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AUTOIONIZATION of WATER H2O(l) + H2O(l) <==> H3O+(aq) + OH¯(aq) Kw = [H3O+] [OH¯] Kw = water autoionization constant=1.0 x 10-14 @ 25 °C From the chemical equation just above, it can be seen that H3O+ and OH¯ concentrations are in the molar ratio of one-to-one. This means that [H3O+] = [OH¯]. Therefore the values of [H3O+] and [OH¯] can be determined by taking the square root of Kw. Hence, both [H3O+] and [OH¯] equal 1.00 x 10¯7 M in pure water. Kw = [H3O+] [OH¯] 1.0 x 10-14= [x][x] x=1.00 x 10¯7 M Result #1: The pH of pure water is 7 By definition, pH = -log [H3O+] The pH of pure water then equals -log 10¯7, which is 7. At other temperatures neutrality changes:
Result #2: If the pH or the pOH is known, the other can be found. Take the negative logarithm of each side of the Kw equation as follows: - log Kw = -log [H3O+] + -log [OH¯] -log 1.00 x 10¯14 = -log [H3O+] + -log [OH¯] Note the use of the add sign on the right side of the equation. The result is usually written as: pKw = pH + pOH = 14 This is an extremely important equation. Learn it well. Result #3: If the [H3O+] or the [OH¯] is known, the other can be found. Simply divide Kw by the known value to get the other. Suppose [H3O+] is known, then: [OH¯] = Kw / [H3O+] Suppose [OH¯] is known, then: [H3O+] = Kw / [OH¯] Result #4: If one variable ( [H3O+] or [OH¯] ) changes value (either up or down), the other variable will change in the opposite direction. The change in values will still preserve this fundamental equality: Kw = [H3O+] [OH¯] Suppose [H3O+] became larger, therefore the [OH¯] becomes smaller. Back to Acid Base Links |
