Do Now: Identify the acids and bases in the following equations: NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq) NH3(aq) + H3O+(aq) ↔ NH4(aq) + H2O(l) CH3COO-(aq) + H2O(l) ↔ CH3COOH(aq) + OH-(aq) CH3COOH(aq) + OH-(aq) ↔ CH3COO-(aq) + H2O(l)
Kw, [OH-] & [H3O+] 201 Chemistry
Kaupapa / Learning Outcomes Understand what Kw is and use it in calculations. Know how to calculate pH of acids & bases
Dissociation constant of water, Kw Pure water is in equilibrium 2H2O(l) H3O+(aq) + OH(aq) The equilibrium constant is: [H3O+] [OH-] = 1 x 10-14 Water is not included because you can’t have a concentration of a pure substance. Similarly solids are not used in the calculation of equilibrium constants
Kw If [H3O+] [OH-] = 1 x 10-14 then In pure water [H3O+] = [OH-] = 1 x 10-7
Kw, [H3O+] & [OH-] Adding an acid to water increases [H3O+] This decreases [OH-] (Le Chetaliers Principle) so that the product of the two concentrations (Kw) remains unchanged. In other words given an H3O+ or OH- concentration we can use the Kw = [H3O+] [OH-] expression to find an unknown concentration Kw = 1x 10-14
Calculating [H3O+] & [OH-] Example: [H3O+] = 2.5 x 10-3 mol L-1 find the [OH-] Rearrange Kw = [H3O+] x [OH] to find the OH- concentration [OH] = = 4.0 x 10-12 mol L-1
Complete the following table using the previous Kw formula. [H3O+] [OH] acid or base? 2.4 x 10-3 0.0175 0.0036 5.4 x 10-10 4.167 x 10-12 acid 5.71 x 10-13 acid base 2.78 x 10-12 1.85 x 10-5 acid
pH
Calculating pH pH = -log10[H3O+]
Have a Go pH = -log10[H3O+] Calculate the pH if [H3O+] = 0.0040 mol L-1 0.05 mol L-1 0.01 mol L-1 0.371 mol L-1 pH = -log10[H3O+]
Answers pH = -log10[H3O+] Calculate the pH if [H3O+] = 0.0040 mol L-1 2.40 1.30 2.00 0.431 pH = -log10[H3O+]
Calculating [H3O+] from pH [H3O+] = 10-pH
Have a go [H3O+] = 10-pH Calculate the [H3O+] if the pH = 1.08 2.97 12.5 4.67 [H3O+] = 10-pH
Answers [H3O+] = 10-pH Calculate the [H3O+] if the pH = 1.08 2.97 12.5 4.67 0.0832 molL-1 1.07 x 10-3 molL-1 3.16 x 10-13 molL-1 2.14 x 10-5 molL-1 [H3O+] = 10-pH
Calculating the pH of bases Use the [OH-] to calculate [H3O+] [OH-] x [H3O+] = 10-14 [H3O+] = 10-14 / [OH-] Then calculate pH
Have a Go Calculate the pH if [OH-] = 0.0040 mol L-1 0.05 mol L-1
Answers Calculate the pH if [OH-] = 0.0040 mol L-1 0.05 mol L-1 11.6 12.7 13 13.6