pH calculations strong acids complete dissociation HA  H+ + A- What is the pH of a 0.040 M solution of HClO4? HClO4  H+ + ClO4- [H+] = [ClO4-] = 0.040 M pH = - log [H+] pH = 1.40 [H+] from H2O = 1.0 x 10-7 M ignore autoionization if [HA] > 10-7 M

pH calculations strong bases complete dissociation What is the pH of a 0.011 M solution of Ca(OH)2? Ca(OH)2 + H2O  Ca2+ (aq) + 2(OH-) (aq) [OH-] = 2 [Ca2+] = 0.022 M pOH = - log [OH-] pH + pOH = 14.00 pOH = 1.67 pH = 12.33 Na2O + H2O  2 Na+ (aq) + 2 OH- (aq)

pH calculations weak acids incomplete dissociation HA  H+ + A- Ka = [H+]eq [A-]eq = x2 [HA]i - x [HA]eq [HA]eq = [HA]i - x  x2 Ka [HA]i Ka very small assume x << [HA]i if x approximation O.K. < 5% [HA]i assume H+ from HA >> H+ from H2O (1 x 10-7 M)

pH calculations weak acids incomplete dissociation What is the Ka of a 0.10 M HCOOH, pH = 2.38 HCOOH  H+ + HCOO- [H+]eq = 10-2.38 = 4.2 x 10-3 [H+] [HCOO-] Ka = = x2 = (4.2 x 10-3)2 =1.8 x 10-4 0.10 [HCOOH] 0.10 - x [HA] (M) [H+] (M) [A-] (M) 4.2x10-3 Initial 0.10 0.00 0.00 0.10 -x +x +x Change 4.2 % Equil. 0.10 - x x x % ionization

successive iterations x = 7.91 x 10-3 pH = 2.10 Calculate the pH of a 0.100 M HF solution Ka = 6.8 x 10-4 = [H+] [F-] HF  H+ + F- [HF] [HF] [H+] [F-] 6.8 x 10-4 = x2 I 0.100 0.00 0.00 0.100 - x C -x +x +x x = [H+] = 8.24 x 10-3 E 0.100 - x x x pH = 2.08 0.00824 x 100 = 8.25 % 5% assumption not good 0.100 solve quadratic or successive iterations .00790 .00791 x = 7.91 x 10-3 pH = 2.10 .00791

Calculate the pH of a 0.100 M HF solution pH = 2.10 % ionization = [H+] x 100 = 7.91 x 10-3 x 100 = 7.91 % [HF] 0.10 Calculate the pH of a 0.010 M HF solution pH = 2.64 [H+] = 2.3 x 10-3 % ionization = 2.3 x 10-3 x 100 = 23% 0.010 HF (aq) H+ (aq) + F- (aq)  1 mol solute 2 mol solute increase concentration

Find the pH of a 0.0037 M solution of H2CO3 H2CO3  H+ + HCO3- Ka1 = 4.3 x 10-7 HCO3-  H+ + CO32- Ka2 = 5.6 x 10-11 4.3 x 10-7 = x2 x = 4.0 x 10-5 x 100 = 1.10 % 3.7 x 10-3 3.7x10-3 pH = 4.40 [H2CO3] [H+] [HCO3-] I 3.7 x 10-3 0.00 0.00 [CO32-] = C -x +x +x E 3.7x10-3-x x x 3.7 x 10-3 4.0 x 10-5 4.0 x 10-5

Find the pH of a 0.0037 M solution of H2CO3 H2CO3  H+ + HCO3- Ka1 = 4.3 x 10-7 HCO3-  H+ + CO32- Ka2 = 5.6 x 10-11 5.6 x 10-11= (4.0 x 10-5 + y) (y) y = 5.6 x 10-11 = [CO32-] 4.0 x 10-5 - y [HCO3-] [H+] [CO32-] % ionization I 4.0 x 10-5 4.0 x 10-5 0.00 1.4 x 10-4 % C -y +y +y E 4.0 x 10-5-y 4.0 x 10-5 +y y

pH calculations weak bases incomplete dissociation What is the [OH-] of a 0.15 M solution of NH3 Kb = 1.8 x 10-5 NH3 + H2O  NH4+ + OH- [NH4+] [OH-] Kb = = x2 = 1.8 x 10-5 [NH3] 0.15 - x x = 1.64 x 10-3 = [OH-] pOH = 2.79 pH = 11.21 [NH3] (M) [NH4+] (M) [OH-] (M) Initial 0.15 0.00 0.00 1.64x10-3 -x +x +x Change 0.15 Equil. 0.15 - x x x 1.1 %