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Unit 4: Equilibrium

Unit 4: Equilibrium. Chapter 8.2 Weak acids and Bases. Acid Ionization Constants 01. Acid Ionization Constant: the equilibrium constant for the ionization of an acid. HA( aq ) + H 2 O( l ) æ H 3 O + ( aq ) + A – ( aq ) Or simply: HA( aq ) æ H + ( aq ) + A – ( aq ).

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Unit 4: Equilibrium

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  1. Unit 4: Equilibrium Chapter 8.2 Weak acids and Bases

  2. Acid Ionization Constants 01 • Acid Ionization Constant: the equilibrium constant for the ionization of an acid. HA(aq) + H2O(l) æ H3O+(aq) + A–(aq) • Or simply: HA(aq) æ H+(aq) + A–(aq) Chapter 15

  3. Acid Ionization Constants, Ka and Kb ACIDKaCONJ. BASE Kb HF HNO2 C9H8O4 (aspirin) HCO2H (formic) C6H8O6 (ascorbic) C6H5CO2H (benzoic) CH3CO2H (acetic) HCN C6H5OH (phenol) 7.1 x 10 –4 4.5 x 10 –4 3.0 x 10 –4 1.7 x 10 –4 8.0 x 10 –5 6.5 x 10 –5 1.8 x 10 –5 4.9 x 10 –10 1.3 x 10 –10 F– NO2 – C9H7O4 – HCO2 – C6H7O6 – C6H5CO2 – CH3CO2 – CN – C6H5O – 1.4 x 10 –11 2.2 x 10 –11 3.3 x 10 –11 5.9 x 10 –11 1.3 x 10 –10 1.5 x 10 –10 5.6 x 10 –10 2.0 x 10 –5 7.7 x 10 –5 Chapter 15

  4. Strength of Acids and Bases (a) Arrange the three acids in order of increasing value of Ka. (b) Which acid, if any, is a strong acid? (c) Which solution has the highest pH, and which has the lowest? Chapter 15

  5. Equilibria Involving Weak Acids and Bases • Consider acetic acid, HC2H3O2 (HOAc) • HC2H3O2 + H2O  H3O+ + C2H3O2- • Acid Conj. base • (K is designated Ka for ACID) • K gives the ratio of ions (split up) to molecules (don’t split up) • The value of Ka is used to calculate pH of weak acids.

  6. Ionization Constants for Acids/Bases Conjugate Bases Acids Increase strength Increase strength

  7. Equilibrium Constants for Weak Acids Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7

  8. Equilibrium Constants for Weak Bases Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7

  9. Relation of Ka, Kb, [H3O+] and pH

  10. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc] [H3O+] [OAc-] Initial Change Equilib. 1.00 0 0 -x +x +x 1.00-xx x

  11. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 2. Write Ka expression This is a quadratic. Solve using quadratic formula. or you can make an approximation if X is very small!

  12. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3. Solve Ka expression First assume x is very small because Ka is so small. Now we can more easily solve this approximate expression.

  13. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3. Solve Ka approximate expression x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

  14. Acid Ionization Constants • Initial Change Equilibrium Table: Determine the pH of 0.50M HA solution at 25°C. Ka = 7.1 x 10–4. - + H + A æ HA (aq) (aq) (aq) Initial ( M ) : 0.50 0.00 0.00 Change (M): – x + x + x Equilib 0.50 – x x x (M): Chapter 15

  15. Acid Ionization Constants • Calculate the pH of a 0.036 M nitrous acid (HNO2) solution. • What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10–4? • The pH of a 0.060 M weak monoprotic acid is 3.44. Calculate the Ka of the acid. Chapter 15

  16. In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid. (1.78x10-5) From the ionization of acetic acid, CH3COOH = CH3COO- + H+0.100             0.0042     0.0042

  17. pKa = - log Ka • The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentration of 1.0 M (x=0.042; pH=2.38)

  18. R I C E [H+][C2H4NO2–] [x]2 [x]2 Ka = = [HF] [0.010-x] [0.010] HC2H4NO2 H+ + C2H4NO2– c=0.01M Calculate pH and Ka HC2H4NO2 H+ C2H4NO2– 1 1 1 0.010 0 0 -x +x +x 0.010 - x x x = 1.4x10–5 Since x is small 0.010 – x = 0.010 x= 3.74 x 10–5 M, pH = 3.43 =1.4 x 10 – 5

  19. R I C E [H+][Bu–] [0.0004]2 Ka = = [HBu] [.0096] HBu  H+ + Bu– c(HBu = 0.01M and pH = 3.40 Calculate Ka HBu H+ Bu– 1 1 1 0.0100 0 0 -0.0004 +0.0004 +0.0004 0.0096 0.0004 0.0004 [H+] = 10–pH = 10–3.40 = 3.98 x 10–4 = 1.67x10–5

  20. Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table [NH3] [NH4+] [OH-] initial change equilib 0.010 0 0 -x +x +x 0.010 - x xx

  21. Acid Ionization Constants • Percent Dissociation: A measure of the strength of an acid. • Stronger acids have higher percent dissociation. • Percent dissociation of a weak acid decreases as its concentration increases.

  22. R I C E [MH+] [OH–] Kb = = [M] M + H2O  MH+ + OH– c=0.01M, pH =10.10 Kb-? M MH+ OH– 1 1 1 0.010 0 0 -0.00013 +0.00013 +0.00013 0.00987 0.00013 0.00013 pOH = 14 - pH = 14 - 10.10 = 3.90 [OH-] = 10-pOH = 10-3.90 = 1.26 x 10-4 [0.00013] [0.00013] =1.7 x 10-6 [0.00987]

  23. R I C E [x] [x] Kb = = [0.020] NH3 + H2O  NH4+ + OH– c=0.02M; pH - ? NH3 NH4+ OH– 1 1 1 0.020 0 0 -x +x +x 0.020-x x x pOH = -log[OH-] = 3.22 pH = 14 - pOH = 10.78 x2 = 1.8 x 10-5 x= 6.0 x 10-4 [0.020]

  24. Ka summary • Ka follows the pattern of other “K” equations • I.e. for HA(aq) + H2O(l)  H3O+(aq) + A–(aq) • Ka = [H3O+][A–] / [HA] • Notice that H2O is ignored because it is liquid • HA cannot be ignored because it is aqueous • This is different than with Ksp. In Ksp, solids could only be in solution as ions • Acidscanbeinsolutionwhetherionizedornot • The solubility of acids makes sense if you think back to the partial charges in HCl for ex.

  25. Ka summary • Generally Ka tells you about acid strength • Strong acids have high Ka values • A “strong” acid is an acid that completely ionizes. E.g. HCl + H2O  H3O+ + Cl– • A “weak” acid is an acid that doesn’t ionize completely. E.g. HF + H2O  H3O+ + F– • Note: don’t get confused between strength and concentration. 1 M HCN has a smaller [H+], thus a higher pH, than 0.001 M HCl • In general: Ka < 10–3 Weak acid 10–3 < Ka < 1 Moderate acid Ka > 1 Strong acid

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