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Chem. 31 – 4/26 Lecture Announcements I • Due today: – Co/Cr lab – AA resubmission • Last Quiz – Next Wednesday • Exam 2 Results – Results were pretty typical – Ave = 70% (broad spread) Score 90-99 N 4 80s 23 70s 11 60s 19 32-59 16 • Homework Set 3 Posted + solutions to HW3.1 sub set Announcements II • Bonus Points Survey – Will be posting instructions (Read instructions BEFORE doing survey) and link to survey soon • Additional Problem 4 – posted • Today’s Lecture • Chapter 9: Acid – Base Chemistry – – – – Strong acid/strong base problems (generalities) Weak acid problems Weak base problems Buffer solutions (if time) 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. 8) • 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 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+ + AH2O ↔ 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-] 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+ + AInitital 2.0 x 10-4 0 0 Change -x +x +x Equil. 2.0 x 10-4 – x x x 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 Acid – Base Equilibria • Weak Acid Problem – cont.: 1.00E+00 1.00E-01 Assupmption #1 Works 1.00E-02 [HA]o – 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 Where ICE Method Works 1.00E-03 1.00E-04 1.00E-05 Fails 1.00E-06 1.00E-07 1E-10 1E-08 1E-06 0.0001 Ka 0.01 1 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 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. 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 HCO2H and 0.025 M Na+HCO2- solution (ignoring activity) – Go to board to show if ICE approach is needed Acid – Base Equilibria • 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 < pKa < 11) • Ratio of weak acid to conjugate base in range 0.1 to 10 • mM+ concentration range Acid – Base Equilibria • Buffer Solutions: – Since ICE not needed, can just use Ka equation – Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o (always valid) (valid for traditional buffer) – But log version more common – pH = pKa + log([A-]/[HA]) – Also known as Henderson-Hasselbalch Equation Acid – Base Equilibria • Buffer Solutions: – Why are they needed?/useful? – The main reason is to keep the pH constant so that the ratio of species of acids and bases is constants – Some examples: • in water hardness titration, we want [Y4-]/[Y]total constant so sample pH won’t affect results • spectroscopy: Beer’s law only applies to single species (e.g. separate laws for HIn and In-) • chromatography: at pH = 2, benzoic acid is a molecule and retained (reversed-phase HPLC)