1. Chem. 31 – 4/20 Lecture

2. Announcements I Exam 2 –Result: average = 75 –Distribution similar to Exam 1 (except a few more high and low scores) Lab Reports –AA regular report due today –Soda Ash report due 4/27 –Will be posting information about Formal Report soon

3. Announcements II Today’s Lecture –Chapter 8: Acid Base Chemistry Weak Acid Problems Weak Base Problems Buffer Solutions

4. Acid – Base Equilibria (Ch. 8) Weak Acid Problems: –e.g. What is the pH and the concentration of major species in a 2.0 x 10 -4 M HCO 2 H (formic acid, K a = 1.80 x 10 -4 ) solution ? –Can use either systematic method or ICE method. –Systematic method will give correct answers, but full solution results in cubic equation –ICE method works most of the time –Use of systematic method with assumptions allows determining when ICE method can be used

5. Acid – Base Equilibria Weak Acid Problem – cont.: –Systematic Approach (HCO 2 H = HA to make problem more general where HA = weak acid) Step 1 (Equations) HA ↔ H + + A - H 2 O ↔ H + + OH - Step 2: Charge Balance Equation: [H + ] = [A - ] + [OH - ] 2 assumptions possible: ([A - ] >> [OH - ] – assumption used in ICE method or [A - ] << [OH - ]) Step 3: Mass Balance Equation: [HA] o = 2.0 x 10 -4 M = [HA] + [A - ] Step 4: K w = [H + ][OH - ] and K a = [A - ][H + ]/[HA] Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.: [HA], [A - ] [H + ], [OH - ]

6. Acid – Base Equilibria Weak Acid Problem – cont.: –Assumption #1: [A - ] >> [OH - ] so [A - ] = [H + ] –Discussion: this assumption means that we expect that there will be more H + from formic acid than from water. This assumption makes sense when [HA] o is large and K a is not that small (valid for [HA] o >10 -6 M for formic acid) –ICE approach (Gives same result as systematic method if assumption #1 is made) –(Equations) HA ↔ H + + A - Initital 2.0 x 10 -4 0 0 Change - x +x +x Equil. 2.0 x 10 -4 – x x x

7. Acid – Base Equilibria Weak Acid Problem – Using ICE Approach –K a = [H + ][A - ]/[HA] = x 2 /(2.0 x 10 -4 – x) x = 1.2 x 10 -4 M (using quadratic equation) Note: sometimes (but not in this case), a 2 nd assumption can be made that x << 2.0 x 10 -4 to avoid needing to use the quadratic equation [H + ] = [A - ] = 1.2 x 10 -4 M; pH = 3.92 [HA] = 2.0 x 10 -4 – 1.2 x 10 -4 = 8 x 10 -5 M Note:  = fraction of dissociation = [A - ]/[HA] total  = 1.2 x 10 -4 /2.0 x 10 -4 = 0.60

8. Acid – Base Equilibria Weak Acid Problem – cont.: –When is Assumption #1 valid (in general)? –When both [HA] o and K a are high or so long as [H + ] > 10 -6 M –More precisely, when [HA] o > 10 -6 M and Ka[HA] o > 10 -12 –See chart (shows region where error < 1%) –Failure also can give [H + ] < 1.0 x 10 -7 M Assupmption #1 Works Fails

9. Acid – Base Equilibria Weak Base Problem: –As with weak acid problem, ICE approach can generally be used (except when [OH - ] from base is not much more than [OH - ] from water) –Note: when using ICE method, must have correct reaction –Example: Determine pH of 0.010 M NH 3 solution (K a (NH 4 + ) = 5.7 x 10 -10, so K b = K w /K a = 1.75 x 10 -5 ) –Reaction NH 3 + H 2 O  NH 4 + + OH - –Go over on board

10. Acid – Base Equilibria Weak Acid/Weak Base Questions: –A solution is prepared by dissolving 0.10 moles of NH 4 NO 3 into water to make 1.00 L of solution. Show how to set up this problem for determining the pH using the ICE method. –A student is solving a weak base problem for a weak base initially at 1.00 x 10 -4 M using the ICE method and calculates that [OH - ] = 2.4 x 10 -8 M. Was the ICE method appropriate? –The pH of an unknown weak acid prepared to a concentration 0.0100 M is measured and found to be 3.77. Calculate  and K a.

11. Acid – Base Equilibria Buffer Solutions: –A buffer solution is designed so that a small addition of acid or base will only slightly change the pH –Most buffer solutions have a weak acid and its conjugate base both present –Example: Determine pH of a mix of 0.010 M HCO 2 H and 0.025 M Na + HCO 2 - solution (ignoring activity) –Go to board to show if ICE approach is needed

12. Buffer Solutions: –Question: Was the ICE Problem set up needed? –Answer: No. The assumption of x << [HA], [A - ] is valid for all “traditional” buffers –Traditional Buffer Weak acid (3 < pK a < 11) Ratio of weak acid to conjugate base in range 0.1 to 10 mM+ concentration range Acid – Base Equilibria

13. Buffer Solutions: –Since ICE not needed, can just use K a equation –K a = [H + ][A - ]/[HA] = [H + ][A - ] o /[HA] o (always valid) (valid for traditional buffer) –But log version more common –pH = pK a + log([A - ]/[HA]) –Also known as Henderson-Hasselbalch Equation

14. Acid – Base Equilibria Buffer Solutions: –Ways to make buffer solution: Mix weak acid and conjugate base (done in making reference solution for soda ash lab) Add strong base to weak acid (weak acid must be in excess) – this converts some of the weak acid to its conjugate base Add strong acid to weak base (weak base must be in excess) – this converts some of weak base to its conjugate acid