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The dissociation constant is:
k = [H][X]/[HX]

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Q: What is the acid dissociation constant for an acid at equilibrium HX H X?
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What is the acid dissociation constant for an acid at equilibrium HX - H X-?

The dissociation constant is:k = [H][X]/[HX]


What is the acid dissociation constant for an acid at equilibrium HX ----- H (spuare Plus) plus x -?

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When 15.12 ml of 0.1194 M NaOH is added to 25.00 ml of 0.1246 M HX an unknown weak acid the resulting pH is 5.9. What is the Ka of the unknown acid?

I think I'm on the right track but I'm not sure of how it is finished; Since the equation representing the dissociation of HX is as follows; HX(aq) + H2O (l) <--> H3O+(aq) + X-(aq) It can then be shown that the equilibrium constant is found as such; Ka = [H3O+][X-]/[HX] (Note: H2O is not in the equilibrium equation because it is a liquid and further, the solvent) I first found [HX] as follows; [HX] = ((0.1246 M HX) x (25.00 mL) x (1L/1000 mL))/(40.12 mL x 1000 mL/1L) [HX] = 0.07764 M HX I then found the [H3O+] by converting the pH into [H3O+], given that pH = -log[H3O+] which results in; 5.9 = -log[H3O+] [H3O+] = 1.3x10-6 M Now as to how we solve for [X-] I'm not sure if [X-] = [H3O+] or what but if anyone can tell me if I'm completely off or somewhere near where I want to be perhaps they can finish it for you or give you the right answer


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