Chapter 17 Acid-Base Equilibria 81 John A. Schreifels
Chapter 17 Acid-Base Equilibria 81 John A. Schreifels Chemistry 212 Chapter 17-1 Overview Solutions of a Weak Acid or Base Acid ionization equilibria Polyprotic acids Base ionization equilbria Acid-Base properties of Salts Solutions of a Weak Acid or Base with Another Solute Common Ion Effect Buffers Acid-Base Titration Curves 82 John A. Schreifels Chemistry 212 Chapter 17-2 Acid Ionization Equilibria
Weak acids and weak bases only partially dissociate; their strengths are experimentally determined in the same way as strong acids and bases by determining the electrical conductivity. The reaction of a weak acid (or base) with water is the same as discussed in previous section. Consider the reaction: HA(aq) H2O(l) H3O (aq) A (aq) [H3O ][ A ] Ka [HA ] Hydronium ion concentration must be determined from the equilibrium expression. Relative strengths of weak acids can be determined from the value of the equilibrium constant. Large equilibrium constant means strong acid Small equilibrium constant means weak acid E.g. determine which acid is the strongest and which the weakest. Acid HCN HCOOH CH3COOH HF
John A. Schreifels Chemistry 212 Ka 4.9x1010 1.8x104 1.8x105 3.5x104 83 Chapter 17-3 Determining K from pH Ka determined if pH and CHA known. Use the equilibrium expression for the acid. E.g. Determine the equilibrium constant of acetic acid if the pH of a 0.260 M solution was 2.68. Determine [H3O+]; [HA]; and [A]. Initial conc to Equil. Equil. HA(aq)+H2O(l) H3O+(aq) 0.100 M 0 +x x +x 0.100 M x + A (aq) 0 +x +x
Strategy Calculate the [H3O+] from pH; this is x in the table above. The rest of the quantities are obtained from the bottom row. 84 John A. Schreifels Chemistry 212 Chapter 17-4 Calculating Equilibrium Concentrations in Weak acid Solutions pH determined if Ka and Ca known; for the dissociation of acetic acid: CH 3COOH (aq ) H 2O(l) H3O (aq) CH3COO (aq) [H3O ][CH3COO ] Ka [CH COOH ] 3 1.8 x10 5 2H2O(l) H3O (aq) OH (aq) K w [H3O ][OH ] 1.00 x10 14 [H3O+]total = [H3O+]CH3COOH + [H3O+]H2O. [H3O+]total [H3O+]CH3COOH. The total hydronium ion concentration is often equal to the contribution from the weak acid which is usually a lot stronger acid than water. The total hydronium ion concentration is needed for the
equilibrium calculation. John A. Schreifels Chemistry 212 Chapter 17-5 85 pH from Ka and Ca E.g. Calculate the pH of 0.100M acetic acid. Given pKa = 4.76 Initial conc to Equil. Equil. CH3COOH(aq)+H2O(l) H3O+(aq)+ CH3COO (aq) 0.100 M 0 0 +x +x x +x +x 0.100 M x Method I: Substitute into equilibrium equation to get x2 + 1.75x105x 1.75x106 = 0. Solve using quadratic equation (see book). Method 2 Assume x << CHA. Then x = (KaCHA)1/2. Check (confirm assumption to be correct) 1.75 x10 5
x2 0.100 x Analytical concentration should be: Ca = 100x[H3O+] Method 3 method of successive approximations. As in Method 2; then x = (Ka(CHA x1))1/2; repeat if necessary. E.g. Calculate pH of 0.0200M lactic acid if its K a = 8.4x104M. John A. Schreifels Chemistry 212 86 Chapter 17-6 % Dissociated (also called % Ionized) Weak Acids % ionization a useful way of expressing the strength of an acid or base. 100% ionized a strong acid. Only partial ionization occurs with weak acids. HA(aq)+ H2O(l) H3O+(aq)+ A (aq) Initial conc CHA 0 0 +x to Equil. +x x
Equil. CHA x +x +x x % Ionization 100 % CHA E.g. determine the % ionization for 0.100 M, 0.0100 M, 0.00100M HCN if K a = 4.9x1010. Solution: determine x for each and sub into definition above. Check assumptions. % Ionization CHA K a 1/ 2 CHA K a CHA 100% 1/ 2 100% 87
Notice % ionization increases with dilution. John A. Schreifels Chemistry 212 Chapter 17-7 Polyprotic Acids Some acids can donate more than one proton to the solution. Thus a diprotic acid has two protons such as H2S and H2SO4, while a common triprotic acid has three acidic protons that can be donated (H3PO4). First proton easily removed; others much more difficult. Treat Polyprotic acids as if they were monoprotic acids; Use Ka1. Formula Ka1 Ka2 Ka3 H3PO4 7.5x10 3 6.2x10 8 4.8x10 11 H2SO3 1.5x10 2 6.3x10 8 The equilibrium constant for removal of each successive proton is about 105 times the equilibrium constant for removal of the preceeding proton.
2 E.g. determine the pH of 0.100 M H2SO3. Then determine [SO 3. ] 88 John A. Schreifels Chemistry 212 Chapter 17-8 Equilibria:Weak bases (WB) (proton acceptor) Treat bases just like we did the weak acid; except you are calculating [OH]. The general equation that describes the behavior of a base in solution is: [BH ][OH ] B(aq) H2O(l) BH (aq) OH (aq) Kb [B] Set up the equilibrium table as before for the acids and substitute values for all the quanitities in the equilibrium expression. B(aq)+H2O(l) BH+(aq)+ OH (aq) Initial conc.
CB 0 0 +x +x to Equil. x Equil. CB x +x +x Since usually CB is supplied, we have one unknown which we can evaluate using standard equil. equation for weak base. [BH ][OH ] Kb [B] + Remember that x = [OH ] and not [H3O ]. x2 E.g. Calculate the pH of 0.10M NH3(aq). CB x Hint: Expect pH > 7 when with weak base. 89 John A. Schreifels Chemistry 212 Chapter 17-9 Equilibria:Weak bases Structure Many nitrogen containing compounds are basic the amine most important. Most of the amines have a lone pair
of electrons that are available for bonding with an acidic proton (Brnsted-Lowry base). Amines usually have a carbon residue in place of a hydrogen. R2 N R1 H Amine Structure 810 John A. Schreifels Chemistry 212 Chapter 17-10 Relation between Ka and Kb Ka and Kb are always inversely related to each other in aqueous solutions. HA(aq)+ H2O(l) H3O+(aq) + A (aq) Add A (aq)+ H2O(l) HA(aq) +OH (aq) [ A ][H3 O ] Ka [HA ] Kb
[HA ][OH ] [A ] 2H2O(l) H3O+(aq)+OH (aq) Kw = KaKb Inverse relationship explains why Acid Ka conjugate base of very weak acid is 4 HF 3.5x10 relatively strong. HCOOH E.g. given the Kas of the following acid 1.8x104 list their conjugate bases in terms of 8 HOCl 3.5x10 relative strength. HCN John A. Schreifels Chemistry 212 4.9x1010 811 Chapter 17-11 Salts of WA and WB Salt: an ionic substance formed as a result of an acidbase neutralization reaction. Salt of an acid(base) obtained by its neutralization with acid if it is a
base and base if it is an acid. E.g. NaCl is a salt from the reaction of HCl with NaOH. The properties of the salt will depend upon the strengths of the acid and base that formed the salt. E.g.1: determine the acidbase reaction that would produce CH3COONa, NaCN, NH4Cl, (NH4)2CO3. Salts are usually soluble in water because of their ionic character. When they dissolve, they affect the pH of the solution. Depends upon relative strengths of the conjugate acid and base. 812 John A. Schreifels Chemistry 212 Chapter 17-12 Salt of Strong Acid and Strong Base Neutral solution results if the salt is from the reaction of a SA + SB. E.g. NaCl Other cations and anions producing neutral solutions: Li+, Na+, K+, Ca2+, Sr2+, Ba2+ and Cl, Br, I, , NO 3). ClO 4 E.g. what is the approximate pH of the following. NaCl, KCl, LiClO4,
etc.? Salt of WA + SB (basic) and Salt of WB + SA (acidic). Ignore cation (or anion) from SA (base). Conjugate of WA is WB basic solution. Conjugate of WB is WA acidic solution. SA + SB Neutral (very WA & WB) SA + WB Acidic (WA) WA + SB Basic (WB) where SA = Strong Acid; SB = Strong Base John A. Schreifels WA = Weak Acid; WB = Weak Chemistry 212 813 Chapter 17-13 Calculating the pH of Salt of WA or WB (other ion from SA(SB)) Salt of WA: Use Kb of the conjugate base and treat it as a weak base: A(aq) + H2O(l) HA(aq) + OH(aq) E.g. determine the pH of 0.100M NaCH3COO. Ka (CH3COOH) = 1.75x105. E.g. determine the pH of 0.200 M NaCN. Ka(HCN) = 4.9x1010. Salt of WB: Use Ka of conjugate acid and treat as a weak acid: E.g. determine pH of 0.250M NH4Cl. Kb = 1.8x105. E.g. determine pH of 0.100 M N2H5Br. Kb = 1.1x108. John A. Schreifels Chemistry 212
814 Chapter 17-14 Salt of WA + WB Determine Ka and Kb of acidic and basic portions of salt. Largest K dominates to make solution either acidic or basic. E.g. determine if 0.100 M NH4CN is acidic or basic. E.g. 2 predict if 0.100 M C6H5NH3F is acidic or basic. 815 John A. Schreifels Chemistry 212 Chapter 17-15 The Common Ion Effect CommonIon Effect: the change in the equilibrium that results from the addition of an ion that is involved in the equilibrium. E.g. NaOCl is added to 0.100 M HOCl; NH4Cl is added to NH3. Setting up the standard equilibrium table can show the effect. E.g. determine the pH of a solution prepared by mixing 50.0 mL of 0.100 M
HOCl with 50.0 mL of 0.100 M NaOCl (Ka = 3.5x108). Set up equilibrium table after calculating the concentrations of each in the final mixture. Initial concentrations change slightly as a result of a change reaction. HOCl + H2O H3O+ + OCl 0.0500M 0 0.0500M +x +x x +x 0.0500 + x 0.0500 x Solve using either approximations or quadratic equation. Shifts equilibrium towards the basic side. John A. Schreifels Chemistry 212 816 Chapter 17-16 Buffers Buffer solution: a mixture of conjugate acid and base that resists pH changes. Significant buffering capacity occurs when [acid] = [base], pH = pKa. An example of the common ion effect. E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20. Set up equilibrium table.
Ignore the value of x compared to the concentrations of the common ion. pH in buffering region related to the relative amount of conjugate acid and base. Let r [base] then the equilibrium equation is: [acid] K a [H3O ] r K [H3O ] a r 817 John A. Schreifels Chemistry 212 Chapter 17-17 Addition of Acid or Base to a Buffer Upon addition of a SB to the buffer we have: Addition of either acid or base changes ratio of acidic and basic forms. Big changes in pH occur only when nearly all of one species is consumed. Cb Vb r Ca Va Cb Vb E.g. determine r after addition of 5.00 mL of 0.100 M NaOH to
10.00 mL of 0.100 M HOCl. Determine pH if Ka = 3.5x108. E.g. Determine pH of 50.00 mL of phosphate buffer containing equilmolar concentrations (0.200M) of acid/base forms, after 10.00 mL 0.100 M NaOH or 10.00 mL of HCl. pKa2 =7.20 Changes in volume don't affect pH. John A. Schreifels Chemistry 212 818 Chapter 17-18 Henderson-Hasselbalch Equation The effect of r (=[A]/[HA]) on pH is better understood by taking log of both sides of equation between K and conc. To give [A ] pH = pK a + log [HA] nA = pK a + log n HA Called Henderson-Hasselbach equation. Allows us to predict pH when HA/A mixed. When [A] /[HA] = 1 (i.e. [HA]=[A]), pH = pKa E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20. E.g.2 determine the ratio of the concentration of the conjugate acid to concentration of the conjugate base for a weak acid in which the pH was 5.45 and pKa was 5.75. E.g. determine the pH of a solution consisting of 0.100 M NH3 and
0.150 M NH4Cl. John A. Schreifels Chemistry 212 819 Chapter 17-19 Neutralization Reactions Neutralization Reaction: the reaction of an acid with a base to produce water. Extent of reaction nearly quantitative (except if both acid and base are weak. SASB: E.g. HNO3 + NaOH NaNO3 + H2O SA produces: H3O+ SB produces: OH Overall reaction: H3O+ + OH 2H2O K w1 = 1/Kw = 1.00x1014 WASB: thought of as two step reaction. E.g. HOCl + NaOH NaOCl + H2O K = ? HOCl H+ + OCl H+ + OH H2O HOCl + OH H2O + OCl Ka = 3.5x10 8 K w1= 1.00x10+14 K = Ka K w1= 3.5x106 820 Large equilibrium constant means reaction nearly quantitative. John A. Schreifels Chemistry 212
Chapter 17-20 Neutralization Reactions WB + SA and WA + WB WB + SA SA produces H3O+ ions; use base as is. E.g. NH3 + HCl NH+ Cl or 4 NH3 + H2O NH4 + OH Ka = 1.8x10 5 H3O+ + OH 2H2O K w1= 1.00x10+14 NH3 + H3O+ NH4 + H2O K = Ka K w1= 1.8x109 Conclusion: Quantitatively generate product (nearly). WA + WB: initially undissociated species dominates. HOCl + H2O H3O+ + OCl Ka = 3.5x10 8 NH3 + H2O NH + OH 4 +
H3O + OH 2H2O Kb = 1.8x10 5 NH3 + HOCl NH4 + OCl K = KaKb K w1 = 63 K w1= 1.00x10+14 Conclusion: Reaction will sometimes, but not always, be quantitative. E.g. determine the extent of reaction when di methyl amine (Kb = 5.4x104) reacts with either HF (Ka = 3.5x104) or HOCl (Ka = 3.5x108). John A. Schreifels Chemistry 212 821 Chapter 17-21 pH Titration Curves pH Titration of 0.100 M HA with Titration curve: plot of pH of 0.100 M NaOH the solution as a function of 14 12 the volume of base (acid) 10 added to an acid (base).
8 6 WA Sharp rise in curve is 4 SA equivalence point. 2 0 pH at equivalence point is 7.0 0 10 20 30 40 for SA but higher for WA. Volume Base Added, mL Equivalence point can be used to determine the concentration of the titrant. E.g. the equivalence point for 15.00 mL of an acid occurred when 25.00 mL of 0.075 M NaOH was 822 added. What was the molarity of the acid? John A. Schreifels Chemistry 212 Chapter 17-22 SASB Titrations Base removes some acid and pH increases. Let nb = moles of base added na,r = moles of acid remaining na,r = na nb = CaVa CbVb Moles of hydronium ion same as moles of acid remaining. n H3O+ = na,r;
+ n [H3O ] H3 O V a + Vb Ca Va Cb Vb V a + Vb Valid until very close to equivalence point. Equivalence point(EP): pH = 7.00 Beyond EP: pH due only to base added (i.e. excess base). Use total volume. E.g. Determine pH of 10.0 mL of 0.100M HCl after addition of 5.00, 823 10.0 and 15.0mL of 0.100M NaOH. John A. Schreifels Chemistry 212 Chapter 17-23 Titration of SB with SA Acid removes some of the base and pH is changed by amount of base removed. Let na = moles of acid added nb,r = moles of base remaining nb,r = CbVb CaVa Moles of hydroxide ion same as moles of base remaining. nOH = nb,r; [OH ]
n OH V a + VB C V Ca Va b b V a + VB Valid until EP. EP: pH = 7.00 Beyond EP: pH due only to excess acid. Use total volume. E.g. Determine pH of 10.0 mL of 0.100M NaOH after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. John A. Schreifels Chemistry 212 824 Chapter 17-24 WA with SB Titration As above base removes some of the acid and pH is changed by amount of acid removed. Let nb = moles of base added nHA = moles of acid remaining nHA = CHAVHA CbVb
nA = nb = CbVb n pH pK a log A nHA pK a log Cb Vb Ca Va Cb Vb Up to equivalence point moles of hydronium ions must be determined from equilibrium expression. Equivalence point: pH = pH of salt of WA Beyond Equivalence point: Use amount of excess base to determine pH. E.g. determine pH of 10.0 mL of 0.100M HA after addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH. Ka = 1.75x105. John A. Schreifels Chemistry 212 825 Chapter 17-25 WBSA Titrations Acid removes some of the base and decreases the pH. Let na = moles of acid added
nb,r = moles of base remaining nb,r = CbVb CaVa nBH+ = na = CaVa n pOH pK b log BH nB Ca Va pK b log Cb Vb Ca Va Moles of hydroxide ions must be determined from equilibrium expression. Valid until EP. EP: pH = pH of salt of weak base. Beyond EP: pH due only to presence of acid added after endpoint (i.e. excess acid) as seen for strong base. Volume correction needed as above (total volume). E.g. Determine pH of 10.0 mL of 0.100M B after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. Kb = 1.75x105. John A. Schreifels Chemistry 212 826 Chapter 17-26
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