Chapter 16 Acids and Bases

Chapter 16 Acids and Bases - Buffer : a solution of a weak...

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Unformatted text preview: Buffer : a solution of a weak acid or base and its conjugate base or acid. Buffers resist changes in pH. HA + OH- H 2 O + A- weak acid reacts with any added OH- A- + H 3 O + HA + H 2 O weak base reacts with any added H 3 O + for HA + H 2 O H 3 O + + A- K a = [H 3 O + ] = so pH = p K a + log [H 3 O + ][A- ] [HA] K a [HA] [A- ] [HA] [A- ] In a buffer system with HA and A- both present in large concentration, [HA] C a C a is the analytical concentration of the acid [A- ] C b C b is the analytical concentration of the base Henderson-Hasselbalch equation pH = p K a + log C b C a Calculate the pH of a solution of 0.50 M HCN and 0.20 M NaCN, K a = 4.9 x 10-10 Calculate the pH of a solution of 0.20 M NH 3 and 0.30 M NH 4 Cl K b = 1.8 x 10-5 Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, K a = 4.9 x 10-10 Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, K a = 4.9 x 10-10 pH = -log(4.9 x 10-10 ) + log = 9.09 0.15 0.25 In-Class Homework #8 Oct 1, 1997 1. Calculate the pH of 0.50 M sodium acetate, CH 3 COONa. K a for acetic acid is equal to 1.8 x 10-5 2. Bonus. Name that tune.(Thank you Mary Peyton) Verify.au The Henderson-Hasselbalch equation pH = p K a + log can be rewritten as pH = p K a + log n b n a C b C a Determine the number of moles of CH 3 COONa that must be added to 250 mL of 0.16 M CH 3 COOH in order to prepare a pH 4.70 buffer. K a = 1.8 x 10-5 n a = MV = (0.16 M )(0.250 L) = 0.040 mol pH = p K a + log 4.70 = -log(1.8 x 10-5 ) + log-0.04 = log n b n a n b 0.040 mol n b 0.040 mol 0.91 = n b = 0.036 mol n b 0.040 mol How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40? K a = 4.9 x 10-10 How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40? K a = 4.9 x 10-10 Answer: 0.031 mol NaCN In-Class Homework #9 Oct 3, 1997 1. Calculate the pH of a buffer that is 1.50 M acetic acid and 1.25 M sodium acetate. K a for acetic acid is 1.8 x 10-5 2. Bonus. Name that movie tune. Verify.au Calculate the initial and fnal pH when 10 mL oF 0.100 M HCl is added to (a) 100 mL oF water, and (b) 100 mL oF a buFFer which is 1.50 M CH 3 COOH and 1.25 M CH 3 COONa Initial pH of water is 7.00 [HCl] = [H 3 O + ] = C HCl = 0.0091 M pH = -log(0.0091) pH = 2.04 (0.110 L) (0.010L)(0.100 M ) n a = (0.100 L)(1.50 M ) = 0.150 mol n b = (0.100 L)(1.25 M ) = 0.125 mol p K a = -log(1.8 x 10-5 ) = 4.74 pH = 4.74 + log = 4.66 0.125 0.150 n(H 3 O + ) = (0.010 L)(0.100 M ) = 0.0010 mol CH 3 COO- + H 3 O + CH 3 COOH + H 2 O s 0.125 mol 0.001 mol 0.150 mol R-0.001 mol -0.001 mol +0.001 mol f 0.124 mol 0 mol 0.151 mol pH = 4.74 + log = 4.65 pH changes -0.01 units (0.124) (0.151) Calculate the fnal pH when 10 mL oF 0.100 M NaOH is added to 100 mL oF a buFFer which is 1.50 M CH 3 COOH and 1.25 M CH 3 COONa Ka = 1.8 x 10-5 Calculate the fnal pH when 10 mL oF 0.100 0....
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Chapter 16 Acids and Bases - Buffer : a solution of a weak...

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