Chem 31 1127 Lecture Announcements Labs Due Wednesday
Chem. 31 – 11/27 Lecture
Announcements • Labs Due Wednesday – GC • Today’s Lecture – 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” (p. H of weak acid in water) • The weak base problem
The Systematic Method 2 nd Example • An aqueous mixture of Cd. Cl 2 and Na. SCN is made – Initial concentrations are [Cd. Cl 2] = 0. 0080 M and [Na. SCN] = 0. 0040 M – Cd 2+ reacts with SCN- to form Cd. SCN+ K = 95 – HSCN is a strong acid – Ignore any other reactions (e. g. formation of Cd. OH+) – Ignore activity considerations – Go through steps 1 through 5
The Systematic Method 3 rd Example • A student prepares a solution that contains 0. 050 mol of Ag. NO 3 and 0. 0040 mol NH 3 in water with a total volume of 1. 00 L. The Ag. NO 3 is totally soluble, NH 3 is a weak base, and Ag+ reacts with NH 3 to form Ag(NH 3)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 Na. OH – Key point is the charge balance equation: - for strong acid HX, [H+] = [X-] + [OH-] - If [X-] >> [OH-], then [H+] = [X-] – for strong base Na. OH, [H+] + [Na+] = [OH-]
The Systematic Method General Comments • Effects of secondary reactions – e. g. Mg. CO 3 dissolution – Additional reactions increase solubility – Secondary reactions also can affect p. H (CO 32+ H 2 O will produce OH- while Mg 2+ + H 2 O 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 p. H and the concentration of major species in a 2. 0 x 10 -4 M HCO 2 H (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 (HCO 2 H = HA to make problem more general where HA = weak acid) • Step 1 (Equations) HA ↔ H+ + AH 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: 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 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] = 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; p. H = 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. : – 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
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 p. H of 0. 010 M NH 3 solution (Ka(NH 4+) = 5. 7 x 10 -10, so Kb = Kw/Ka = 1. 75 x 10 -5) – Reaction NH 3 + H 2 O NH 4+ + OH- – Go over on board
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 to set up this problem for determining the p. H 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 p. H of an unknown weak acid prepared to a concentration 0. 0100 M is measured and found to be 3. 77. Calculate a and Ka.
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