Acid-Base Equilibria

transcript

Acid-Base EquilibriaThe reaction of weak acids with water,

the reaction of weak bases with water,

always results in an equilibrium!!

The equilibrium constant for the reaction of a weak acid with water is Ka

Acid-Base Equilibria

eg. HF(aq) + H2O(l)

Ka =[H3O+] [F-]

H3O+(aq) + F-

Keq = ?

For any weak acid

Why is H2O(l) omitted from the Ka expression?

Ka =[H3O+] [conjugate base]

[weak acid]

the equilibrium constant for the reaction of a weak base with water is Kb

HS-(aq) + H2O(l)

H2S(aq) + OH-(aq)

For any weak base

Kb =[OH-] [conjugate acid]

[weak base]

Write the expression for Kb for S2-(aq)

ANSWER:

S2-(aq) + H2O(l)

Kb =[OH-] [HS-]

HS-(aq) + OH-

5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq)

HF(aq) + H2O(l) H3O+(aq) + F-

][F ]OH[K

Ka = 6.6 x 10-4

x]- [0.100

[x] [x]10 x 6.6 4-

x 2 = (0.100)(6.6 x 10-4)

x 2 = 6.6 x 10-5

x = 8.1 x 10-3 mol/L

1st try - Ignore x

2nd try– Include x

0.0081] - [0.100

[x] [x]10 x 6.6 4-

x 2 = (0.0919)(6.6 x 10-4)

x 2 = 6.0654 x 10-5

x = 7.8 x 10-3 mol/L

3rd try– Include new x

0.0078] - [0.100

[x] [x]10 x 6.6 4-

x 2 = (0.0922)(6.6 x 10-4)

x 2 = 6.0852 x 10-5

x = 7.8 x 10-3 mol/L

[H3O+] = 7.8 x 10-3 mol/L

5.b) find [H3O+] for 0.250 mol/L CH3COOH(aq

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-

COOH][CH

]COO[CH ]OH[K

Ka = 1.8 x 10-5

x]- [0.250

[x] [x]10 x 1.8 5-

x 2 = (0.250)(1.8 x 10-5)

x 2 = 4.5 x 10-6

x = 2.1 x 10-3 mol/L

1st try - Ignore x

2nd try– Include x

0.0021] - [0.250

[x] [x]10 x 1.8 5-

x 2 = (0.2479)(1.8 x 10-5)

x 2 = 4.462 x 10-6

x = 2.1 x 10-3 mol/L

[H3O+] = 2.1 x 10-3 mol/L

pH of a weak acid

Step #1: Write a balanced equation

Step #2: ICE table OR assign variables

Step #3: Write the Ka expression

Step #4: Check (can we ignore dissociation)

Step #5: Substitute into Ka expression

pH of a weak acid

eg. Find pH of 0.100 mol/L HF(aq).Step #1: Write a balanced equation

HF(aq) + H2O(l) H3O+(aq) + F-

Step #2: Equilibrium Concentrations

Let x = [H3O+] at equilibrium

[F-] = x

[HF] = 0.100 - x

Step #3: Ka expression

Ka =[H3O+] [F-]

Step #4: Check (can we ignore dissociation)

dissociation (- x) may be IGNORED

= 151 (0.100)

6.6 x 10-4

Acid dissociation CANNOT beIGNORED in this question.

[weak acid]

If > 500

We have to use the – x

Step #5: Substitute into Ka expression

x]- [0.100

[x] [x]10 x 6.6 4-

x2 = 6.6 x 10-5 - 6.6 x 10-4 x

x2 + 6.6 x 10-4 x - 6.6 x 10-5 = 0

a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5

QuadraticFormula!!

4acbbx

)10x4(1)(-6.6)10x(6.610x6.6x

-52-4-4

0.00026410x6.6x

mol/L0.0078x Ignore

negative roots

a) Find the [H3O+] in 0.250 mol/L HCN(aq)

Check: 4.0 x 108

x = 1.24 x 10-5

[H3O+] = 1.24 x 10-5

b) Calculate the pH of 0.0300 mol/L HCOOH(aq)

Check: 167

x = 2.24 x 10-3

pH = 2.651

Try these:

Practice1. Formic acid, HCOOH, is present in the sting of

certain ants. What is the [H3O+] of a 0.025 mol/L solution of formic acid? (0.00203 mol/L)

2. Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid.

( [H3O+] = 3.87 x 10-3 pH = 2.413 )

3. What is the percent dissociation of the vinegar in 2.?

% diss = 0.466 %

Practice4. A solution of hydrofluoric acid has a molar

concentration of 0.0100 mol/L. What is the pH of this solution?

( [H3O+] = 0.00226 pH = 2.646 )

5. The word “butter” comes from the Greek butyros. Butanoic acid gives rancid butter its distinctive odour. Calculate the [H3O+] of a 1.0 × 10−2 mol/L solution of butanoic acid. (Ka = 1.51 × 10−5 )

(3.89 x 10-4 mol/L)

pH of a weak base same method as acids calculate Kb

ignore dissociation if

K x K Ka b w KK

pH of a weak baseCalculate the pH of 0.0100mol/L Na2CO3(aq)

pH of a weak baseCalculate the pH of 0.500 mol/L NaNO2(aq)

Calculating Ka from [weak acid] and pH

eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the numerical value of the Ka for this acid.

- Equation- Find [H3O+] from pH

- Subtract from [weak acid]- Substitute to find Ka See p. 591 #6 & 8

C6H5COOH(aq) + H2O(l) H3O+(aq) + C6H5COO-

[H3O+] = 10-2.68 = 0.00209 mol/L

[C6H5COOH] = 0.072 – 0.00209

= 0.06991 mol/L

Find Ka

Ka =(0.00209)(0.00209)

(0.06991)= 6.2 x 10-5

[C6H5COO-] = 0.00209 mol/L

Calculating Ka from [weak acid] and pH

eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the % dissociation for this acid.

See p. 591 #’s 5 & 6[H3O+] = 10-2.68

= 0.00209 mol/L

100%xacid] [weak

]O[Hdiss % 3

= 2.9 %

100% x 0.072

0.00209

a) 0.250 mol/L chlorous acid, HClO2(aq); pH = 1.31

0.012 19.5%b) 0.150 mol/L cyanic acid, HCNO(aq); pH =

2.150.00035 4.7%

c) 0.100 mol/L arsenic acid, H3AsO4(aq); pH = 1.70

0.0050 20%d) 0.500 mol/L iodic acid, HIO3(aq); pH = 0.670

0.160 42.8%

Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid:

More Practice: Weak Acids:

pp. 591, 592 #’s 6 -8 Weak Bases:

p. 595 #’s 11 - 16 (Kb’s on p. 592)

1.a) 1.4 x 10-10

b) 0.0014 %2.a) 2.5 x 10-9

b) 0.0080 %

3.a) 1.6 x 10-9

b) 0.0080 %4.a) 2.7 x 10-9

b) 0.042 %

Acid-Base Stoichiometry

Solution Stoichiometry (Review)

1. Write a balanced equation

2. Calculate moles given ( OR n = CV)

3. Mole ratios

4. Calculate required quantity

OR OR m = nM

eg. 25.0 mL of 0.100 mol/L H2SO4(aq) was used to neutralize 36.5 mL of NaOH(aq). Calculate the molar concentration of the NaOH solution.

H2SO4(aq) + NaOH(aq) → H2O(l) + Na2SO4(aq)22

nH2SO4 =

nNaOH =

CNaOH =

Acid-Base Stoichiometry

pp. 600, 601 – Sample Problems

p. 602 #’s 17 - 20

Dilution Given 3 of the four variables Only one solution CiVi = CfVf

Stoichiometry Given 3 of the four variables Two different solutions 4 step method

Excess Acid or BaseTo calculate the pH of a solution produced by mixing an acid with a base:

write the B-L equation (NIE) calculate the moles of H3O+ and OH-

subtract to determine the moles of excess H3O+ or OH-

divide by total volume to get concentration calculate pH

eg. 20.0 mL of 0.0100 M Ca(OH)2(aq) is mixed with 10.0 mL of 0.00500 M HCl(aq).

Determine the pH of the resulting solution.

ANSWER:

Species present:

Ca2+OH- H3O+ Cl- H2O

0.0200 mol/L0.0200 L

0.00500 mol/L0.0100 L

NIE: OH- + H3O+ → 2 H2O

n = CV 4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+

3.5 x 10-4 mol excess OH-

= 0.01167 mol/L

[OH-] = 0.01167 mol/L

pOH = 1.933

pH = 12.067

totalV

L0.0300

mol10x3.5 4

Indicators An indicator is a weak acid that

changes color with changes in pH

To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color

HIn(aq) + H2O(l) H3O+(aq) + In-

Colour #1 Colour #2

HIn is the acid form of the indicator. Adding H3O+ causes colour 1 (LCP)

Adding OH- removes the H3O+ & causes colour #2

methyl orange

HMo(aq) + H2O(l) H3O+(aq) + Mo-

red yellow

bromothymol blue

HBb(aq) + H2O(l) H3O+(aq) + Bb-

yellow blue

Acid-Base Titration (p. 603 → ) A titration is a lab technique used to

determine an unknown solution concentration.

A standard solution is added to a known volume of solution until the endpoint of the titration is reached.

Acid-Base Titration

The endpoint occurs when there is a sharp change in colour

The equivalence point occurs when the moles of hydronium equals the moles of hydroxide

The colour change is caused by the indicator added to the titration flask.

Acid-Base Titration

An indicator is a chemical that changes colour over a given pH range

(See indicator table) A buret is used to deliver the standard

solution

Acid-Base Titrationstandard solution - solution of known

concentration

primary standard - a standard solution which can be made by direct weighing of a stable chemical.

Titration Lab – pp. 606, 607

Multi-Step Titrations (p. 609 - 611)

Polyprotic acids donate their protons one at a time when reacted with a base.

eg. Write the equations for the steps that occur when H3PO4(aq) is titrated with NaOH(aq)

H3PO4(aq) + OH-(aq)

H2PO4-(aq) + OH-

HPO42-

(aq) + OH-(aq) 47

Multi-Step TitrationsH3PO4(aq) + OH-

(aq) → H2PO4-(aq) + H2O(l)

H2PO4-(aq) + OH-

(aq) → HPO42-

(aq) + H2O(l)

HPO42-

(aq) + OH-(aq) PO4

3-(aq) + H2O(l)

H3PO4(aq) + 3 OH-(aq) PO4

3-(aq) + 3 H2O(l)

Multi-Step Titrations

Write the balanced net ionic equations, and the overall equation, for the titration of Na2S(aq) with HCl(aq).

p. 611 #’s 21.b), 22, & 23

LAST TOPIC!! Titration Curves

Properties / Operational Definitions Acid-Base Theories and Limitations

Arrhenius – H-X and X-OHModified – react with water → hydroniumBLT – proton donor/acceptor (CA and CB)

Writing Net Ionic Equations (BLT) Strong vs. Weak pH & pOH calculations Equilibria (Kw, Ka, Kb) Titrations/Indicators/Titration Curves Dilutions and Excess Reagent questions

Acids and Bases

Step #2: ICE table

[HF] [H3O+] [F-]

0.100 mol/L 0 0

-x +x +x

0.100 - x x x

(aq) + H2O(l) HCO3-(aq) + OH-

0.0100mol/L CO23-

[PO43-] [HPO4

2-] [OH-]

0.0100 mol/L 0 0

-x + x + x

0.0100 - x x x52

eg. 20.0 mL of 0.0100 M Ca(OH)2(aq) is mixed with 10.0 mL of 0.00500 M HCl(aq).

Determine the pH of the resulting solution.

ANSWER:

Ca(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + CaCl2(aq)

nbase = 0.0002 mol

→ needs 0.004 mol HCl

nacid = 0.00005 mol

→ needs 0.000025 mol Ca(OH)2

Limiting reactant

Excess reactant

Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us

WorkSheet #9 answers:

1. 0.210 mol/L

2. a) 22.5 mL

b) 24.7 mL

c) 4.8 mL

3. 31.5 mL

4. a) 0.0992 mol/L

b) 0.269 mol/L

c) 0.552 mol/L

WorkSheet #10 answers:

1. pH = 13.0002. [H3O+] = 4.12 x 10-2 mol/L

[OH-] = 2.43 x 10-13 mol/L3. pH = 13.1254. pH = 7

p. 586 #’s 1 – 456

A primary standard is a pure substance that is stable enough to be stored indefinitely without decomposition, can be weighed accurately without special precautions when exposed to air, and will undergo an accurate stoichiometric reaction in a titration.

15. pH of 0.297 mol/L HOCl

HOCl(aq) + H2O(l) H3O+(aq) + OCl-(aq)

Let x = [H3O+] at equilibrium

[OCl-] = x

[HOCl] = 0.297 - x

Ka =[H3O+] [OCl-]

[HOCl]

Check:

dissociation (- x) may be IGNORED

= 1.02 x 107 (0.297)

2.9 x 10-8

[0.297]

[x] [x]10 x 2.9 8-

X = 9.28 x 10-5 pH = 4.03

0.484 mol/L0.07000 L

0.125 mol/L0.02500 L

16. NIE: OH- + H3O+ → 2 H2O

0.03388 mol OH- 0.003125 mol H3O+

0.030755 mol excess OH-

[OH-] = 0.3237 mol/L pOH = 0.490

pH = 13.51060

17. Ignore dissociation

[OH-] = 0.0146 mol/L

% diss = 2.92 %

18. Vave = 10.975 mL

nNaOH = 0.001262 mol

nH2SO4 = 0.000631 mol

C = 0.0252 mol/L

19. Kb = 3.93 x 10-4

% diss = 6.27

Acid-Base Equilibria

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