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BUFFERS

BUFFERS. What is a buffer?. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. http://www.youtube.com/watch?v=g_ZK2ABUjvA&feature=related

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BUFFERS

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  1. BUFFERS

  2. What is a buffer? A buffer is a solution that resists changes in pH when small amounts of acid or base are added. http://www.youtube.com/watch?v=g_ZK2ABUjvA&feature=related Buffers are very important in biological systems. For example Blood plasma has a pH of 7.4 and contains the buffer H2CO3/HCO3- and a change in pH of 0.4 could be fatal, without the buffer in the blood a glass of orange juice could be fatal!

  3. Buffer Composition: A buffer is composed of a weak acid and one of its salts: e.g. CH3COOH/ NaCH3COO HF/ NaF weak acid/ conjugate base • A buffer can also be composed of a weak base and one of its salts • E.g. NH3/ NH4Cl • Weak base/conjugate acid

  4. How do buffers work? • http://www.youtube.com/watch?v=8EZ2cL2tFPQ&feature=related • Example: CH3COOH (0.10 M)/CH3COO-(0.25 M) • Solution Equilibrium • CH3COOH  CH3COO- + H+ • Ka = (CH3COO-)(H+)/(CH3COOH) • (H+) = Ka (CH3COOH) / (CH3COO-) • (H+) =1.8 x 10-5 (0.10 M)/(o.25M) • (H+) =7.2 x 10-6 M • pH = 5.14

  5. Add a strong acid to the buffer: • Add H + to the buffer equilibrium: • CH3COOH  CH3COO- + H+ • Le Chatelier predicts that the extra H+ will be absorbed by the acetate ion and the equilibrium will shift to the left  to produce more weak acid. • The conjugate base gobbles up the extra H+ ions.

  6. Add strong base to the buffer: • Add OH- to the buffer equilibrium: • CH3COOH  CH3COO- + H+ • Le Chatelier predicts that the extra OH- will be absorbed by the H +, producing water, and the equilibrium will shift to the right  to produce more conjugate base. • The Hydorgen ions gobble up the extra OH - ions.

  7. In general: • HA  H+ + A- Buffer (weak acid/ conjugate base) • Add H+ A- gobbles up extra H+ and buffer equilibrium shifts to the left. • Add OH- H+ gobbles up extra OH- and the buffer equilibrium shifts to the right.

  8. A- + H2O  HA + OH-Buffer (weak base/ conjugate acid) • Add H+OH- gobbles up the extra H+ and the buffer equilibrium shifts right. • Add OH- HA reacts with the extra OH- and the buffer equlibrium shifts to left.

  9. Henderson Hasselbalch • HA  H+ + A- • Ka = (H+)(A-)/(HA) • Rearrange (H+) = Ka (HA)/(A-) • Take –log -log(H+)=-log Ka (HA)/(A-) • pH = -log Ka - log (HA) /(A-) • pH = pKa - log (HA)/(A-)

  10. Calculate the pH of a buffer! • What is the pH of a buffer that contains 0.1 mol/L of ethanoic acid and 0.10 mol/L of sodium ethanoate? • Ka = 1.78 x 10-5 • pH = pKa - log (ethanoic acid) /(ethanoate ion) • pH = pKa - log (0.10M/0.10M) • pH = pKa(weak acid and conjugate base are the same concentration) • pH = -log (1.78 x 10-5) = 4.750

  11. Add HCl to the buffer! • What is the pH after adding 10. mL of 1.0 M HCl to a liter of the buffer? • CH3COOH  CH3COO- + H+ • 0.10M 0.1o M • Expect the pH to go down but not too much! • pH = pKa - log (CH3COOH)/(CH3COO-) • CH3COO- will go down as it combines with the acid • CH3COOH will go up as equilibrium shifts to the left.

  12. How many moles of acid have been added to the buffer? n = cxv = 10.0 mLx 1.0 M = 10. mmol = 0.010 mol • (CH3COOH) increases to 0.10 moles/L + 0.010 mol = 0.11mol/ L • (CH3COO-) decreases to 0.10 mol/L – 0.010 mol = 0.09 mol/L • Total volume is now 1.00 L + 0.01 L =1.01L • pH = 4.75 - log (0.11mol/1.01L/0.09mol/1.01L) • pH = 4.75 – log (.11/.09) = 4.66

  13. pH went down, but not too much! • To compare add 10.0 mL of 1.0 M HCl (aq) to a liter of water and calculate the pH! • (H+) = n/V = 0.01 mol/1.01L= 9.90 x 10 -3 mol/L • pH = 2 • Buffers really work!!!!

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