Chem Chem 103103Lecture 2b

Acids and Bases3

Last time:Last time:

Today:Today:

1. Acids and Bases 2:a) self ionization of waterb) Relative strengths of acids/conjugatesc) Ka and Kb equilibriad) pH scale

1. Summary of previous lecture2. Calculating pH: 5 scenarios3. Acid base titration

No office hours this FridayI will not be holding office hours this coming Friday.

Ch 103 Seating arrangementIn preparation for group work and midterm exams:

Seating arrangement enforced for extra credit work.(if in wrong row, no extra credit)

If your last name starts with please sit in row

A, B or C A (front)

D, E, F, G, H, I, J B

K, L, M C

N, O, P, Q, R, Sa D

Si, T, U, V, W, X, Y, Z E

Summary (so far…)[H3O+] [ OH-] = 1.0 x 10-14

pH = -log[H3O+] and pOH = -log[OH-]

pH + pOH = 14.00

Ka = [H+][A-]/[HA] Kb = [HA][OH-]/[A-]

KaKb = 1.0 x 10-14

[H+] = 10-pH [OH-] = 10-pOH

pKa + pKb = 14.00

Sample problem

A solution has a pH of 2.10. Determine the following:

a) [H+]=?

[H+] = 10-2.10 = 7.9x10-3 M

b) pOH =?

pOH = 14.00 - pH = 14.00 -2.10 = 11.90

c) [OH-] = ?

[OH-] = 10-pOH = 10-11.90 = 1.3 x 10-12 M

Calculating pH of a solution: 5Calculating pH of a solution: 5scenariosscenarios

5 general scenarios involving pH calculations. MASTER thesecalculations. (simple monoprotic acids/bases)

1) Strong acid solutions -use: pH = - log[acid]

2) Strong base solutions -use: pH = 14.00+ log[base]

3) Pure weak acid sol’ns -use: Ka equilibrium, ICE calculation of [H+]

4) Pure weak base sol’ns -use: Kb equilibrium, ICE calculation of [OH-]

5) Buffer solutions: -use Ka equil. ICE calc’n of [H+]

OR, use buffer equation: pH = pKa + log {[base]/[acid]}

Calculations for salt solutionsCalculations for salt solutions

A “salt” is a soluble ionic compound. It dissociates. It’s anelectrolyte. You must recognize the acid or base!

E.g. NaCH3CO2 ---> Na+ + CH3CO2- ;

CH3CO2- is acetate, a base. (what’s its conjugate acid?)

NH4Cl ---> NH4+ + Cl- ; NH4

+ is ammonium, an acid.

In above examples, Na+ and Cl- act like “spectator ions”.(N.b. These don’t affect pH and can be ignored).

Remember: HCH3CO2 is acetic acid; NH3 is ammonia.

Salt solutionsSalt solutions……

NH3 is a weak base. It is NOT a salt.

NH4Cl is a salt.

Is a solution of NH4Cl an acidic or basic solution?

NH4+ is conjugate acid of NH3 so NH4

+ is weak acid.

Cl- is conjugate base of HCl a strong acid.

Therefore, Cl- is so weak it can be ignored.

Thus the solution is a weak acid solution.

NetNet ionic equations and ICEionic equations and ICE

You are also expected to know how to write andrecognize net ionic equations.

e.g. What is the Ka equilibrium for a NH4Cl solution?

NH4+ + H2O <==> NH3 + H3O+

Or sometimes “ignore” H2O: NH4+

<===> NH3 + H+

Master I.C.E. calculations involving Ka and Kb equilibria.

I=initial, C=change, E= equilibrium

(1) pH(1) pH of strong acid solutionof strong acid solution

1) Strong acid solutions -use: pH = - log[acid]

For example: What is the pH of a 0.020 M HCl solution?

Answer:

HCl is a strong acid, it dissociates 100% (I.e. completely):

[HCl]F = 0.020 M = [H3O+]

So, pH = - log [H3O+] = -log [HCl]F = -log(.020)=1.70

Voila!

(2) pH of strong base solution(2) pH of strong base solution

2) Strong base solutions -use: pH = 14.00 + log[base] orpH=14.00-pOH

Ex. What is pH of 5.2x10-4M NaOH solution?

Answer: Since NaOH is strong base:

[NaOH] = [OH-] = 5.2 x10-4M

Easier to determine pOH first:

pOH = -log[OH-] = -log(5.2x10-4) = 3.28

pH = 14.00 - 3.28 = 10.72 (makes sense?)

Yes because it is a basic pH

(3) pH of weak acid solution(3) pH of weak acid solution3) Pure Weak acid solutions -use: Ka equilibrium, ICEcalculation of [H+]

Ex. What is pH of 2.5x10-2M acetic acid (HA, Ka=1.8x10-5)?

Answer: Use Ka equilibrium:

HA < = = => H+ + A-

initial: 0.025 0 0

change: -x +x +x

equilibrium 0.025-x x x

Ka = ; = 1.8x10-5 => x2 +1.8x10-5x - 4.5x10-7 =0

Use quadratic eq’n: x = 6.6x10-4 :pH = -log(6.6 x10-4) = 3.18

!

[H+][A

"]

[HA]

!

x2

(0.025 - x)

Approximation using 5% rule:Approximation using 5% rule:Again: HA < = = => H+ + A-

From ICE: 0.025-x x x

5% rule: assume that x << 0.025.

Like last slide: = 1.8x10-5 ; drop x from denom.

≈ 1.8x10-5 => x ≈ 6.7x10-4 ≈ [H+]

=> pH = -log(6.7x10-4 ) = 3.17 (compare to 3.18 in last slide)

5% rule-valid? Calculate:

N.b. : = 2.7% < 5% indeed.

!

x2

(0.025)

!

x2

(0.025 - x)

!

x

Nx100%

!

6.7x10-4

0.025x100%

(4) pH of weak base solution(4) pH of weak base solution

(4) Pure weak base solutions -use: Kb equilibrium, ICEcalculation of [OH-]

What is the pH of 0.10 M ammonia? (Kb = 1.8x10-5M)

Answer: Use Kb equil. NH3 + H2O<==>NH4++OH-

0.10-x x x

Kb = 1.8x10-5M= = 1.8x10-5; 5%: ≈

x2 ≈ (1.8x10-5)(0.10) => x =1.3 x 10-3 =[OH-]

pOH = 2.87 => pH = 14.00-2.87 = 11.13 (makes sense?)!

x2

(0.10 - x)

!

x2

(0.10 - x)

!

x2

(0.10)