentrations of H3O+, A-, and HA (already given to us, 0.150M) and plug them into the equation to find Ka.

We have the pH, which can be used to find the concentration of H3O+, using the formula:

[H3O+] = 10^(-pH)

So in this case, [H3O+] = 10^-4.35 = 4.467x10^5 M

From the chemical equation, HA + H2O --> A- + H3O+, we can see that for every H3O+ produced there is also one A- produced; in other words, there's a 1:1 ratio between them. So if 4.467x10^5 M of H3O+ is present, that means 4.467x10^5 M of A- is present as well.

We have our three concentrations:

[H3O+] = 4.467x10^5 M

[A-] = 4.467x10^5 M

[HA] = 0.150 M (from the problem)

Plug them into the Ka equation:

Ka = ([A-]*[H3O+]) / [HA] = ((4.467x10^5 M)*(4.467x10^5 M))/0.150 = 1.330x10^-8

So the Ka of the acid is 1.330x10^-8 (no units).