CH204 Spring 2008 Cheat Sheet for Experiment 9. Acid-Base Equilibria 1Part 3. The Ionization Constant of Acetic Acid Solution: 10 ml of 1.0 M CH3COOH Measured: pH To be determined: Ka Calculations: 1. x==−+ pH310]OH[ 2. Determine [CH3COO–] and [CH3COOH] from the following: CH3COOH + H2O CH3COO– + H3O+ 1.0 0 0 initial –x x x change (1.0 – x) x x equilibrium 3. )0.1(]COOHCH[]COOCH][OH[K2333axx−==−+ Solution: 20 ml of 1.0 M NaCH3COO + 10 ml of 1.0 M CH3COOH CH3COOH + H2O CH3COO– + H3O+ (reversible) NaCH3COO → CH3COO– + Na+ (to completion) Measured: pH To be determined: Ka Calculations: 1. pH310]OH[−+= 2. Determine [CH3COO–] in a new volume of 30 ml (dilution problem) 3. Determine [CH3COOH] in a new volume of 30 ml (dilution problem) 4. ]COOHCH[]COOCH][OH[K333a−+= Part 4. Buffering Capacity Solution: 30 ml of DI H2O + 4 ml of 2.0 M HCl HCl + H2O → H3O+ + Cl– (to completion) To be determined: pH (calculated) Calculations: 1. Determine [HCl] in a new volume of 34 ml (dilution problem) 2. [H3O+] = [HCl] 3. pHcalc = –log [H3O +]CH204 Spring 2008 Cheat Sheet for Experiment 9. Acid-Base Equilibria 2Solution: 30 ml of DI H2O + 4 ml of 2.0 M NaOH NaOH → Na+ + OH– (to completion) To be determined: pH (calculated) Calculations: Use the same approach as shown above for the (H2O + 2.0 M HCl) solution. Remember, that [H3O+]×[ OH–] = 10–14 Solution: 30 ml of 1.0 M NaCH3COO + 30 ml of 1.0 M CH3COOH = 60 ml of the buffer To be determined: pH (calculated) Calculations: 1. Determine [CH3COO–] in a new volume of 60 ml (dilution problem) 2. Determine [CH3COOH] in a new volume of 60 ml (dilution problem) 3. ]COOHCH[]COOCH[logpKpH33a−+= Use Ka = 1.76×10–5 Solution: 30 ml of the buffer (prepared previously) + 4 ml of 2.0 M HCl To be determined: pH (calculated) Calculations: 1. Determine the initial number of moles of CH3COOH in solution: a 2. Determine the initial number of moles of CH3COO– in solution: b 3. Determine the number of moles of HCl added: x 4. H3O+ + CH3COO– → CH3COOH + H2O x b a initial –x –x x change 0 (b – x) (a + x) equilibrium 5. )()(logpKCOOHCH of molesCOOCH of moleslogpKpHa33axaxb+−+=+=− Use Ka = 1.76×10–5 Solution: 30 ml of the buffer (prepared previously) + 4 ml of 2.0 M NaOH OH– + CH3COOH → CH3COO– + H2O Calculations: Use the same approach as shown above for the (buffer + 2.0 M HCl)
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