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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, K

_{a}, 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 + H

_{2}O --> H

_{3}O + 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:

K

_{a}= [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:

K

_{a}= (0.001 mol/L)(X) / (0.1 mol/L)

The balanced equation shows that [H

^{+}] = [X

^{-}] = 1:1

Therefore, K

_{a}= (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.