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Previously in Chem104: plant pigments do acid/base chemistry it’s just equilibrium new names: K a , K b for same K expressions. Today in Chem104: the concept of K w the concept of the K w circle p-functions (pH, pK a , pK w ) pH scale P-functions in calcs.
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Previously in Chem104: • plant pigments do acid/base chemistry • it’s just equilibrium • new names: Ka, Kb for same K expressions • Today in Chem104: • the concept of Kw • the concept of the Kw circle • p-functions (pH, pKa, pKw) • pH scale • P-functions in calcs
Dual Personality of Water: both an acid and a base Amphoteric (or amphiprotic) The Autoionization Reaction of water. 2 H20 H3O+ + OH- Keq =
Because Kw = [H3O+][OH-] = 10-14 and Kw is a constant: then the solution is neutral when [H3O+] = [OH-] = 10-7 M [H3O+] [OH-] Kw
But, if [H3O+] > [OH-], [H3O+] > 10-7 M, the solution is acidic [H3O+] > 10-7 M [OH-] < 10-7 M Kw
Or if [OH-] > [H3O+], [OH-] > 10-7 M, the solution is basic [H3O+] < 10-7 M [OH-] > 10-7 M Kw Because Kw = [H3O+] x [OH-] must be 10-14
Because all these exponential numbers are a pain: i.e. Kw = [H3O+][OH-] = 10-14 [H3O+] = [OH-] = 10-7 M (in neutral water) Ka = 10-10 we will use: The P-Function p of (a number) = -log10(a number) so “pH” means p of [H3O+], or -log10[H3O+] so the pH of neutral water, where [H3O+] = 10-7 M is pH 7
The P-Function and K’s p of (Keq) = -log10Keq = pKeq p of (Ka) = -log10Ka = pKa p of (Kw) = -log10Kw = pKw Examples: Keq = 10-10 the pKeq = 10 Ka = 1.20 x 10-5 the pKeq = 4.92 Kw = 10-14 the pKw = 14
The P-Function simplifies exponential numbers Ka = 1.20 x 10-5 , pKeq = 4.92 [H3O+] = 4.20 x 10-5M, pH = 4.376 [OH-] = 6.66 x 10-6M, pOH = 5.176 (note number of sig figs) Examples converting from p-function: if pH = 7.47, [H3O+] = 10-7.47 M= 3.39 x 10-8M, if pKa = 4.92, Ka = 10-4.92 = 1.20 x 10-5
P-Function simplifies a large range of numbers: graphically 10 1 10-110-310-5 10-7 10-9 10-11 10-13 10-14 [H3O+], M converts to a simpler scale -1 0 1 3 5 7 9 11 13 pH Note that on a p-scale, the smaller the p-number, the larger the actual number
Working in P-Functions can simplify problems Recall Kw = [H3O+][OH-] = 10-14 Apply the P-function to each side p of Kw = p of [H3O+][OH-] = p of 10-14 -log Kw = -log ( [H3O+][OH-] )= -log 10-14 -log Kw = -log [H3O+] + ( -log [OH-] ) = -log 10-14 pKw = pH + pOH = 14
Now apply this equation: pKw = pH + pOH = 14 to this picture [H3O+] [OH-] Kw = 10-14
Now apply this equation: pKw = pH + pOH = 14 to this picture pH pOH pKw = 14
When the solution is acidic [H3O+] > 10-7 M, pH < 7 : pH is small pH < 7 pOH > 7 pKw Because pKw = pH + pOH must be 14
Fill in the blanks! pOH is _______ pKw pH is _______ When the solution is ____________ [H3O+] __10-7 M, pH ___ 7 : pH is ________