# Acid-Base Equilibria - Ka constant

• ctamasi

#### ctamasi

Ok, I'm having a hard time understanding the concept behind finding the dissociation constant. The question is:

A 0.1 mol/L aqueous solution of weak monoprotic acid has a hydrogen ion concentration of 0.001 mol/L. The value of the ionization, Ka, for this acid is:

a) 10-6
b) 10-2
c) 10-3
d) 10-5

Ok, I'm not necessarily interested in the answer, I just need a really good explanation of how to go about solving this question.

I'm assuming that I would first write out a balanced equation, however, the question is asking for the dissociation constant, therefore I'm not sure if the equation should be:

HX + H2O --> H3O + X

or

HX --> H+ + X- <-- my bet is on this equation because it says nothing about reacting with water.

The HX represents the weak acid, and X represents its anion.

The dissociation constant can be found as:

Ka = [H+][X-] / [HX]

However, this is where I get lost. If I had to go further, I would substitute the concentrations given, into the equation:

Ka = (0.001 mol/L)(X) / (0.1 mol/L)

The balanced equation shows that [H+] = [X-] = 1:1

Therefore, Ka = (0.001 mol/L)(0.001 mol/L) / (0.1 mol/L) = 10-5

This last little bit of the questions was somewhat of a guess, but some guidance would be greatly appreciated! Thanks in advance.