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.