Chem. 31 – 11/22 Lecture

Announcements Labs Due Today Today’s Lecture Formal Part A IC Part 1 11/22 Today’s Lecture Chapter 8 – Advanced Equilibrium Failure of the ICE method with two equilibria The systematic method and its six steps Chapter 9 – Acid/Base Equilibria The “weak acid problem” (pH of weak acid in water)

The Systematic Method Solubility of MgCO3 – Why did it fail? MgCO3  Mg2+ + CO32- x x Equil. (in ICE) So x = (Ksp)1/2 = 1.87 x 10-4 M (neglecting ionic strength effects) Problem is both ions can react further: CO32- + H2O  HCO3- + OH- And HCO3- + H2O  H2CO3 + OH- Also, Mg2+ + OH-  MgOH+ And Mg2+ + CO32-  MgCO3 (aq) Finally, we also have H2O  H+ + OH- re-establishing equilibrium Each additional reaction results in greater dissolution To properly solve problem we must consider 6 reactions not just 1 Measured “[CO32-]” from titration = [CO32-] + 0.5[OH-] + 0.5[HCO3-] + [MgCO3] + 0.5[MgOH+] The “further” reactions makes [Mg2+] ≠ [CO32-], so ICE method fails (or needs modification by ICE tables for other reactions) Actual solubility is greater than ICE method finds [Mg2+]total = solubility ~ 3.3 x 10-4 M (from systematic approach) Predicted HCl needed = 3.3 mL (vs. ~3.5 mL) These calculations didn’t include activity which would lead to a ~10% increase in solubility (~3.6 mL HCl needed). In 0.1 M NaCl, I get 6.1 mL HCl needed (close to that observed) enhancements: (% over rxn 1 only) 90% 0% 9% 16%

The Systematic Method The Six Steps Write out all relevant reactions Write a “Charge Balance Equation” Write “Mass Balance Equations” Write out all equilibrium equations Check that the number of equations (in 2 to 4 above) = (or maybe >) the number of unknowns (undefined concentrations) Solve for the desired unknown(s) by reducing the equations to one equation with one unknown. Then solve for remaining unknowns Note: the emphasis of teaching the systematic method is steps 1 to 5. Step 6 will be reserved for “easy” problems with 2 to max 3 unknowns

The Systematic Method pH of 5.0 x 10-8 M HCl Demonstrate Method on Board

The Systematic Method Conceptual Approach to Mass Balance Equations With every source of related species, there should be one mass balance equation (or one set for ionic compounds) Example: Solubility of AgCl in water with 0.010 M 1,10-phenathroline (Ph) Reactions: 1) AgCl(s)  Ag+ + Cl- 2) Ag+ + 2Ph  Ag(Ph)2+ Mass Balance equations: if only rxn 1) [Cl-] = [Ag+] w/ rxn 2) [Cl-] = [Ag+] + [Ag(Ph)2+] 1,10-phenathroline Ag+ Ph Ph Ph Ph Ag+ Ag+ Cl- Cl- Ag+ Ag+ Cl- Cl- 2nd Mass Balance Equation: [Ph]o = 0.010 M = [Ph]Total = [Ph] + 2[Ag(Ph)2+] AgCl(s) Notes: with rxn 1) only, 2 Ag+s = 2 Cl-s; with rxn 2) also, 3 Cls = 2 Ags + 1 Ag(Ph)2 Initially 4 Phs, then 2 Phs + one complex containing 2 Phs (so total # of Phs remains constant)

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 2nd 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 11

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-] 12

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 13

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 14

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 15