If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Autoionization of water

## Want to join the conversation?

• At , how is it that Kw is used for determining [OH-] for lemon juice. Doesn't Kw apply to water? Shouldn't we use the Ka of lemon juice here?
• If you need to know pH, the Ka of lemon juice is irrelevant. It can be used to find the hydronium ion concentration, but since that's already been given, we don't need to concern ourselves with it. Kw specifically concerns the concentration of hydronium ions and hydroxide ions in an aqueous solution in a given temperature. Lemon juice is an aqueous solution, so we can use Kw
• Does Kahn Academy teach the trig function tricks for the MCAT, since we cannot use a calculator?
• When would Ka be Kw? Thanks in advance for anyone who answers!
• Ka is for an Acid. Water, being amphoteric, can react both as an acid and as a base. Kw is therefore a special case of Ka.
• Will Kw always be 1.0 X 10^-14?
• At temperatures greater than 25°C the degree of self-ionization will increase and at temperatures below it,it will decrease.
Hence Kw will be different at different temperatures.
• In the "Relationship between Ka and Kb" video, Jay says that Ka * Kb = Kw. But in this video he says that Ka = Kw at equilibrium, so does Kb = 1 at equilibrium?
• Yup! I just explained it in my answer to kristenmpitts1 under this video
(1 vote)
• How do we know to use Kw (1.0x10^-14) instead of some other Ka value when solving the lemon juice problem?
• My guess is that he skipped a few steps to make it simpler to understand. If you have an acid in an aqueous solution, to determine the pH, you first need to figure out which gives off more H+ ions - the water or the acid. It may intuitively seem as if the acid should give up more H+ than water, but some acids are very weak acids and barely give up any H+, while others are present in such low concentrations in a solution that they do not end up giving up much H+ ions. In these cases, one would use the Kw of water instead of some other Ka value, as the water is the main proton donor; this makes calculations much easier to do.
• Do you know of any reactions where there is H3O but no OH? (or the vice versa). And if there is, how do you know when to use Kw? Also, thanks so much for posting these videos, they are great!
• If whatever reactants yields H3O as one of its products, we have to use Ka values because the reaction is acting like an acid.
For example: HCl (aq) + H2O (l) → H3O (aq) + Cl (aq) : the hydrochloric acid (HCl) is a strong acid that dissociates into hydroxide (H3O) molecules and chlorine molecules. Since H3O is one of the products due to HCl donating its hydrogen proton, we use the Ka value of HCl in our equation.
Ka = (1.3 * 10^6) = [H3O][Cl] / [HCl]

Similarly, if a reaction yielded an OH-* ion, then we'd use the *Kb (base-dissociation constant) value in our equation. Keep in mind that Kb = Kw/Ka :)
• I'm confused about the autoionization constant. I thought it only described the relative concentrations of OH and H30 in water.. It can be used for other liquids like lemon juice too?
• We call it as dissociation constant(Ka) and it varies from solution to solution.
(1 vote)
• How does Ka = Kw = 1x10^-14 for water at equilibrium, if Ka X Kb = Kw? I would have thought Ka = 1x10^-7 and Kb = 1x10^-7.
• Kb is used for bases, so in this case the base is the conjugate base of the acid. A normal acidic equation would be HA -> H+ + A-, where A- is the conjugate base. To calculate the Kb, we would switch the products and reactants to A- + H2O = HA + OH- (the water comes from the fact that this is an aqueous solution).

Therefore
Ka = [H+][A-] / [HA]
Kb = [OH-][HA] / [A-]

Ka * Kb = [H+][A-] / [HA] * [OH-][HA] / [A-]
By crossing out common factors we get
Ka * Kb = [H+] * [OH-]
remember that Kw = [H+] * [OH-], so Ka*Kb = Kw

Now, just do the same for water:

acid version:
H2O + H2O -> H3O+ + OH-
Ka = [Product] / [Reactant] = [OH][H3O+]

base version:
OH- + H20 = H20 + OH-
Kb = [Product] / [Reactant] = [OH-] / [OH-] = 1

Ka * Kb = Kw
[OH][H3O+] * 1 = Kw
Ka = [OH][H3O+] = Kw
Ka = Kw

Note that H+ and H3O+ are considered interchangeable in an aqueous solution as all the H+ ions will end up binding to the H2O molecules in the solution to form H3O+.
(1 vote)
• if Ka = [H3O+][OH-], logically then Kb = [OH-][H3O+], so Ka = Kb = 10^-14. In that case, Ka x Kb = 10^28. But we know that Ka x Kb = Kw = 10^-14. I'm confused