Chem. 31 – 11/27 Lecture

Announcements Labs Due Wednesday Today’s Lecture GC Chapter 8 – Advanced Equilibrium The systematic method and its six steps – more practice Generalities about using the systematic method Chapter 9 – Acid/Base Equilibria The “weak acid problem” (pH of weak acid in water) The weak base problem

The Systematic Method 2nd Example An aqueous mixture of CdCl2 and NaSCN is made Initial concentrations are [CdCl2] = 0.0080 M and [NaSCN] = 0.0040 M Cd2+ reacts with SCN- to form CdSCN+ K = 95 HSCN is a strong acid Ignore any other reactions (e.g. formation of CdOH+) Ignore activity considerations Go through steps 1 through 5

The Systematic Method 3rd Example A student prepares a solution that contains 0.050 mol of AgNO3 and 0.0040 mol NH3 in water with a total volume of 1.00 L. The AgNO3 is totally soluble, NH3 is a weak base, and Ag+ reacts with NH3 to form Ag(NH3)2+. Assume the Ag+ does not react with water or OH-. Go through the first 5 steps of the systematic method.

The Systematic Method Stong Acid/Strong Base Problems When do we need to use the systematic approach? when more than 1 coupled reaction occur (unless coupling is insignificant) examples: 4.0 x 10-3 M HCl. 7.2 x 10-3 M NaOH Key point is the charge balance equation: for strong acid HX, [H+] = [X-] + [OH-] If [X-] >> [OH-], then [H+] = [X-] for strong base NaOH, [H+] + [Na+] = [OH-]

The Systematic Method General Comments Effects of secondary reactions e.g. MgCO3 dissolution Additional reactions increase solubility Secondary reactions also can affect pH (CO32- + H2O will produce OH- while Mg2+ + H2O will produce H+) Software is also available to solve these types of problems (but still need to know steps 1 → 5 to get problems solved)

Acid – Base Equilibria (Ch. 9) Weak Acid Problems: e.g. What is the pH and the concentration of major species in a 2.0 x 10-4 M HCO2H (formic acid, Ka = 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 7

Acid – Base Equilibria Weak Acid Problem – cont.: Systematic Approach (HCO2H = HA to make problem more general where HA = weak acid) Step 1 (Equations) HA ↔ H+ + A- H2O ↔ 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: Kw = [H+][OH-] and Ka = [A-][H+]/[HA] Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.: [HA], [A-] [H+], [OH-] 8

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 Ka 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 9

Acid – Base Equilibria Weak Acid Problem – Using ICE Approach Ka = [H+][A-]/[HA] = x2/(2.0 x 10-4 – x) x = 1.2 x 10-4 M (using quadratic equation) Note: sometimes (but not in this case), a 2nd 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: a = fraction of dissociation = [A-]/[HA]total a = 1.2 x 10-4 /2.0 x 10-4 = 0.60 10

Acid – Base Equilibria Weak Acid Problem – cont.: When is Assumption #1 valid (in general)? When both [HA]o and Ka 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 11

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 NH3 solution (Ka(NH4+) = 5.7 x 10-10, so Kb = Kw/Ka = 1.75 x 10-5) Reaction NH3 + H2O  NH4+ + OH- Go over on board 12

Acid – Base Equilibria Weak Acid/Weak Base Questions: A solution is prepared by dissolving 0.10 moles of NH4NO3 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 a and Ka. 13