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pH Calculations of Strong Acids and Bases: Complete Dissociation

Learn about pH calculations for strong acids like HClO4 and strong bases like Ca(OH)2, understanding dissociation, autoionization, Ka values, and percent ionization.

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pH Calculations of Strong Acids and Bases: Complete Dissociation

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

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

  3. pH calculations weak acids incomplete dissociation  = x2 HA H+ + A- Ka = [H+]eq [A-]eq [HA]i - x [HA]eq [HA]eq = [HA]i - x  x2 Ka [HA]i Ka very small assume x << [HA]i if x approximation x 100 O.K. < 5% [HA]i x x 100% = percent ionization [HA] H+ from HA H+ assume >> from H2O (1 x 10-7 M)

  4. pH calculations weak acids incomplete dissociation What is the Ka of a 0.10 M CHOOH, pH = 2.38  H+ + CHOO- [H+]eq = 10-2.38 = 4.2 x 10-3 CHOOH [H+] [CHOO-] = x2 = (4.2 x 10-3)2 =1.8 x 10-4 Ka = 0.10 [CHOOH] 0.10 - x [HA] (M) [H+] (M) [A-] (M) 4.2x10-3 x 100 0.00 0.00 Initial 0.10 0.10 -x +x +x Change 4.2 % Equil. 0.10 - x x x % ionization

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

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

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

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

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

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