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Acid Base Follow-up

Acid Base Follow-up. pH. There are four main relationships that you need to be able to use for pH. pH + pOH = 14 pH = -log[H 3 O + ] and pOH = -log[OH - ] 10 -pH = [H 3 O + ] and 10 -pOH = [OH - ] 1x10 -14 = [H 3 O + ] [OH - ]

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Acid Base Follow-up

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  1. Acid Base Follow-up

  2. pH • There are four main relationships that you need to be able to use for pH. • pH + pOH = 14 • pH = -log[H3O+] and pOH = -log[OH-] • 10-pH = [H3O+] and 10-pOH = [OH-] • 1x10-14 = [H3O+] [OH-] • It is also important to understand that the concentrations of the hydronium and hydroxide ions give information about the amount of acid in solution.

  3. Try these: • What is the hydrogen ion concentration of a solution with the pOH of 6.43? • (by the way, a pOH of 6.43 has two significant figures, why?) • What is the pH of a 0.288M calcium hydroxide solution? • What is the resulting pH of a solution produced from the reaction between 25.0mL of a hydrochloric acid solution with a pH of 3.25 and 32.7mL of a sodium hydroxide solution with a pOH of 2.32?

  4. Solutions • pH + pOH = 14 14 - 6.43 = 7.57 = pH • 10-pH = [H3O+] 10-7.57 = 2.7x10-8M = [H3O+] • 0.288M Ca(OH)2 = 0.576M OH- • 1x10-14 = [H3O+] [OH-] • 1x10-14 = [H3O+] 0.576M = 1.73x10-14M • pH = -log[H3O+] = -log1.73x10-14 • pH = 13.76

  5. Last One! • pH = 3.25 [H3O+] = 5.62x10-4M • 5.6x10-4Mx0.025L = 1.41x10-5moles H+ • pOH = 2.32 [OH-] = 4.79x10-3M • 4.79x10-3Mx0.0327L = 1.57x10-4moles OH- • Reaction is HCl + NaOH → NaCl + H2O • So 1.41x10-5 moles of H+ will neutralize the same amount of OH- resulting in 1.43x10-4 moles of OH- leftover. • [OH-] = 2.48x10-3M • pOH is 2.61 so pH is 11.39

  6. Relating Ka to pH • When you set up the equilibrium expression for the dissolving of an acid, one of your products is the H3O+. • You can use pH to determine that concentration and then use that to determine the other concentrations. • Ka = [H3O+][A-] [HA]

  7. Question • What is the pH of a 0.15M solution of acetic acid with a Ka of 1.8x10-5?

  8. Answer • HC2H3O2 + H2O ↔ H3O+ + C2H3O2- • ICE • The –x can be ignored for the change of acetic acid concentration since the Ka is so much smaller than 0.15M

  9. Sooooo…… • 1.8x10-5 = [x][x] 0.15 • 2.7x10-6 = x2 • X = [H+] = 1.6x10-3M • pH = -log [H+] = 2.78

  10. pH of Weak Acid Mixture • Calculate the pH of a mixture of 1.00 M HCN, 5.00 M HNO2 and the equilibrium concentration of [CN-1] • HCN Ka = 6.2 x 10-10 • HNO2Ka = 4.0 x 10-4 • H20 Kw = 1 x 10-14 How do you approach this?

  11. Calculate pH • Ka = 4.0 x 10-4 = [H+][NO2-] [HNO2] HNO2 H+ + NO2- I 5.00 0 0 C -x +x +x E 5.00 – x x x

  12. Calculate pH • Ka = 4.0 x 10-4 = x2 = x2 5.00 - x 5 x = [H+] = 4.5 x 10-2 M pH = - log [H+] = 1.35 Now calculate [CN-], you know [HCN] and [H+]

  13. Calculate [CN-] • Ka = 6.2 x 10 –10 6.2 x 10 –10 = [CN-][H+] = [CN-][4.5 x 10-2] [HCN] 1.00 Solve for [CN-] = 1.4 x 10-8 M

  14. Percent Dissociation • % dissociation = [amount disassociated] x 100 [initial concentration] In the HF example [H+] = 1.27 x 10-2 M x 100 = 1.27% [HF] 1.00 M

  15. 1.00 M HC2H3O2 Left side of room Write on board 0.100 M HC2H3O2 Right side of room Write on board Calculate the Percent Dissociation

  16. Percent Dissociation • For solutions of weak acids: • the more dilute the solution • The greater the percent dissociation General Proof Suppose you have acid HA, with [HA]0 Dilute it to 1/10 th initial concentration Q = (x/10)(x/10) = x2 = 1/10 Ka [HA]/10 10 [HA] Since Q < Ka, the reaction moves to the right And you get a greater percent dissociation

  17. [OH-] [HBz] [Bz-] Weak base equilibria Example The Bz-(aq) formed in the benzoic acid solution is a weak conjugate base. Determine the pH of a 0.10 M sodium benzoate solution NaBz(aq), at 25 oC. The Kaof Benzoic Acid is 6.5x10-5. Bz-(aq) + H2O(l) HBz(aq) + OH-(aq) Kb =

  18. Relationship Between Ka and Kb • What happens when you multiply Ka and Kb together? • Ka x Kb = [H+][OH-] = Kw = 1.0 x 10-14 • And, just like pH + pOH = 14.00 for strong acids/bases at standard temperature… • pKa + pKb = pKw = 14.00

  19. Weak base equilibria Example The Kb value is related to the Ka value by the equation Kax Kb = Kw = 1.0 x 10-14 Proof: [H+] [Bz-] [OH-] [HBz] = [H+] [OH-] [HBz] [Bz-] Kb= Kw /Ka= 1.0 x 10-14 / 6.5 x 10-5 = 1.5 x 10-10

  20. Weak base equilibria Bz - OH- HBz Initial conc., M 0.10 0.00 0.00 Change, DM -x+x +x Eq. Conc., M 0.10 - x x x [OH-] = [HBz] = x We’ll assume that [HBz] and [OH-] are negligible compared to [Bz -], since the value of the Ka << [Bz -]. (1.5 x 10-10 << 0.10 M)

  21. x2 0.10 Weak base equilibria Solve the equilibrium equation in terms of x Kb = 1.5 x 10-10 = x = (1.5 x 10-10 )(0.10) = 3.9 x 10-6 M pOH = - log (3.9 x 10-6 M) = 5.4 pH = 14 - pOH = 8.6

  22. Acidic and Basic Salts • Certain ions in solution can exhibit acid/base properties. • For example, consider the weak base, ammonia: • NH3 + H2O  NH4+ + OH- • What if you dissolve the salt, ammonium sulfate, in water? • (NH4)2SO4 2 NH4+ + SO42- • Because NH4+ is the conjugate acid of NH3, when this salt dissolves in water, the solution will become slightly acidic.

  23. Anions and Water • An anion that is a conjugate base of a weak acid raises the pH of a solution: • X- + H2O  HX + OH- • Example – sodium acetate in water • Ionic: Na+ + CH3COO- + H2O  Na+ + CH3COOH + OH- • Net ionic: CH3COO- + H2O  CH3COOH + OH- • The Kb of this reaction can be found using the Ka of acetic acid (Ka x Kb = Kw)

  24. Cations and Water • An cation that is a conjugate acid of a weak base that contains hydrogen lowers the pH of a solution: • HX+ + H2O  X + H3O+ • Example – ammonium chloride • Ionic: NH4+ + Cl- + H2O NH3 + Cl- + H3O+ • Net ionic: NH4+ + H2O  NH3 + H3O+ • The Ka of this reaction can be found using the Kb of ammonia (Ka x Kb = Kw)

  25. Some Rules • An anion that is the conjugate base of a strong acid will not affect the pH of a solution. (Ex: Br- from HBr) • An anion that is the conjugate base of a weak acid will cause an increase in pH (CN- from HCN) • A cation that is the conjugate acid of a weak base will cause a decrease in pH (NH4+ from NH3)

  26. Some Rules • Alkali metal cations and Ca2+, Sr2+, and Ba2+ will not affect pH (they are conjugate acids of strong bases). • Other metals (Al3+, etc.) will cause a decrease in pH. • When a solution contains both a cation and anion that will affect pH, the ion with the larger equilibrium constant (Ka or Kb will have the greater influence on pH).

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