This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 HendersonHasselbalch equation pH = p K a + log C b C a Calculate the pH of a solution of 0.20 M NH 3 and 0.30 M NH 4 Cl K b = 1.8 x 105 Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, K a = 4.9 x 1010 Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, K a = 4.9 x 1010 pH = log(4.9 x 1010 ) + log = 9.09 0.15 0.25 The HendersonHasselbalch 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 105 n a = MV = (0.16 M )(0.250 L) = 0.040 mol pH = p K a + log 4.70 = log(1.8 x 105 ) + log0.04 = 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 1010 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 1010 Answer: 0.031 mol NaCN 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 105 ) = 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 R0.001 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 105 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 Answer: pH = 4.67 Titration : an analysis based on measuring the amount of substance (the titrant ) required to react with a measured amount of analyte the substance whose concentration is being determined...
View
Full
Document
This note was uploaded on 02/21/2012 for the course CHEM 112 taught by Professor Chen during the Spring '09 term at South Carolina.
 Spring '09
 CHEN
 pH

Click to edit the document details