of Applications Aqueous Equilibria
Buffers
A Very Important Buffer
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Human blood is a buffered solution
Source: Visuals Unlimited
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What is Needed to Make a Buffer?
A. A weak acid B. A weak base C. A strong acid and a strong base D. A weak acid and a weak base E. Blood, sweat, and tears
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What Solutes Make Blood a Buffer?
CO2 (aq) + H2O = HCO3- + H+ H2PO4- = HPO4= + H+ Proteins: RNH3+ = RNH2 + H+ The total phosphate concentration is about 20 mM; bicarbonate about 24 mM. The pH is buffered at 7.4.
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Pure water at pH 7.00
What happens if we add 0.01 mol of NaOH (0.4 g) to 1 L of pure water?
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What is the pH of 0.01 M NaOH? A. 10-2 B. 2 C. 7 D. 9 E. 12
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When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.00!
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An Acetate Buffer
Suppose we have 1 L of acetate buffer, containing 0.5 M. acetic acid and 0.5 M. sodium acetate? Henderson-Hasselbalch Equation: pH = pKa + log([base]/[acid]) pH = 4.74 = pKa of acetic acid. [H+] = 1.8 x 10-5 = Ka of acetic acid
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A digital pH meter shows the pH of the buffered solution to be 4.74
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How does it respond to base?
If we add 0.01 mole NaOH (0.4 g) to 1 L of this buffer, how does the pH change? Remember that the neutral water pH changed from 7 to 12.
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When a strong acid or base is added to a buffered solution, deal first with the stoichiometry of the resulting reaction. After the stoichiometric calculations are done, then consider the equilibrium calculations. This procedure can be represented as follows:
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OH- ions are not allowed to accumulate but are replaced by A- ions.
The strong base OH- reacts with the strongest available acid, in this case acetic acid, forming acetate ion.
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When the OH- is added, the concentrations of HA and Achange, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant.
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Acetate Buffer
Initially, [HOAc] = [OAc-] = 0.50 M. Add 0.01 mol OH-. [HOAc] 0.50 0.01 = 0.49 M [OAc-] 0.50 + 0.01 = 0.51 M pH = pKa + log([base]/[acid]) pH = 4.74 + log(0.51/0.49) = 4.76 Change = +0.02 rather than +5.
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8a15
Le Chatelier strikes again!
Note that these buffer calculations describe another example of Le Chateliers Principle. Suppose we have 0.10 M HF, Ka = 7.2 x 10-4, pKa = 3.14. HF = H+ + F What is the pH of this solution? Ka = [H+][F-]/[HF] = x2/(0.10 x) x = 0.012 M; pH = 1.91 If we add F-, it represses the ionization. There is less H+ and a higher pH.
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8a16
Le Chteliers principle for the dissociation of HF
HF (aq) = H+ (aq) + F- (aq)
It shouldnt be a surprise that adding a base (F-) to a solution reduces the acidity.
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8a17
If [HF] = [F-] = 0.10 M,
Ka of HF = 7.2 x 10-4 What is the pH?
A. pH = 1.91 B. pH = 3.14 C. pH = 4.0 D. pH = 7.2
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8a18
Can you explain buffers to a History major?
Adding the base OH- to a buffer converts the acid HA to the base A-. Adding the acid H+ to a buffer converts the base A- to the acid HA. So if theres just as much total base or acid as there was before adding the OHor H+, how come the pH doesnt change much?
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8a19
Which buffer is better?
If we wish to hold the pH of a solution at 7.00, which buffer would be most effective? (pKa of acid form) A. Acetic acid/acetate (4.84) B. Formic acid/formate (3.75) C. H3PO4/H2PO4- (2.12) D. H2PO4-/HPO4= (7.21) E. HPO4=/PO43- (12.67)
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Which buffer is better?
If we wish to hold the pH of a solution at 7.0, which buffer would be most effective? A. 0.01 M H2PO4-/HPO4= B. 0.05 M H2PO4-/HPO4= C. 0.1 M H2PO4-/HPO4= D. 0.2 M H2PO4-/HPO4= E. 0.5 M H2PO4-/HPO4=
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Titration
Titration curves of weak acids and bases illustrate the buffering effect.
Source: American Color
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Volume of 0.1 M NaOH added (mL) to 50 mL 0.2 M HNO3
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Figure 8.1: The pH Curve for the same Titration
Strong acid + strong base: no buffering at midpoint.
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pH Curve for Titration of 50.0 mL 0.10 M Acetic Acid with 0.10 M NaOH
There is a buffering region around halfequivalence Note: The pH is not 7 at the equivalence point. Why not? .
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Buffer region
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pH Curves for the Titrations of Acids of Varying Strengths
Halfway to the equivalence point, [HA] = [A-], so pH = pKa, from the H-H equation.
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pH Curve for Titration of NH3 with HCl
Buffer region
Weak base + strong acid, so pH at equivalence is on the acidic side.
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The indicator phenolphthalein is pink in base and colorless in acid.
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Structure of Phenolphthalein
- H+
-
All acid-base indicators are themselves acids or bases, with the unusual property that the conjugate forms have different colors due to a structural change.
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The useful pH ranges for several common indicators
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pH Curve for Titration of HCl with NaOH
The titration curve is so steep between pH 4 and 10 that either indicator will give a satisfactory result.
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pH Curve for Titration of Acetic Acid with NaOH
The pH at the end point is 8.72. Methyl red will give a very diffuse color change around 40 mL rather than the correct 50 mL. Phenolphthalein will work well.
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Choice of Indicator
Which indicator would be best for showing the end point in a titration of a weak base with a strong acid? Phenolphthalein, pKa = 9 Bromthymol blue, pKa = 7 Bromcresol green, pKa = 5 Whichever doesnt clash with your outfit.
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A. B. C. D.
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Solubility Equilibria
For dissolution of a covalent substance, the reaction is simply Whatever (s) = Whatever (aq) The equilibrium expression is just K = [Whatever]. This is a pretty trivial case.
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Solubility Equilibria: MX (s) = M+ + X For dissolution of a solid salt MX, Ksp = [M+][X-] (normal Keq expression) Your PSY major friend looks at this and asks if a more soluble salt always has a numerically larger Ksp than a less soluble salt. What do you tell her?
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8a35
How Soluble is BaSO4?
Ksp for BaSO4 = 1.5 x 10-9 BaSO4 (s) = Ba2+ (aq) + SO42- (aq) A. 1.5 x 10-9 M. B. 2.25 x 10-18 M. C. 3.9 x 10-4 M. D. 3.9 x 10-5 M. E. 0.75 x 10-9 M.
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How Soluble is Barium Sulfate?
Ksp for BaSO4 = 1.5 x 10-9 BaSO4 (s) = Ba2+ + SO42 Solubility = [Ba2+] = [SO42-] = x x2 = 1.5 x 10-9, so x = 3.9 x 10-5 M = 39 M Low solubility allows BaSO4 to be used for X-ray imaging without causing barium poisoning.
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An X ray of the Colon using Barium Sulfate Suspension for Visualization
Source: Photo Researchers, Inc.
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How Soluble is PbCl2?
PbCl2 (s) = Pb2+ + 2 Cl Ksp = 1.6 x 10-5 A. 1.6 x 10-2 M. B. 2.0 x 10-3 M. C. 4.0 x 10-3 M. D. 1.6 x 10-5 M.
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8a39
How Soluble is PbCl2?
PbCl2 (s) = Pb2+ + 2 ClKsp = 1.6 x 10-5 Ksp = [Pb2+][Cl-]2 = (x)(2x)2 = 1.6 x 10-5 4x3 = 1.6 x 10-5 x3 = 4.0 x 10-6 x = 1.6 x 10-2 = 0.016 M. or 16 mM
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The Trouble with Ksp Calculations
What anions form insoluble salts? Review from Chapt. 4: Some halides (Ag+, Pb2+, Hg22+) Some sulfates (Ba2+, Pb2+, Ca2+) These are salts of strong acids, and the dissolution reaction is fairly well represented by M+X- = M+ + X-, so Ksp calculations are fairly valid in these cases.
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The Trouble with Ksp Calculations
Other mostly insoluble salts: Anion OHS= CO3= PO33Copyright Houghton Mifflin Company. All rights reserved.
Conj. Acid H2O HSHCO3HPO3=
Acid pKa 14 19 12 14
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The Trouble with Ksp Calculations
Ksp calculations assume that all that happens is M+X- (s) = M+ + X- . What really happens when you dissolve a salt of a weak acid? Upon dissolving, X- + H2O = HX + OH-, for which Kb = Kw/Ka. If your chemical are assumptions invalid, your mathematical conclusions wont be valid either.
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The Trouble with Ksp Calculations
Example: CdS (s): Ksp = 1 x 10-28 If solubility = Ksp, then it = 1 x 10-14 M. If we recognize that the main reaction involves acid-base reaction, CdS (s) + H2O = Cd2+ + HS- + OH-, K = KspKw/Ka2 = (10-28)(10-14)/10-19 = 10-23 Solubility = [Cd2+] = 2.2 x 10-8 M.
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The Trouble with Ksp Calculations
In addition to complications of acidbase reactions of basic anions, several other phenomena can perturb results of Ksp calculations: Stable ion-pairs may exist in solution Complex ions may form, such as AgX2 In cases of carbonates and sulfides, CO2 and H2S may evaporate from solutions. Hydroxides may precipitate.
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Solubility in a Buffered Solution
What is the solubility of CdS (Ksp = 1.0 x 10-28) in a solution buffered at pH 6? We have already calculated that the solubility in water is about 2.16 x 10-8 M., with the main reaction being
CdS (s) + H2O = Cd2+ + HS- + OH In this solution, [OH-] = [Cd2+] = 2.2 x 10-8 M, so pH = 7.66.
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Solubility in a Buffered Solution
If the pH is held at 6 by buffering, do you expect the CdS to be more soluble or less soluble than in pure water? More soluble Less soluble About the same I have no clue.
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A. B. C. D.
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Solubility in a Buffered Solution
The pKa values for H2S are 7 and 19. What will be the principal form of sulfide at pH 6? H-H: pH = pKa + log([base]/[acid]) 6 = 7 + log([HS-]/[H2S]) So log([HS-]/[H2S]) = -1; [H2S]/[HS-] = 10 and H2S will account for 90% of the dissolved sulfide.
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Solubility in a Buffered Solution
The principal reaction upon dissolving CdS will be: CdS (s) + 2 H+ Cd2+ + H2S Ksp CdS(s) Cd2+ + S= S= + H+ HSK = 1/K2 HS- + H+ H2S K = 1/K1 Overall K = Ksp/K1K2 = (10-28)/(10-7)(10-19) = 10-2 Solubility = [Cd2+] = [H2S] = x
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Solubility in a Buffered Solution
[Cd2+][H2S]/[H+]2 = 10-2 x2/(10-6)2 = 10-2 x2 = 10-14, so x = 10-7 M at pH 6 This is about 5 times more soluble than the value (2.16 x 10-8 M.) for water. At pH 5, solubility = 10-6 M. At pH 4, solubility = 10-5 M. At pH 1, solubility = 0.01 M. Many sulfides are quite soluble at pH 1.
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Formation of Complex Ions
Most transition metal cations form complex ions by associating with ligands in solution. Ligands include anions, :NH3, and H2O: Ag+ + Cl- = AgCl (s) But AgCl (s) + Cl- = AgCl2-, so Ag+ is more soluble in a solution with a large excess of Cl-.
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Formation of Complex Ions
AgCl is also soluble in fairly concentrated aqueous ammonia. AgCl (s) = Ag+ + Cl- Ksp = 1.6 x 10-10 Ag+ + NH3 = Ag(NH3)+ K1 = 2.1 x 103 Ag(NH3)+ + NH3 = Ag(NH3)2+ K2 = 8.2 x 103 Solubility of AgCl in water: 1.3 x 10-5 M. What is solubility in 1M NH3?
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Formation of Complex Ions
For the reaction AgCl(s) + 2 NH3 = Ag(NH3)2+ + Cl-, K = KspK1K2 = (1.6 x 10-10)(2.1 x 103)(8.2 x 103) = 2.8 x 10-3 We assume that little Ag(NH3)+ or Ag+ exists in solution; so [Ag(NH3)2+] = x x2/(1 2x)2 = 2.8 x 10-3 x = 0.042 M. (>> 1.3 x 10-5 in water).
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8a53
Open-Heart Surgery
How does chemistry come into the surgical picture? Life is chemistry; chemistry is life.
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Open-Heart Surgery
During open-heart surgery, the heart is stopped by chilling, and the patient is sustained on a heart-lung machine. After the surgical repairs are complete, the surgeon must close and restart the heart. A significant danger is that air bubbles will be left behind when the heart is closed up.
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Open-Heart Surgery CO2 Flooding
Bubbles lead to air aneurysms: they get stuck in tiny arteries and block blood flow. If this happens in the lungs or brain, the consequences can be deadly. In order to alleviate these problems, the practice of carbon dioxide flooding is widely practiced nowadays.
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Open-Heart Surgery CO2 Flooding
The surgeon passes a large volume of gaseous CO2 over the incision before closing. But CO2 is a suffocant! Why does the surgeon do this? How does it help? The gas fills the cavity, displacing air. Why?
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8a57
Open-Heart Surgery CO2 Flooding
At a given temperature and pressure, what is the relationship between gas density and molar mass M? A. ~ M B. ~ M2 C. ~ 1/M D. Theres no relationship.
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8a58
Open-Heart Surgery CO2 Flooding
PV= nRT = (m/M)RT m = mass
Density = m/V = M(P/RT) This also follows from Avogadros principle. CO2 is 44/29 as dense as air, so it settles into the surgical cavity, displacing air.
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8a59
Open-Heart Surgery - Gas Solubilities
At 1 atm and 35C in pure water: N2: 0.016 g/kg (0.57 mM) O2: 0.035 g/kg (1.1 mM) CO2: 1.17 g/kg (26.6 mM) Why is CO2 more soluble?
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8a60
Open-Heart Surgery - Gas Solubilities
At 1 atm and 35C in pure water: N2: 0.016 g/kg (0.57 mM) O2: 0.035 g/kg (1.1 mM) CO2: 1.17 g/kg (26.6 mM) Why is CO2 more soluble? CO2 + H2O = HCO3- + H+ What is the pH of water saturated with 1 atm. CO2 at 25 C? Ka = 4.3 x 10-7
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Open-Heart Surgery CO2 Flooding
What is the pH of water saturated with 1 atm. CO2 at 25 C? Ka = 4.3 x 10-7 and [CO2] = 26.6 mM. About 4 About 5 About 7 Above 7
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A. B. C. D.
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Open-Heart Surgery CO2 Flooding
Of course, blood, being buffered, cannot drop to such a low pH. Keeping the pH around 7 makes the CO2 more soluble. In effect, the CO2 dissolves in the blood and tissues at the surgical site and is transported, when circulation is restored, to the lungs, where it is exhaled.
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8a63
Open-Heart Surgery CO2 Flooding
Blood in the lungs is at slightly lower pH because binding of oxygen to hemoglobin releases some H+. The lower pH makes CO2 less soluble, so it is efficiently exhaled. Displacing air by CO2 at the surgical site assures that gas remaining in the closed heart will be soluble and will therefore not form bubbles.
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8a64
Problem 8-135
What is the pH of aq. NH4+ X- ( with X- a weak base)? Why isnt it dependent on concentration? NH4+ + X- = NH3 + HX Keq = [NH3][HX]/[NH4+][X-] We find that pH = {pKa(NH4+) + pKa(HX)}/2
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Problem 8-135
A. B. C. D. So what is the pH of ammonium acetate? pKa (NH4+) = 9.25; pKa (HOAc) = 4.75 4.75 9.25 4.50 7.00
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Problem 8-135
At pH 7, what is the ratio of [NH3]/[NH4+]? (Use H-H Equation) 178 1/178 2.25 1/2.25
A. B. C. D.
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Problem 8-135
[acetate]/[acetic acid] is also 178/1. Almost all the material in an ammonium acetate solution is in the form of the ions NH4+ and C2H3O2- . This is also true of substances with both acidic and basic groups in the same molecule: they exist predominantly in +/- forms called zwitterions.
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8a68
Amino Acids
Amino acids are essential nutrients, from which our proteins are assembled. Although commonly written in uncharged forms, they are actually zwitterions under normal conditions:
CH3 O H2N OH H3N
+
CH3 O O
-
alanine
as commonly drawn
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as it exists
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Self-Neutralizing Species
The equation pH = (pK1 + pK2)/2, which we derived for salts of weak acids + weak bases, also applies to solutions of partially neutralized polyprotic acids. They have the same property of selfneutralization: 2 H2PO4- = H3PO4 + HPO4=. For this reaction, K = KA2/KA1 10-5. pH = (pKA1 + pKA2)/2 = (2.12 + 7.21)/2 = 4.67.
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8a70
Ink Jet Chemistry
How is it possible to spray water-based ink onto paper and make a permanent document which doesnt smudge when touched by moist fingers? An application of acid-base chemistry. The ink is at pH 8-10. The dye is in the form of a water-soluble ammonium salt, {Dye}-CO2- NH4+ .
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8a71
Ink Jet Chemistry
When the dye is sprayed onto the paper, the acid in the paper neutralizes some of the salt to form the waterinsoluble {Dye}-COOH. The remaining ammonium ions are in equilibrium with ammonia, which evaporates when the ink is exposed to air, fixing the image on the paper. {Dye}-CO2- NH4+ = {Dye}-CO2H + NH3
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8a72
Pro-Jet Fast Black 2
O O
-
NH4
+
+ NH4
O
-
N N O NH N O
+ NH4
O NH2
O
-
S O
This black dye combines into one molecule groups that absorb all three primary colors.
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8a73