__________________________________________

1 / 14

# __________________________________________ - PowerPoint PPT Presentation

## __________________________________________

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Class Monday, Oct 4, 2004 __________________________________________

2. Exam 1 __________________________________________ Exam 1 will be given starting at 1:00 PM on Wednesday, Oct 6. You may work how ever long is needed; after finishing the exam, go to lab. The class will not meet at the 12:00 hour on that day.

3. Homework Assignment 04 __________________________________________ ...is from Chapter 7. Problems assigned are: 7-1, 7-5, 7-6,7-7, 7-11, 7-17, 7-20, 7-21 Friday, Oct 8 Prepare on regularly sized paper, one side only with multiple pages stapled.

4. Class Monday, Oct 4, 2004 __________________________________________ • Comments regarding Homework and Lab Reports • The mean value is reported with the number of significant digits possessed by the least significant data. • The standard deviation s and the confidence interval CI is reported with the same number of decimal places as the mean value; both are treated like as + /  quantities to the mean value.

5. Fraction Dissociation __________________________________________ Fraction of dissociation as a weak acid is diluted. As the concentration decreases, the fraction dissociation increases. A- = [A] / ([A] + [HA]) Values of A- range from 0 – 1 Some authors use % dissociation; % dissociation = A- x 100%

6. Fraction Dissociation __________________________________________ • Dissociation is often determined by making conductivity measurements, which may be made with the apparatus shown. The electrical conduction is depends on the number of ions present in solution. • For strong electrolytes, there is a high population of ions, and thus high conductance. • For weak electrolytes, the population of ions is low, and low electrical conductance.

7. Weak Base Problems __________________________________________ Example, page 168. What is the pH of a 0.037 M solution of the weak base cocaine?

8. What is the pH of a 0.0372 M solution of the weak base cocaine? __________________________________________ • Let B represent the base, cocaine. Then BH+ is its conjugate acid form. • Re-write the dissociation reaction as B + H2O < === > BH+ + OH • Kb = 2.6 x 106 =[BH+] [OH] /[B] • Grid: B + H2O < === > BH+ + OH • Initial 0.0372 0 0 • Change -x x x • Equil. (0.0372 – x) x x • Substitute the equilibrium values into the Kb expression • Kb = 2.6 x 106 = [BH+] [OH] / [B] = x2 / (0.0372 – x) • Assume x << 0.0372; equation reduces to • 2.6 x 106 = x2 / 0.0372; x2 = 9.67 x 108; x = 3.11 x 104 = [OH] • 8) Is this assumption (made in 7) reasonable? 3.11 x 104 << 0.0372? YES! • pOH = log10(3.11 x 104) = 3.51; pH = 14.0 – 3.51 = 10.49 • What is the fraction dissociation? • 3.11 x 10-4 / 0.0372 = 0.0083

9. Start Chapter 9 - Buffers __________________________________________ • pH buffers are solutions that resist changes in pH whenever small amounts of either acid or base are added, or if the solution is diluted. • pH buffersare composed of a mixture of a weak electrolyte and its conjugate pair, that is, either • A weak acid and its conjugate base, or • A weak base and its conjugate acid

10. pH Buffers __________________________________________ • Examples of pH buffers include: • Acetic acid and sodium acetate • Lactic acid and sodium lactate • H3PO4 and KH2PO4 • NaH2PO4 and Na2HPO4 • Na2HPO4 and Na3PO4 • NH3 and NH4Cl • CH3NH2 and CH3NH3Cl • Note that a pH buffer solution MUST contain a weak electrolyte and its conjugate pair.

11. The Henderson-Hasselbalch Equation __________________________________________ The major equation for the calculation of the pH of buffer solution is the Henderson-Hasselbalch Equation; it is derived from the defined Kafor the acid form of the acid- base pair. Ka = [H+] [A] / [HA] Taking the –log10 of this expression and rearranging, we obtain pH = pKa + log10 ([A] / [HA]), or broader, pH = pKa + log10 ([base form] / [acid form]) This is known as the Henderson-Hasselbalch Equation.

12. The pH Buffer Solution __________________________________________ • The pH buffer solution works because if • Acid is added, the H+ ion reacts with the conjugate base form, forming additional amount of the acid form. On the other hand, • The addition of base removes the proton (H+) from the acid form, forming increased amounts of the basic form. • In either event, it is the logarithm of the ratio of • Base : Acid that establishes the new pH of the solution. The ratio of Base : Acid must change by a factor of ten in order to show a change of 1 pH unit in the pH of the solution.

13. Problem - Henderson-Hasselbalch Equation __________________________________________ What is the pH of a solution prepared by mixing together 250 mL of 0.1000 M acetic acid and 250 mL of 0.2000 M sodium acetate? The Ka for acetic acid is 1.75 105. Answer: pH = 5.06

14. End Monday, Oct 4, 2004 __________________________________________