EQUILBRIUM OF Acids and Bases Chapter 17 1 Water AUTOIONIZATION H2O can function as both an ACID and a BASE. Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC 2 Water 3 OH- H3O+ Autoionization Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC In a neutral solution [H3O+] = [OH-] so [H3O+] = [OH-] = 1.00 x 10-7 M [H3O+], [OH-] and pH A common way to express acidity and basicity is with pH pH = - log [H3 + O ] In a neutral solution, [H3O+] = [OH-] = 1.00 x 10-7 at 25 oC pH = -log (1.00 x 10-7) = - (-7) = 7 4 [H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution? [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00 General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7 5 [H3O+], [OH-] and pH If the pH of Coke is 3.12, it is ________. log [H3O+] = - pH Take antilog and get [H3O+] = 10-pH [H3O+] = 10-3.12 = 7.6 x 10-4 M 6 pX Scales In general pX = -log X pOH = - log [OH-] pH = - log [H+] pKw = 14 = pH + pOH 7 Weak Base Step 3. Calculate pH [OH-] = 4.2 x 10-4 M so pOH = - log [OH-] = 3.37 Because pH + pOH = 14, pH = 10.63 8 Equilibria Involving Weak Acids and Bases 9 Aspirin is a good example of a weak acid, Ka = 3.2 x 10-4 10 Weak Acids and Bases Acid Conjugate Base acetic, CH3CO2H CH3CO2-, acetate ammonium, NH4+ NH3, ammonia bicarbonate, HCO3CO32-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%). Weak Acids and Bases 11 acetic acid, CH3CO2H (HOAc) HOAc + H2O D H3O+ + OAcAcid Conj. base + [H3O ][OAc ] -5 Ka 1.8 x 10 [HOAc] (K is designated Ka for ACID) [H3O+] and [OAc-] are SMALL, Ka << 1. Equilibrium Constants for Weak Acids 12 Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7 13 0.0001 M 0.003 M 0.06 M 2.0 M a pH meter, Screen 17.9 Calculations with Equilibrium Constants pH of an acetic acid solution. What are your observations? Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH Step 1. ICE table. [HOAc] I C E [H3O+] [OAc-] 14 Equilibria Involving A Weak Acid [HOAc] I 1.00 C -x E 1.00-x [H3O+] [OAc-] 0 0 +x +x x x Note that we neglect [H3O+] from H2O. 15 Equilibria Involving A Weak Acid 16 Step 2. Write Ka expression + ][OAc- ] 2 [H O x 3 Ka 1.8 x 10-5 = [HOAc] 1.00 - x This is a quadratic. Use quadratic formula or method of approximations (see Appendix A). HOWEVER Equilibria Involving A Weak Acid Assume x is very small because Ka is so small. + [H3O ][OAc ] -5 Ka 1.8 x 10 = [HOAc] -5 Ka 1.8 x 10 = 2 x 1.00 - x 2 x 1.00 Now we can more easily solve this approximate expression. 17 18 Equilibria Involving A Weak Acid Step 3. Solve Ka approximate expression 2 x -5 Ka 1.8 x 10 = 1.00 x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = -log (4.2 x 10-3) = 2.37 Equilibria Involving A Weak Acid 19 For many weak acids [H3O+] = [conj. base] = [Ka • Co]1/2 where C0 = initial conc. of acid Useful Rule of Thumb: If 100•Ka < Co, then [H3O+] = [Ka•Co]1/2 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO2H. HCO2H + H2O D HCO2- + H3O+ Ka = 1.8 x 10-4 Approximate solution [H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.4 Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M pH = 3.5 20 Weak Bases Equilibrium Constants for Weak Bases 22 Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7 Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O D NH4+ + OH- Kb = 1.8 x 10-5 Step 1; ICE table [NH3] I C E [NH4+] [OH-] 23 Weak Base 24 Step 1. ICE table [NH3] [NH4+] [OH-] 0 0 I 0.010 C -x +x +x E 0.010 - x x x Weak Base 25 Step 2. Solve the equilibrium expression 2 + ][OH- ] [NH x -5 4 Kb 1.8 x 10 = = [NH3 ] 0.010 - x Assume x is small (100•Kb < Co), so x = [OH-] = [NH4+] = 4.2 x 10-4 M [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M The approximation is valid ! 26 Acids Conjugate Bases 27 Relation of Ka, Kb, [H3O+] and pH 28