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 HA  H + + A - Acid Proton Conjugate base Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion)

HNO 2 (aq) + H 2 O(l) NO 2 - (aq) H 3 O+(aq) Acid Base Conjugate base Conjugate acid NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) BaseAcid Conjugate base Conjugate acid

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

Dissociation Constants: Strong Acids AcidFormula Conjugate Base KaKa Perchloric HClO 4 ClO 4 - Very large Hydriodic HI I - Very large Hydrobromic HBr Br - Very large Hydrochloric HCl Cl - Very large Nitric HNO 3 NO 3 - Very large Sulfuric H 2 SO 4 HSO 4 - Very large Hydronium ion H 3 O + H 2 O 1.0

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

Dissociation Constants: Weak Acids AcidFormula Conjugate Base KaKa Iodic HIO 3 IO x Oxalic H 2 C 2 O 4 HC 2 O x Sulfurous H 2 SO 3 HSO x Phosphoric H 3 PO 4 H 2 PO x Citric H 3 C 6 H 5 O 7 H 2 C 6 H 5 O x Nitrous HNO 2 NO x Hydrofluoric HF F x Formic HCOOH HCOO x Benzoic C 6 H 5 COOH C 6 H 5 COO x Acetic CH 3 COOH CH 3 COO x Carbonic H 2 CO 3 HCO x Hypochlorous HClO ClO x Hydrocyanic HCN CN x

PROBLEM: Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. a.Hydrochloric acid (HCl) b.Acetic acid (HC 2 H 3 O 2 ) c.The ammonium ion (NH 4 + ) d.The anilinium ion (C 6 H 5 NH 3 + ) e.The hydrated aluminum(III) ion [Al(H 2 O) 6 ] 3+

Self-Ionization of Water H 2 O + H 2 O  H 3 O + + OH - At 25 , [H 3 O + ] = [OH - ] = 1 x K w is a constant at 25  C: K w = [H 3 O + ][OH - ] K w = (1 x )(1 x ) = 1 x

Calculating pH, pOH pH = -log 10 (H 3 O + ) pOH = -log 10 (OH - ) Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH

pH and pOH Calculations

pH Scale

Problem: Calculate [H + ] or [OH - ] as required for each of the following 25 o C, and state whether the solution is neutral, acidic, or basic. a.1.0 x M OH - b.1.0 x M OH - c.10.0M H +

Do Now: Please review your notes on yesterday’s acid/base problems. Also, get out your homework and notebook in preparation for today’s exciting continuation of Chapter 14! Objective: Describe equilibria involving strong and weak bases. Explain the method for calculating pH for basic solutions. Describe the dissociation equilibria of acids with more than one acidic proton. Homework: Acid/Base problem set

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? Step #1: Write the dissociation equation HC 2 H 3 O 2  C 2 H 3 O H +

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? Step #2: ICE it! HC 2 H 3 O 2  C 2 H 3 O H + I C E x +x x xx

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? Step #3: Set up the law of mass action HC 2 H 3 O 2  C 2 H 3 O H xxx E

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? Step #4: Solve for x, which is also [H + ] HC 2 H 3 O 2  C 2 H 3 O H xxx E [H + ] = 3.0 x M

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? Step #5: Convert [H + ] to pH HC 2 H 3 O 2  C 2 H 3 O H xxx E pH = -log(3.0 x ) = 2.52

Percent Dissociation/Ionization = Amount dissociated (mol/L) Initial concentration (mol/L) X 100% 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 HC 2 H 3 O 2 or 0.1 M HC 2 H 3 O 2 ? So….Calculate the percent dissociation for the previous problem. 60%? JK…How about 0.6%? Better?

Aspirin, is a weak organic acid whose molecular formula is HC 9 H 7 O 4. An aqueous solution of aspirin has a total volume of mL and contains 1.26 g of aspirin. The pH of the solution is found to be 2.60 Calculate the Ka for aspirin Step #1: Write the dissociation equation HC 9 H 7 O 4  C 9 H 7 O H +

Step #3: ICE it! HC 9 H 7 O 4  C 9 H 7 O H + I C E x +x.02 - x xx Step #2: Determine concentration of acid! 1.26 g x 1 mol.350L g = 0.02M HC 9 H 7 O 4

Step #5: Set up the law of mass action.02 - xxx E HC 9 H 7 O 4  C 9 H 7 O H + Step #4: Determine the [H + ] pH = 2.60[H + ] = [H + ] = 2.5 x M Ka =.02 - x x2x2 Ka = Ka = 3.6 x 10 -4

The calculation of concentration and pH for weak acids is more complex than for strong acids due to: A.The incomplete ionization of weak acids. B.The low Ka value for strong acids. C.The more complex atomic structures of strong acids. D. The low percent ionization of strong acids. E. The inconsistent K b 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) H 3 O (aq) + A - (aq) A conjugate acid base pair for this reaction is: A.HA and HOH B.HA and A - C.HOH and A - D.H3O + and A - E.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 A.Have an equilibrium lying far to the left. B.Yield a weak conjugate base when reacting with water C.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 ) from its concentration, the contribution of water ionizing (Kw = 1.0 x ) is usually ignored because A.Hydrofluoric acid is such a weak acid. B.Hydrofluoric acid can dissolve glass C.The ionization of water provides relatively few H + ions. D.The [OH - ] for pure water is unknown. E.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 A.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 M. The pH of this same solution must be A.1.0 x B.5.00 C.7.00 D.9.00 E Answer: D Referring to pH + pOH = at 25 o C, if [OH - ] = 1 x M, pOH = 5, pH = =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 A.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 A B C D.1.6 E.16.5 Answer: B [HA] o = 0.100M [H + ] = [A - ] = 0.004M Ka = [H + ][A - ]/[HA] = (0.004 x 0.004) / = (4 x )(4 x ) / (10 -1 ) = 1.6 x 10 -4

Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ? Step #1: Write the equation for the reaction NH 3 + H 2 O  NH OH -

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ? Step #2: ICE it! I C E x +x x xx NH 3 + H 2 O  NH OH -

A Weak Base Equilibrium Problem Step #3: Set up the law of mass action xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ? NH 3 + H 2 O  NH OH -

A Weak Base Equilibrium Problem Step #4: Solve for x, which is also [OH - ] xxx E [OH - ] = 3.0 x M NH 3 + H 2 O  NH OH - What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ?

A Weak Base Equilibrium Problem Step #5: Convert [OH - ] to pH xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ? NH 3 + H 2 O  NH OH -

ANOTHER TIP OF THE DAY K a x K b = K w

Dissociation of Strong Bases  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 MOH(s)  M + (aq) + OH - (aq)

Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH OH - K b = 4.38 x 10 -4

K b for Some Common Weak Bases BaseFormula Conjugate Acid KbKb Ammonia NH 3 NH x Methylamine CH 3 NH 2 CH 3 NH x Ethylamine C 2 H 5 NH 2 C 2 H 5 NH x Diethylamine (C 2 H 5 ) 2 NH (C 2 H 5 ) 2 NH x Triethylamine (C 2 H 5 ) 3 N (C 2 H 5 ) 3 NH x Hydroxylamine HONH 2 HONH x HydrazineH 2 NNH 2 H 2 NNH x Aniline C 6 H 5 NH 2 C 6 H 5 NH x Pyridine C 5 H 5 N C 5 H 5 NH x Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH OH - K b = 4.38 x 10 -4

K b for Some Common Weak Bases BaseFormula Conjugate Acid KbKb Ammonia NH 3 NH x Methylamine CH 3 NH 2 CH 3 NH x Ethylamine C 2 H 5 NH 2 C 2 H 5 NH x Diethylamine (C 2 H 5 ) 2 NH (C 2 H 5 ) 2 NH x Triethylamine (C 2 H 5 ) 3 N (C 2 H 5 ) 3 NH x Hydroxylamine HONH 2 HONH x HydrazineH 2 NNH 2 H 2 NNH x Aniline C 6 H 5 NH 2 C 6 H 5 NH x Pyridine C 5 H 5 N C 5 H 5 NH x Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

SALTS

Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a strong acid KCl, KNO 3 NaCl NaNO 3 Both ions are neutral Neutral These salts simply dissociate in water: KCl(s)  K + (aq) + Cl - (aq)

Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a weak acid NaC 2 H 3 O 2 KCN, NaF Cation is neutral, Anion is basic Basic C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH- base acid acid base The basic anion can accept a proton from water:

Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH 4 Cl, NH 4 NO 3 Cation is acidic, Anion is neutral Acidic NH 4 + (aq)  NH 3 (aq) + H + (aq) Acid Conjugate Proton base The acidic cation can act as a proton donor:

Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH 4 C 2 H 3 O 2 NH 4 CN Cation is acidic, Anion is basic See below  IF K a for the acidic ion is greater than K b for the basic ion, the solution is acidic  IF K b for the basic ion is greater than K a for the acidic ion, the solution is basic  IF K b for the basic ion is equal to K a for the acidic ion, the solution is neutral

Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is a highly charged metal ion; Anion is from strong acid Al(NO 3 ) 2 FeCl 3 Hydrated cation acts as an acid; Anion is neutral Acidic Step #1: AlCl 3 (s) + 6H 2 O  Al(H 2 O) 6 3+ (aq) + Cl - (aq) Salt water Complex ion anion Step #2: Al(H 2 O) 6 3+ (aq)  Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) Acid Conjugate base Proton

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

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 NaC 2 H 3 O 2. Ka for HC 2 H 3 O 2 is 1.8 x Good luck

Effect of Structure on Acid-Base Properties