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CHEM100_11

CHEM100_11. Lecture #30. Chapter 16. Acid-base equilibrium. Quiz#2: November 25. Chapters to study for the November quiz: 4.4, 20.2 redox reactions 5.1-5.4, 5.6- thermochemistry 10.1-10.8 - gases 15- chemical equilibrium 16.1-16.5 - pH, pOH, strong acids and bases.

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CHEM100_11

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  1. CHEM100_11 Lecture #30 Chapter 16. Acid-base equilibrium Quiz#2: November 25 Chapters to study for the November quiz: 4.4, 20.2 redox reactions 5.1-5.4, 5.6- thermochemistry 10.1-10.8- gases 15- chemical equilibrium 16.1-16.5- pH, pOH, strong acids and bases X-mas exam: December 15, 7 p.m. Student Chemical Society Tutorials: Every Monday and Tuesday , NH244 from 7pm-8:30pm Dr. Orlova PS-3026 Ph: 867-5237 gorlova@stfx.ca Helping hrs at PS-3026: Tu & Th 1-2 pm; Wd 4-5 pm

  2. Add small amount of NaOH CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + H2O (l) Small increase in CH3COO- and small decrease in CH3COOH The CH3COOH/CH3COO- ratio is nearly the same  [H+] is the same  pH is nearly the same Add more and more NaOH  all CH3COOH will be spent: The buffer is broken!!!

  3. Buffer Capacity and pH • Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH. • Buffer capacity depends on the composition of the buffer. • The greater the amounts of conjugate acid-base pair, the greater the buffer capacity. • The pH of the buffer depends on Ka. To calculate pH let us take ‘–log’ of the above equation

  4. Human blood is a buffer, pH = 7.35-7.45 pH <6.8 or >7.8 - death Acid- dicarbonate buffer H+ (aq) + HCO3- (aq)  H2CO3 (aq)  H2O (l) + CO2 (g) Normal blood plasma has [HCO3-] = 0.024M [H2CO3] = 0.012M So the buffer has higher capacity to consume H+ and lower capacity to consume OH-

  5. excess Hemoglobin-H+ + O2 (g) hemoglobin-O2 (aq) + H+ (aq) Antifreeze: ethylene glycol (sweet) HO-CH2-CH2-OH Liver oxidize ethylene glycol to glycolic acid: HO-CH2-CH2-COOH Glycolic acid enters blood stream and breaks the blood buffer Low blood pH is called ACIDOSIS Medical treatment- alcohol (ethanol)

  6. 2 types of problems: 1. Calculate pH of a buffer composed from known concentration (Ka is taken from reference literature) 2. Find concentration of components to be mixed to build a buffer of certain pH. Ka is taken from reference literature.

  7. Henderson- Hasselbalch equation Simplification (initial concentrations instead of equilibrium) is valid for buffers!

  8. Type 1 Buffer is prepared by dissolving 25.5 g CH3COONa in 0.550 M solution of CH3COOH to make 500.0 mL of the buffer. Calculate pH. Ignore change in volume Ka = 1.8 x 10-5 Use Henderson -Hasselbalch eq. instead of reaction table! Simplification (initial molarities instead of equilibrium) is valid for buffers! Moles CH3COO- = CH3COONa = 25.5 g x 1mol/82.04g = 0.311 mol M CH3COO- = 0.311mol/0.500L = 0.622 M pKa = -log 1.8 x 10-5 = 4.74 pH = 4.74 + log (0.622/0.550) = 4.8

  9. Type 2 Example #2 • 1. How many grams of CH3COONa must be dissolved in 0.300L of 0.25 M • CH3COOH to make a buffer with pH =5.09. • Calculate pH of the buffer after adding 0.002mol HCl • Calculate pH of the buffer after adding 0.002 mol of NaOH • Assume that volume is constant

  10. pH = pKa + log {[CH3COO-]/[CH3COOH]} Ka = 1.8 x 10-5 pKa = -log Ka = 4.74 5.09 -4.74 = log [CH3COO-] – log [0.25] log [CH3COO-] = 0.35 – 0.602 = -0.252 [CH3COO-] = 10-0.252 = 0.56 M 0.56 mol/L x 0.300L = 0.168 mol Gram CH3COONa: 0.168 mol x82 g/mol = 14 g

  11. Before rxn CH3COOH(aq) + OH- (aq) CH3COO- (aq) + H2O (l) 0.168 mol 0.075 mol 0.002mol 0.073 0.0 0.170 After Add 0.002mol of strong base, NaOH, to the initial buffer pH = 4.74 + log [0.17]/[0.073] = 5.11 (pH of the buffer was 5.09) Put 0.002 mol (a few drops) of NaOH in 0.3L of water 0.002/0.3 = 0.0067 pOH = -log 6.7 x10-3 = 2.17 pH = 14 -2.17 =11.83 Go to lecture 33

  12. Before rxn 0.168 mol 0.002 0.075 After rxn 0.166 0 0.077 2. pH after adding strong acid: CH3COO- (aq) + H+ (aq)  CH3COOH (aq) Find how many moles of the base was consumed, then solve again H-H eq Mol CH3COO- = 0.168 mol Mol CH3COOH= 0.25M x 0.3L = 0.075mol Mol H+ added = 0.002 CH3COO- (aq) + H+ (aq)  CH3COOH (aq) pH = 4.74 + log [0.166]/[0.077] = 5.07

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