ACID-BASE EQUILIBRIA (Cont.) Chapter 15

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Chapter 15

ACID-BASE EQUILIBRIA (Cont.)

1 Summary : pH calculations of strong acids and strong bases

1. At relatively high [S.A] or high [S.B.] , i.e. ≥ 10 -6 M, pH is calculated from [S.A.] or [S.B.]

2. In very dilute [S.A] or [S.B.] , i.e. ≤ 10 -8 M, dissociation of water is more important, so the pH is 7 .

Question: How do we know if a given acid is strong or weak?

Know the following by heart: You have to be able to name them too.

There are only six (6) strong acids to remember:

HCl, HBr, HI, H 2SO 4, HNO 3, HClO 4 Perchloric acid

There are only six (6) strong bases to remember:

LiOH, NaOH, KOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2

2 Determining the pH of solutions

Solutions of weak acids (W.A.) Ionization is incomplete; Equilibrium is set up

Ionization (dissociation) equation:

Ka + - HA (aq) + H2O(l) H3O (aq) + A (aq)

Ka + - Or simply: HA (aq) H (aq) + A (aq)

where K a is the acid ionization (or dissociation) constant

[H +] [A -] K = a The Ka for W.A. is < 1 . [HA] 5

3 - Cont.

Determining the pH of Weak Acids

+ Example : What is the H 3O concentration and pH of a 0.10 M -8 solution of hypochlorous acid, HOCl, Ka = 3.5 x 10 ?

Note that the K a is a small number (i.e. HOCl is a weak acid. Start with the equilibrium equation:

Ka + - HOCl + H 2O H 3O + OCl ⇔ + - or HOCl (aq) H (aq) + OCl (aq)

HOCl is W.A.; only a fraction of 0.10 M HOCl will dissociate, i.e. we can set up our ICE table as:

+ - HOCl + H 2O(l) H3O + OCl

I: 0.10 0 0 C:– x + x + x E: (0.10 – x) x x 8

4 The K a expression is:

- + (x)(x) [OCl ] [H 3O ] Ka = = [HOCl] (0.10 – x)

x2 3.5 x 10 -8 = (0.10 – x)

Assuming x << 0.10 (since K a is very small) and solving for x, we get: -5 + - x = 5.9 2 x 10 M = [H 3O ] = [OCl ] Check: Since 5.9 x 10 -5 << 0.10 our assumption is valid. We calculate pH as: pH = - log (5.9 x 10 -5) = 4.23 2 9

Weak Acid Equilibria – Cont. Exercise : Calculate the pH of a 1.50 x 10 -2 M formic acid,

HCO 2H. (pK a = 3.745). Answer: pH = 2.808

HINT : You must first determine K a from pK a. How?

5 11

Determining K a from Measured pH Example : A student measured the pH of a 0.10 M solution of 0 formic acid, HCOOH, to be 2.38 at 25 C. Calculate K a for 0 -4 formic acid at 25 C. Answer: K a = 1.8 x 10

6 13

Percent Ionization of a Weak Acid

The percent ionization of a weak acid HA refers to the percentage of HA in ionized form, H + (and A -)

[H O+] % ionization = 3 eq x 100

[HA] initial

Example : What is percent of HOCl in ionized form in the previous example? {Recall that [H +] = 5.92 x 10 -5 M and initial [HOCl] = 0.10 M}

Answer: 0.059 % ionized

7 Weak Base Equilibria: Hydrolysis

Recall that B-L bases are H + acceptor. In aqueous solutions, a weak base “grabs” a H + from water, forming its conjugate acid and OH -. This reaction is called hydrolysis . For a monobasic species, B, the hydrolysis equation is:

Kb + - B + H 2O BH + OH Conj. acid Hydroxide Weak base (H + acceptor) ion The equilibrium constant expression for this reaction is:

[BH +] [OH -] Kb = [B] 15

Hydrolysis – Cont. Exercise: Write the hydrolysis reaction of each of the following bases: 2- 2- Carbonate, CO 3 Sulfate, SO 4 - Phosphate, PO 3- Bicarbonate, HCO 3 4

Kb CO 2- + H O HCO - + OH - Note: Add a +1 charge 3 2 3 to the base to determine Kb charge of conjugate acid - - HCO 3 + H2O H2CO 3 + OH

Kb 2- - - SO 4 + H 2O HSO 4 + OH

Kb PO 3- + H O HPO 2- + OH - 4 2 4 16

8 Determining pH of Solutions of Weak Bases

Ammonia is the most commonly used weak base. The hydrolysis of ammonia is written as:

Kb + - NH 3 + H 2O NH 4 + OH Weak base

The K b expression for NH 3 can be written as: + - [NH 4 ][OH ] Kb = [NH 3] Exercise: Calculate the concentration of OH - in a 0.15 M -5 0 solution of NH 3. K b for NH 3 is 1.8 x 10 at 25 C. What is the pH of this solution? 17

Kb + - NH 3 + H 2O NH 4 + OH WB I: 0.15 0 0 C:– y + y + y E: (0.15 – y) y y

Kb and pH can be expressed as:

y2 and pH = 14.00 – pOH Kb = - (M B – y) = 14.00 – (-log [OH ]

or pH = 14.00 - log (y) 18

9 Relationship Between K a and K b Recall: Conjugate species = related by one H (as H +) The stronger the acid the weaker its conj. base Similarly, the stronger the base the weaker its conj. acid

>> 1 + - Weakest Strongest HClO 4 + H 2OH3O + ClO 4 acid base >> 1 + - HCl + H 2OH3O + Cl

7.11 x 10 -3 + - H3PO 4 + H 2OH3O + H2PO 4

1.75 x 10 -5 + - HC 2H3O2 + H 2OH3O + C2H3O2

5.70 x 10 -10 Weakest + + Strongest NH 4 + H 2OH3O + NH 3 acid base 19

Relationship Between K a and K b

The K a of an acid is also related to the K b of its conj. base, i.e.

For conjugate acid-base pairs :

Ka x K b = Kw pK a + pK b = 14.00

Exercises: 2- - -11 (1) Calculate the K b for CO 3 if Ka for HCO 3 is 5.6 x 10 . (2) Hydrofluoric acid, HF, has a pK a of 3.17. Calculate K b for fluoride ion, F -. -4 -11 Answers: (1) K b = 1.8 x 10 ; (2) K b = 1.5 x 10

10 Importance of pH Control

What if pH is not controlled?

Hyperacidic stomach o o

Highly acidic soil => low crop yield

Air pollutants => acid rain Acidic lakes/rivers => fish kills; erosion of statues; vegetation dies

There must be a way to control pH 22

11 Acid rain: a significant problem in the Northeastern U.S.

Damages Caused by Acid Rain

Damage to aquatic life

http://www.sciencemaster.com/ju mp/earth/acid_rain.php

In some sensitive lakes and streams, acidification has completely eradicated fish species , such as the brook trout , leaving these bodies of water barren. In fact, hundreds of the lakes in the Adirondacks surveyed in the NSWS have acidity levels indicative of chemical conditions unsuitable for the survival of sensitive fish species. 24

12 Damage to vegetation and outdoor artwork (limestone)

Ways of Controlling pH

1) Through neutralization reactions = acid-base reactions

Ex. HCl + antacid (a carbonate) => salt + H2O Excess A base stomach acid Always produced in neutralization

Source: C. Baird and W. Gloffke, “Chemistry In Your Life.” New York: Freeman, 2003. (p. 427) 26

13 Writing Neutralization Reactions

1) The metal from the base is always written (or named ) first . 2) The metal from the base replaces the hydrogen of the acid.

NaCl HOH (or H O) HCl + Na OH ______+ ______2 salt

KNO 3 H2O HNO 3 + KOH ______+ ______salt

HCN + NaOH NaCN ______+ ______H2O salt

Ways of Controlling pH (Cont.)

2) Through the action of buffers

Substances that resist drastic changes in pH. HOW? They consist of a mixture of a weak acid and its conjugate base (i.e. an acid-base pair).

Ex. The pH of blood is maintained by a biological buffer, a mixture of carbonic acid and bicarbonate ions .

acid base component component

14 Ways of Controlling pH: Action of Buffers (Cont.)

Hypoventillation :

More CO 2 in lungs, more carbonic acid in Hyperventillation : blood = acidosis Less CO 2 in lungs, less carbonic acid in blood = alkalosis

• During alkalosis (blood pH rises) the acid component (H 2CO 3) of our biological buffer neutralizes the excess base and restores the pH to around 7.4, the normal pH of blood.

- • During acidosis (blood pH drops), the base component (HCO 3 )

neutralizes the excess acid and restores the pH to around 7.4. 29

Acid-Base Properties of Salts

15 Acid-Base Properties of Salts

Recall: Acids and bases can be electrically neutral (ex. HCl, + 2- NH 3) or charged (ex. NH 4 , CO 3 ) Implication : Salts , ionic compounds formed from neutralization reactions, can be acidic, basic or neutral solutions.

The anion or the cation of a salt, or both, can react with water (= hydrolysis reaction) The pH of salt solution depends on the strengths of the original acids and bases

Salts – Cont.

Q. What is the pH of an aqueous solution of a salt? The pH of the salt depends on the strengths of the original acids and bases

Acid Base Salt pH Example of salt Strong Strong 7 (Neutral) NaCl, KNO 3, CaBr 2 Weak Strong > 7 (Basic) Na 2CO 3, K 3PO 4 Strong Weak < 7 (Acidic) NH 4Cl, CH 3NH 3Br Weak Weak Depends on NH ClO, (NH ) CO which is stronger 4 4 2 3

16 Salts – Cont. Predicting whether a salt solution will be acidic , basic or neutral :

1. Break the salt down into its cation and anion.

2. Identify (+) and (-) ions as either: Spectator ion* (SI) Weak acid (WA) Weak base (WB) A spectator ion , SI, does not react with water (i.e. it does not hydrolyze) and does not change the pH of water because it is neither acidic nor basic.

3. Add the two effects to decide on effect on solution pH

Exercise: Show that Na 2CO 3 hydrolyzes in water to form a basic solution. Work: 2 Na + (1) Ionize Na 2CO 3 -2 CO 3

(2) ID ions as SI, WA or WB

Na + = SI Comes from SB NaOH; no reaction with water! Na 2CO 3 -2 CO 3 = WB (There’s no acidic H!) Will hydrolyze to form OH - -2 Hydrolysis of CO 3 : -2 - - CO 3 + H 2O HCO 3 + OH Base (will grab H+ from water) Basic! 34

17 Exercise: Show that ammonium chloride hydrolyzes in water to form an acidic solution.

Work: + NH 4 = WA NH 4Cl Cl - SI – comes from a strong acid, HCl

+ Hydrolysis of NH 4 :

+ + NH 4 + H2O NH 3 + H3O

Acidic!

Exercise: 1) Predict whether each of the following salts will be acidic, basic or neutral in water. Show your work, including applicable hydrolysis equations.

SrCl 2

NaCHO 2

AlBr 3

NH 4NO 3

18 Exercise 2) Calculate the pH of each of the following aqueous

solutions: (a) 0.100 M NaCHO 2, (b) 0.100 M AlBr 3 and (c) 0.100 M NH 4NO 3 WORK: - + (a) pH of 0.100 M CHO 2 (Na is S.I.)

W.B. = will hydrolyze to form OH-

- - CHO 2 (aq) + H 2O (l) HCHO 2 (aq) + OH (aq) I 0.100 0 0 C- x + x + x E (0.100 – x) x x

Next we need Kb => use K a (HCHO 2) and Kw

-4 37 Table: K a (HCHO 2) = 1.80 x 10

WORK – Cont. - (a) pH of 0.100 M CHO 2

−14 - 1.00 x 10 −11 Table: K b (CHO 2 ) = − = 5.56 x 10 1.80 x 10 4

- - Hydrolysis : CHO 2 (aq) + H 2O (l) HCHO 2 (aq) + OH (aq)

2 − x K = 5.56 x 1011 = Assume x << 0.100 b (0.100 - x)

− − x = 2.36 x 106 M = [OH ] pOH = 5.63; pH = 8.37

19 WORK: - (b) pH of 0.100 M AlBr 3 (Br is S.I. – comes from HBr, SA)

3+ 3+ Al is a WA which exists as Al(H 2O) 6 in water (See K a table) 3+ -5 Ka for Al(H 2O) 6 = 1.4 x 10

3+ 3+ + Dissociation: Al(H 2O) 6 (aq) + H 2O(l) Al(H 2O) 5(OH) (aq) + H3O (aq)

I 0.100 0 0 C- x + x + x E (0.100 – x) x x

2 − x K = 1.4 x 105 = Assume x << 0.100 a (0.100 - x)

−3 + x = 1.18 x 10 M = [H3 O ]

pH = 2.93 39

WORK: - (c) pH of 0.100 M NH 4NO 3 (NO 3 is S.I. – comes from HNO 3, SA)

+ NH 4 is a WA (CA of WB NH 3; From K a and K b table, -5 Kb of NH 3 = 1.76 x 10 )

−14 1.00 x 10 −10 Ka (NH+ ) = −5 = 5.68 x 10 4 1.76 x 10

+ + Ionization: NH 4 (aq) + H 2O (l) NH 3 (aq) + H3O (aq)

I 0.100 0 0 C- x + x + x E (0.100 – x) x x 2 −10 x Ka = 5.68 x 10 = Assume x << 0.100 (0.100 - x) 40

20 −6 + x = 7.54 x 10 M = [H3 O ] pH = 5.12

Diprotic Acids and Bases

General formulas:

H2A = fully acidic form HA - = intermediate form; amphoteric

A2- = fully basic or fully deprotonated form

21 Equilibria involved : Diprotic Acids and Bases

Diprotic Acid , H 2A K +→a1 − + + First dissociation: HAHO2 2← HA HO 3 K −+→a2 −2 + + Second dissociation: HA HO2← A HO 3

Dibasic species , A 2- K 2−+→b1 − + − First hydrolysis: A H2 O← HA OH K − +→b2 + − Second hydrolysis: HA HO2← HAOH 2

Q. How do we calculate K b1 and K b2 from K a values?

- Note that H 2A and HA species in the K a1 expression both appear - 2- in the K b2 expression. Similarly, the conjugates HA and A in the Ka2 expression both appear in the K b1 expression.

Thus,

Proof: K +→a1 − + + HAHO2 2← HA HO 3 + K − +→b2 + − HA HO2← HAOH 2

K 2 →w ++ − H2 O← H 3 O OH Kw = K a1 x Kb2

22 pH Calculation : Diprotic Acids and Bases

1. The fully acidic form, H 2A

Approximation: In a solution of H 2A (Ex. 0.050 M H 2SO 3), nd the 2 dissociation is usually negligible that H2A behaves as a monoprotic acid . Also, [A 2-] ≈ 0 M.

Calculation of pH and [species] K +→a1 − + + HAHO2 2← HA HO 3 Equil: F-x x x

x2 K = a1 F− x

pH Calculation: Fully acidic form (H 2A) – Cont.

Problem 1: Find the pH of a 0.050 M H 2SO 3 solution. -2 -8 Ka1 = 1.23 x 10 ; K a2 = 6.6 x 10 K +→a1 − + + HSO232 HO← HSO 33 HO Equil: 0.050-x x x

2 −2 x K =1.23x 10 = x cannot be ignored a1 − (0.050x ) since K a1 isn’t too small

2− 2 − 4 x+1.23x 10 x − 6.1510 x Solve for x using quadratic equation =−2 = + = − x1.9410x M [ HO3 ][ HSO 3 ] pH =1.71

23 pH Calculations : Diprotic systems – Cont.

2. The fully basic form, A 2-

2- Approximation: In a solution of A (Ex. 0.050 M Na 2SO 3), the 2 nd hydrolysis is usually negligible that A2- behaves

as a monobasic species. Also, [H 2A] ≈ 0 M. Calculation of pH and [species] K 2−+→b1 − + − A H2 O← HA OH Equil: F-y y y y2 K = pOH = -log ( y) b1 F− y pH = 14 - pOH

Recall: K b1 = Kw/K a2

Problem 2: Find the pH of 0.050 N Na 2SO 3 solution: -2 Ka1 = 1.23 x 10 ; -8 Ka2 = 6.6 x 10

Answer: pH = 9.94; [OH -] = 8.7 x 10 -5 M

24 pH Calculations : Diprotic systems – Cont.

3. The intermediate (amphoteric) form, HA - HA - can act as an acid or a base Q. What is the predominant species in a solution of HA -?

Compare K a2 and K b2 equilibria: K − +→a2 2− + + Dissociation: HA HO2← A HO 3 K Hydrolysis: − +→b2 + − HA H2 O← H 2 A OH

- 2- HA will dissociate/hydrolyze to form A and H 2A

No pH calculation involving HA -

The intermediate form, HA - (Cont.)

Calculation of pH and [species]

Where

K1= K a1 K2 = K a2 F = F HA- pH = -log [H +]

Quick check: pH = ½ (pK 1 + pK 2)

2- + Solve for [H 2A] and [A ] using [H ] above and K 1 & K 2 equilibria

25 Trends in Acidity

1. Effect of bond strength on binary acids, HX Acidity increases with decreasing bond strength (top to bottom within a group) Easier to pull out H as H +

2. Effect of electronegativity of X in binary acids, HX Acidity increases with increasing electronegativity (left to right across the table) H-X bond becomes more polar; easier to pull out H + 52

26 Oxyacids = bonded to H, X and O Acidity increases with increasing number of O bonded to X

X = Cl Increasing Increasing acidity