Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations

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1 Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations Sec 1 The Common-Ion Effect: The dissociation of a weak electrolyte decreases when a strong electrolyte that has an ion in common with the weak electrolyte is added to the solution. A. Examples include: 1) a solution containing a WA + soluble salt of the WA 2) a solution containing a WB + a soluble salt of the WB Suppose we have a weak acid and a soluble salt of that acid. CH 3 COOH NaCH 3 COO CH 3 COOH CH 3 COO + H + Since NaCH 3 COO provides a high concentration of acetate ions (common ions), adding it to the solution By Le Chatelier The result is that (shifts the equilibrium far to the left) (H+ concentration is drastically decreased) B. How does this equilibrium reaction act as a buffer system? + NH 3 + H 2 O <---> NH 4 + OHanswer: If an acid is added NH 3 will accept a proton, formed NH + 4. As ammonia increases, the equilibrium shifts to the left, forming more ammonia. If a base is added, the increase in hydroxide shifts the reaction to the left. Ex. #1: Find the ph of a solution containing M HNO 2 (K a = 4.5 x 10 4 ) and 0.10 M KNO 2. Sec 2. Buffered Solutions ( buffers ) A. contain a weak conjugate acid-base pair B. contain an acidic species and a basic species that do NOT consume each other through neutralization C. Buffer solutions resist changes in ph because they react with both hydronium and hydroxide ions. Ex #2: For a CH 3 COOH and-ch 3 COO buffer, you could use 1

2 For an NH 4 + -and-nh 3 buffer, you could use D. When an acid is added to a salt solution... What happens when an acid is added to a salt solution instead of just pure water. How does the presence of a salt affect the dissociation of the acid? Problem: A. Calculate the ph of a solution that is 0.10 M acetic acid AND 0.10 M sodium acetate. First, let's write the reaction for the dissociation of the salt, sodium acetate and identify the concentrations of each of the components: NOTE: The concentrations of the ions are the same as the concentration of the salt because the mole ratio is 1:1. Now, let's set up the REACTION for the dissociation of the acetic acid. Now, we set up our concentration CHART. Remember, however, that the initial concentrations for the conjugate base, C 2 H 3 O 2 - is no longer 0 because of the salt: [HC 2 H 3 O 2 [H 3 O + ] [C 2 H 3 O 2 - Initial: 0.10 M M Equilibrium: 0.10 M M + ASSUME is negligible when subtracted and added: 2

3 CHECK the assumption: Since the acid dissociates less than 5%, is negligible when added and subtracted M 1.8 x 10-5 M 0.10 M In a sample of this buffer, [CH 3 COOH] and ] [CH 3 COO-] are about 5000 x greater than [H 3 O + ]. If a small amount of hydronium is added, it will be quickly converted to acetic acid. If a small amount of hydroxide is added, it will neutralize hydronium, but CH 3 COOH will dissociate and maintain the ph fairly constant. B. Considering the same system in A, suppose mol of NaOH is added to this buffer solution. What will the ph of the resulting solution be? Step 1: Stoich part HC 2 H 3 O 2 + H 2 O H 3 O+ + C 2 H 3 O 2 - buffer 0.10 M 1.7 x M after M? 0.11 M NaOH addition Step 2: Equilibrium part [H 3 O+] (.11) = 1.7 x 10-5 (0.090) [H 3 O+] = 1.7 x 10-5 (.090) = 1.4 x 10-5 ph = -log 1.4 x 10-5 = 4.85 (.11) 1. Addition of.010 mol of NaOH changes the ph by only 0.15 units. (refer to ph= 4.7 in Part A) 2. A similar addition of pure water changes the ph by 5 units (from 7 to 12). 3

4 Ex. #3: Find the ph of a buffer that is 0.12 M lactic acid, HC 3 H 5 O 3 (K a = 1.4 x 10 4 ) and 0.10 M sodium lactate. (answer: 3.8) Sec 2. Addition of Strong Bases to Buffer -- Reactions between strong acids/bases and weak bases/acids proceed to completion. -- We assume the strong acid/base is completely consumed. Sec 3 Buffer Capacity: the amount of acid or base the buffer can neutralize before the ph begins to change appreciably. 2 calculations needed: 1) the ph of a solution where the buffer capacity has been exceeded; 2) choose among alternatives the best buffer system for a given ph. Problem 1: Calculate the ph of the M HCOOH/0.100 HCOONa a buffer used after the addition of 10.0 ml of 6.00 M NaOH to the original buffered solution volume of ml. Strategy: 1) determine the moles of weak acid and base existing after the addition of the SB (stoich part); 2) calculate the ph of the buffered solution after you have your new equilibrium conc. (equilibrium part) 1) stoichiometry part mmoles HCOOH initial = mmol/ml x ml = 125 mmol mmoles HCOO - initial = mmol/ml x ml = 50.0 mmol mmol OH-added = 6.00 mmol/ml x 10.0 ml = 60.0 mmol HCOOH + OH- ----> HCOO- I (mmol) F (mmol) after addition of NaOH, [HCOOH] = 65.0 mmol/510 ml = M (500+10) equilibrium concentration [HCOO-] = 110 mmol/510 ml = M 2) equilibrium part The acid dissociation of HCOOH is the significant factor here. 4

5 Ka = [H+][HCOO-] [HCOOH] 1.8 x 10-4 = [H+][0.216] => [H+] = 1.1 x 10-4 ph = 4.0 (5% rule holds) [0.128] Ex.#4: A buffered solution of ph 4.74 contains 0.30 mol CH 3 COOH (K a = 1.8 x 10 5 ) and 0.30 mol NaCH 3 COO. Calculate the ph after mol NaOH is added. Ignore volume changes. Alternately, you could use the Henderson-Hasselbalch equation: ph pka log A HA poh pkb log HB B where pka = -log Ka where pk b = -log K b Problem 2: Calculate the poh and ph of a solution that is 0.20 M in NH 3 and 0.10 M in NH 4 Cl. NH H 2 O <----> NH 4 + OH-.20 M 0.10 M -x +x +x.20-x 0.10+x x K b = 1.7 x 10-5 (given) pk b = -log 1.7 x 10-5 pk b = 4.77 (Use H-H formula): poh = pk b + log (salt) where pk b = -log K b base poh = log (.10)/(.20) = = 4.47 ph = = 9.53 Problem 3: What is the ph a solution containing 2.00 mol of ammonia and 3.00 mol of ammonium chloride in a volume of 1.00 L? K b = 1.81 x ways to solve the problem 5

6 Way 1: Do ICE and the K b expression: NH H 2 O <----> NH 4 + OH- K b = [ NH +1 4 ][ OH-] = ( x) (x) = 1.81 x 10-5 [NH 3 ] (2.00-x) assume x small: 1.81 x 10-5 = 3.00 x 2.00 x = 1.21 x 10-5 poh = ph = Way 2: Henderson/Hasselbach equation: (for acids) ph = pka + log [A-] HA] OR (for bases) poh = pkb +log [HB + ] [B] = log = ph = Sec4 Acid-Base Titrations & Indicators A. Most acid-base indicators are weak organic acids (HIn) or weak organic bases (InOH). In represents complex organic groups. B. Bromothymol blue is an acid-base indicator that is a weak acid and has a Ka of about 1 x HIn + H 2 O <---> H 3 O+ + In- (yellow) (blue) Ka = [H 3 O+] [ In-] = 1 x 10-7 [HIn] Problem: The yellow color of the bromothymol blue indicator can be seen when the acid form (Hin) is 6.3 x more concentrated that the ionized form (In-). The blue color can be seen when In- is 4 x more concentrated than Hin. What is the ph range over which this color change occurs? Yellow color H 3 O+ = Ka [HIn] = 1 x 10-7 (6.3) [In] H 3 O+ = 6.3 x 10-7 ph = -log 6.3 x 10-7 = 6.2 blue color H 3 O+ = Ka [HIn] = 1 x 10-7 [In-] 4.0 H 3 O+ = 2.5 x 10-8 M ph = 7.6 6

7 1. Yellow color is seen at a ph at 6.2 or below. This blue color is seen at a ph of 7.6 or above. 2. The pt at which the color change for the indicator occurs in a titration is the endpt. 3. The equivalence pt of a titration is the pt at which chemically equivalent amounts of A and B have reacted. 4. Ideally, the endpt and the equivalence pt should coincide in a titration. Sec 5 Strong Acid Strong Base Titrations H+ + OH > H 2 O The reverse reaction of the autoionization ph curve for HCl titrated with NaOH of water is K = 1/K w K = 1/ 1 x This reaction is certainly complete; This type of titration is less an equilibrium problem than a stoich one. ph 7 *Any indicator whose color change begins and ends along the vertical line is okay. -- phenolphthalein (ph ) base = acid = -- methyl red (ph ) base = acid = ml of NaOH added Problem 1: What is the ph of a solution containing ml of a M HCl to which ml of a M solution of NaOH has been titrated? Step 1: calculate the # of moles HCl and NaOH initially present 0.100mol/L x L = moles HCl *MV = moles mol/l x L = moles NaOH HCl + NaOH ----> NaCl + H 2 O start.00250mol mol 0 mol change mol mol mol after rxn mol 0 mol mol Step 2: Calculate the M of acid (or base) both are STRONG! HCl = mol =.0430 M L (total volume) 7

8 [H 3 O+] = 4.30 x 10-2 M ph = -log 4.30 x 10-2 M = *After ml of M NaOH has been titrated (the equivalence pt), the ph is 7.00 because the NaOH exactly neutralizes the HCl. Ex # 5: Calculate the ph after the 4.00 ml of M HCl have been added to ml of M NaOH. (ans: ph =13.1) Ex. #6: Find ph when ml of 0.10 M HNO 3 are mixed with ml of 0.10 M KOH. (ph = 10.3) Weak Acid Strong Base Titrations The equivalence point is when, say, 50.0 ml of 0.10 M NaOH have been added to 50.0 ml of 0.10 M CH 3 COOH, but ph is > 7 at that point because... p H 7 ph curve for CH 3 COOH titrated with NaOH ml of NaOH added EX #7: (in 5 steps) A ml of M nitrous acid (pka = 3.35) was titrated with M NaOH. Calculate the ph of the solution after the following quantities of base have been added to the acid solution: A) 0.00 ml B) 25.00mL, C) ml, D) ml ph curve for H 2 CO 3 Titration curves for polyprotic acids (H 2 CO 3 ) look something like -- they have... ph 8

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