Dissociation constants?

Question

I just need help with three problems.

1. A 0.75M solution has a pH of 2.0. Find Ka.

2. A 0.55M solution of a new acid, HX, has an equilibrium [H+]=4.1*10^-2. What is the value of Ka?

3. A base has a Kb=5.5*10^-12, if the concentration of a base is 0.36M, what is the equilibrium value of the hydroxide ion concentration?

I just really need to know what to do in these types of problems.

Answer 1

In the first 2 problems, you'll be using the acid dissociation reaction for a generic weak acid:

HA + H2O <--> A- + H3O+

For this reaction, the equilibrium constant expression, Ka is:

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

When you begin with a pure weak acid in water, you need to recognize that [H3O+] = [A-], since whenever you for an A- ion, you also form a hydronium ion.

So, if you let [A-] = x, the expression for Ka reduces to:

Ka = x^2 / ([HA]-x)

If the acid is quite weak, then you can usually ignore x when compared to [HA], and this is easy to solve.

So, in your first problem, you are give the pH= 2, so [H3O+] = 0.01. So, you have everything you need to calculate the Ka.

In your second problem, you can do exactly the same thing. You might be able here, too, to ignore the x compared to [HX], but I might think about including that term in the denominator.

For your third problem, just begin with the base dissociation equation, and approach it the same way.

Send me a message through here if this doesn't help.