Acids and Bases

Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors

Acid Dissociation Acid Proton Conjugate base HA  H+ + A- Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

HNO2(aq) + H2O(l) NO2-(aq) H3O+(aq) Acid Base Conjugate base Conjugate acid NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Base Acid Conjugate acid Conjugate base

Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large Ka or small Ka? Reactant favored or product favored?

Dissociation Constants: Strong Acids Formula Conjugate Base Ka  Perchloric   HClO4   ClO4-   Very large   Hydriodic   HI   I-   Hydrobromic   HBr   Br-   Hydrochloric   HCl   Cl-   Nitric   HNO3   NO3-   Sulfuric   H2SO4   HSO4-   Hydronium ion   H3O+   H2O   1.0 

Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large Ka or small Ka? Reactant favored or product favored?

Dissociation Constants: Weak Acids Formula Conjugate Base Ka  Iodic   HIO3   IO3-   1.7 x 10-1   Oxalic   H2C2O4   HC2O4-   5.9 x 10-2   Sulfurous   H2SO3   HSO3-   1.5 x 10-2   Phosphoric   H3PO4   H2PO4-   7.5 x 10-3   Citric   H3C6H5O7   H2C6H5O7-   7.1 x 10-4   Nitrous   HNO2   NO2-   4.6 x 10-4   Hydrofluoric  HF   F-   3.5 x 10-4   Formic   HCOOH   HCOO-   1.8 x 10-4   Benzoic   C6H5COOH   C6H5COO-   6.5 x 10-5   Acetic   CH3COOH   CH3COO-   1.8 x 10-5   Carbonic   H2CO3   HCO3-   4.3 x 10-7   Hypochlorous   HClO   ClO-   3.0 x 10-8   Hydrocyanic   HCN   CN-   4.9 x 10-10 

PROBLEM: Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. Hydrochloric acid (HCl) Acetic acid (HC2H3O2) The ammonium ion (NH4+) The anilinium ion (C6H5NH3+) The hydrated aluminum(III) ion [Al(H2O)6]3+

Self-Ionization of Water H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

pH Scale

Relationship between pH and pOH Calculating pH, pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

pH and pOH Calculations

Problem: Calculate [H+] or [OH-] as required for each of the following solutions @ 25oC, and state whether the solution is neutral, acidic, or basic. 1.0 x 10-5M OH- 1.0 x 10-7M OH- 10.0M H+

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1: Write the dissociation equation HC2H3O2  C2H3O2- + H+

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #2: ICE it! HC2H3O2  C2H3O2- + H+ I C E 0.50 - x +x +x 0.50 - x x x

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #3: Set up the law of mass action HC2H3O2  C2H3O2- + H+ E 0.50 - x x x

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #4: Solve for x, which is also [H+] HC2H3O2  C2H3O2- + H+ E 0.50 - x x x [H+] = 3.0 x 10-3 M

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #5: Convert [H+] to pH HC2H3O2  C2H3O2- + H+ E 0.50 - x x x pH = -log(3.0 x 10-3) = 2.52

Percent Dissociation/Ionization Amount dissociated (mol/L) = X 100% Initial concentration (mol/L) So….Calculate the percent dissociation for the previous problem. 60%? JK…How about 0.6%? Better? Tip of the day! For a given weak acid….the percent dissociation INCREASES as the acid becomes more dilute. Larger Percent dissociation for which one? 1.0 M HC2H3O2 or 0.1 M HC2H3O2?

HC9H7O4  C9H7O4- + H+ HANDOUT!!! DO ALL WORK ON THIS!!! Aspirin, is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin has a total volume of 350.0 mL and contains 1.26 g of aspirin. The pH of the solution is found to be 2.60 Calculate Ka for aspirin.culate the Ka for aspirin Step #1: Write the dissociation equation HC9H7O4  C9H7O4- + H+

I C E HC9H7O4  C9H7O4- + H+ Step #2: Determine concentration of acid! 1.26 g x 1 mol .350L 180.15 g = 0.02M HC9H7O4 Step #3: ICE it! HC9H7O4  C9H7O4- + H+ I C E .02 - x +x +x .02 - x x x

HC9H7O4  C9H7O4- + H+ E Step #4: Determine the [H+] pH = 2.60 [H+] = 2.5 x 10-3 M Step #5: Set up the law of mass action HC9H7O4  C9H7O4- + H+ E .02 - x x x 2 x2 .0025 Ka = Ka = Ka = 3.6 x 10-4 .02 - x .02-.0025

Dissociation of Strong Bases MOH(s)  M+(aq) + OH-(aq) Strong bases are metallic hydroxides Group I hydroxides (NaOH, KOH) are very soluble Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble pH of strong bases is calculated directly from the concentration of the base in solution

Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O  BH+ + OH- (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you? Base Formula Conjugate Acid Kb Ammonia   NH3  NH4+  1.8 x 10-5   Methylamine  CH3NH2  CH3NH3+  4.38 x 10-4   Ethylamine  C2H5NH2  C2H5NH3+  5.6 x 10-4   Diethylamine  (C2H5)2NH  (C2H5)2NH2+  1.3 x 10-3   Triethylamine   (C2H5)3N   (C2H5)3NH+  4.0 x 10-4   Hydroxylamine  HONH2   HONH3+    1.1 x 10-8   Hydrazine H2NNH2  H2NNH3+    3.0 x 10-6   Aniline  C6H5NH2   C6H5NH3+    3.8 x 10-10   Pyridine  C5H5N   C5H5NH+    1.7 x 10-9 

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #1: Write the equation for the reaction NH3 + H2O  NH4+ + OH-

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #2: ICE it! NH3 + H2O  NH4+ + OH- I C E 0.50 - x +x +x 0.50 - x x x

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #3: Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #4: Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #5: Convert [OH-] to pH NH3 + H2O  NH4+ + OH- E 0.50 - x x x

ANOTHER TIP OF THE DAY Ka x Kb = Kw

The calculation of concentration and pH for weak acids is more complex than for strong acids due to: The incomplete ionization of weak acids. The low Ka value for strong acids. The more complex atomic structures of strong acids. D. The low percent ionization of strong acids. E. The inconsistent Kb value for strong acids. Answer: A Strong acids ionize 100%, therefore, the [H+] is equal to the initial concentration of the strong acid. This is not the case for weak acids. Since not all of the acid ionizes, it is necessary to determine both how much of the acid forms H+ and how much is left in molecular form.

The general reaction of an acid dissolving in water may be shown as HA (aq) + HOH (l) H3O (aq) + A-(aq) A conjugate acid base pair for this reaction is: HA and HOH HA and A- HOH and A- H3O+ and A- HA and H3O+ Answer: B The conjugate acid differs from its conjugate base by a proton, (H+). The acid form has the proton (HA in this example) and the base form has lost the proton (A- in this case).

Strong acids are those which Have an equilibrium lying far to the left. Yield a weak conjugate base when reacting with water Have a conjugate base which is a stronger base than water D. Readily remove the H+ ions from water. E. Are only slightly dissociated (ionized) at equilibrium. Answer: B Strong acids give up their protons (H+ ions) easily. If their conjugate base form were strong, then the acid formed would hold strongly to the H+, which would not be a characteristic of a strong acid.

When calculating the pOH of a hydrofluoric acid solution (Ka = 7.2 x 10-4) from its concentration, the contribution of water ionizing (Kw = 1.0 x 10-14) is usually ignored because Hydrofluoric acid is such a weak acid. Hydrofluoric acid can dissolve glass The ionization of water provides relatively few H+ ions. The [OH-] for pure water is unknown. The conjugate base of HF is such a strong base. Answer: C By comparing the Ka for HF with Kw (for water), you can see that even though hydrofluoric acid is a very weak acid it is a much stronger acid than is water, so much so that the H+ ion contribution of water may be ignored.

The percent dissociation (percent ionization) for weak acids Is always the same for a given acid, no matter what the concentration. B. Usually increases as the acid becomes more concentrated. C. Compares the amount of acid that has dissociated at equilibrium with the initial concentration of the acid. D. May only be used to express the dissociation of weak acids. E. Has no meaning for polyprotic acids. Answer: C the percent dissociation does change with the concentration of the acid.

The [OH-] of a certain aqueous solution is 1.0 x 10-5M. The pH of this same solution must be 1.0 x 10-14 5.00 7.00 9.00 12.00 Answer: D Referring to pH + pOH = 14.00 at 25oC, if [OH-] = 1 x 10-5M, pOH = 5, pH = 14-5.0 =9.0

In many calculations for the pH of a weak acid from the concentration of the acid, an assumption is made that often takes the form [HA]o – x = [HA]o This Is valid because x is very small compared to the initial concentration of the weak acid. B. Is valid because the concentration of the acid changes by such large amounts. C. Is valid because actual value of x cannot be known. D. Is valid because pH is not dependent upon the concentration of the weak acid. E. Approximation is always shown to be valid and so need not be checked. Answer: A. We can expect x to be very small (as indicated by the low value for Ka for weak acids), so that the concentration of the weak acid does not significantly change from its initial value. As the final step in such problems, you must determine that the change in the initial [HA] is less than 5% for the approximation to be considered valid.

HA is a weak acid which is 4.0% dissociated at 0.100M. Determine the Ka for this acid 0.0040 0.00016 0.040 1.6 16.5 Answer: B [HA]o = 0.100M [H+] = [A-] = 0.004M Ka = [H+][A-]/[HA] = (0.004 x 0.004) / 0.100 = (4 x 10-3)(4 x 10-3) / (10-1) = 1.6 x 10-4

SALTS

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is from a strong base, anion from a strong acid KCl, KNO3 NaCl NaNO3 Both ions are neutral Neutral These salts simply dissociate in water: KCl(s)  K+(aq) + Cl-(aq)

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is from a strong base, anion from a weak acid NaC2H3O2 KCN, NaF Cation is neutral, Anion is basic Basic The basic anion can accept a proton from water: C2H3O2- + H2O  HC2H3O2 + OH- base acid acid base

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH4Cl, NH4NO3 Cation is acidic, Anion is neutral Acidic The acidic cation can act as a proton donor: NH4+(aq)  NH3(aq) + H+(aq) Acid Conjugate Proton base

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH4C2H3O2 NH4CN Cation is acidic, Anion is basic See below IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

Acid-Base Properties of Salts Type of Salt Examples Comment pH of solution Cation is a highly charged metal ion; Anion is from strong acid Al(NO3)2 FeCl3 Hydrated cation acts as an acid; Anion is neutral Acidic Step #1: AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq) Salt water Complex ion anion Step #2: Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq) Acid Conjugate base Proton

Summary of Acid/Base Properties of Salts: Cation From Anion From Solution Strong Base Weak Base Strong Acid Weak Acid Neutral Basic Acidic Must compare K values!

Effect of Structure on Acid-Base Properties

Question: Do I really know this stuff? Answer: Let’s find out! Good luck

Let’s try this: Determine the pH of a 0.10M solution of NaC2H3O2. Ka for HC2H3O2 is 1.8 x 10-5 Good luck

PRACTICE TIME! Please complete problems: #44,50,52, and 64a-d on a separate sheet of paper (Will be collected at end of block!) You may work with other students AT YOUR WORKSTATION!

PRACTICE TIME! pp. 673-75: #28,30,32,36,38,40,44,50 52 and 55. PLEASE BEGIN TO SOLVE THE FOLLOWING PROBLEMS RELATING TO: NATURE OF ACIDS-n-BASES, AUTOIONIZATION OF WATER, pH SCALE AND THE SOLUTIONS OF ACIDS. pp. 673-75: #28,30,32,36,38,40,44,50 52 and 55.

Please complete problems: on a separate sheet of paper PRACTICE TIME! Please complete problems: Pp 675-76 64a-d and # 78,82,84,86,92 and 94 on a separate sheet of paper

PRACTICE TIME! Please complete problems: #44,50,52,58,59 and 64a-d on a separate sheet of paper (Will be collected at end of block!)

PRACTICE TIME! PLEASE BEGIN TO SOLVE THE FOLLOWING PROBLEMS RELATING TO: NATURE OF ACIDS-n-BASES, AUTOIONIZATION OF WATER, pH SCALE AND THE SOLUTIONS OF ACIDS. pp. 673-75: #28-30,32,33-36,38,40,44,50 52,55,58,59, and 64a-d Please complete problems #28-30,32,33-36,38 and 40 on a separate sheet of paper (Will be collected at end of block!)

PRACTICE TIME! pp. 675-76: #78,82,84,86,92 and 94

PRACTICE TIME! pp. 674-75: # 43-46,50,52,55,58, 59,62 and 64(a-d)