CH204 Fall 2008 Cheat Sheet for Experiment 9. Acid-Base Equilibria 1Part 3. The Ionization Constant of Acetic Acid Solution: 10 ml of 1.2 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+ 1.2 0 0 initial –x x x change (1.2 – x) x x equilibrium 3. )2.1(]COOHCH[]COOCH][OH[K2333axx−==−+ Solution: 30 ml of 1.2 M NaCH3COO + 10 ml of 1.2 M CH3COOH CH3COOH + H2O CH3COO– + H3O+ (reversible) NaCH3COO → CH3COO– + Na+ (to completion) Measured: pH To be determined: Ka Assume that the equilibrium molarities of CH3COOH and CH3COO– are the same as their initial molarities. However, keep in mind that the initial molarities decrease after the two solutions are mixed together (they dilute each other) Calculations: 1. pH310]OH[−+= 2. Determine [CH3COO–] in a new volume of 40 ml (dilution problem) 3. Determine [CH3COOH] in a new volume of 40 ml (dilution problem) 4. ]COOHCH[]COOCH][OH[K333a−+= Part 4. Buffering Capacity Solution: 40 ml of DI H2O + 4 ml of 2.0 M HCl HCl + H2O → H3O+ + Cl– (to completion) To be determined: pH (calculated)CH204 Fall 2008 Cheat Sheet for Experiment 9. Acid-Base Equilibria 2Calculations: 1. Determine [HCl] in a new volume of 44 ml (dilution problem) 2. [H3O+] = [HCl] 3. pHcalc = –log [H3O +] Solution: 40 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: 40 ml of 1.2 M NaCH3COO + 40 ml of 1.2 M CH3COOH = 80 ml of the buffer To be determined: pH (calculated) Calculations: 1. Determine [CH3COO–] in a new volume of 80 ml (dilution problem) 2. Determine [CH3COOH] in a new volume of 80 ml (dilution problem) 3. ]COOHCH[]COOCH[logpKpH33a−+= Use Ka = 1.76×10–5 Solution: 40 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: 40 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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