Chemistry 201 Colligative properties

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Colligative properties. • How do we count “solute particles”? • What properties depend on concentration of particles? • What is the role of solute in determining ...
Chemistry 201 Lecture 5

Strong and Weak Acids and Bases NC State University

Introduction to pH • How do we think about acid / base reactions? • What is pH? • How do we calculate the pH of strong acids and bases? Text : Sections 6.0 - 6.3, 6.6

Definitions Standard Bronsted-Lowry definition: an acid is a molecule that can donate a proton.

HA

H+ + A-

A base is a molecule that can donate a hydroxide ion.

BOH

B+ + OH-

The conjugate base of HA is AThe conjugate acid of BOH is B+

Definitions Water can do both of these things:

H 2O

H+ + OH-

Because of this we also define a base as a molecule that can accept a proton.

B + H+

BH+

The conjugate acid of B is BH+

Acidity as an equilibrium One goal is to predict the products of : ? a) HNO2 + CN1-  NO21- + HCN ? b) SO32- + H2O  HSO31- + OH1? c) NH41+ + HS1-  NH3 + H2S

Acidity as an equilibrium d)

HCO31- + HS1-  H2CO3 + S2HCO31- + HS1-  CO32- + H2S cannot both have favorable DGo maybe neither has favorable DGo

e) H2O + H2O  H3O1+ + OH1-

water both an acid and base

Definition of Kw Water is in equilibrium with its ions H3O+ and OH-. H2O + H2O  H3O+ + OH-

The equilibrium constant is expressed as: We do not include the water solvent in this or any other equlibrium constant in aqueous solution. The Value of Kw is Kw = 10-14. This means that the concentration of H3O+ and OH- are both 10-7 in neutral water.

Sample problem: dissociation of water If [H3O1+] = 4.7 x 10-9, in solution, what is [OH1-]?

Solution: (4.7 x 10-9) [OH1-] = 1.0 x 10-14 [OH1-] = 2.1 x 10-6

Definition of Ka Ka is the acid equilibrium constant the describes the concentration of H3O+ in aqueous solution. For acid HA we have

HA + H2O  H3O+ + AThe equilibrium constant is expressed as:

If Ka is large than the acid is mostly dissociated, i.e. it is a strong acid.

Definition of pH and pKa Since the numbers for Kw and Ka often are quite Small, it is convenient to a log scale:

We can do the same for the acid equilibrium constant:

pH in reverse If you know the pH of the solution you can immediately calculate the hydronium ion concentration. or the hydroxide ion concentration. and, of course, these are related by,

Example: pKa of acetic acid An acid dissociation is actually a proton transfer reaction as shown below.

The pKa is a measure of the proton transfer equilibrium

Acid equilibria: strong acids The equilibrium constant for an acid is:

Which we can also write as

If the equilibrium constant is very large then we can assume that the acid is “completely” dissociated. This means that

Base equilibria: strong bases The definition of a strong base is exactly parallel to the definition of a strong acid:

We can define pKb in the same way as pKa. If the base is strong then pKb < 0. Again, this can mean that the base is essentially totally dissociated

Strong acids The list of strong acids is pretty short: HCl, HBr, HI, … not HF H2SO4, HNO3, HClO4….not H2SO3, HNO2, HClO

In reality all acids have a pKa, but with strong acids we can assume that they will be 100% dissociated. That is what we normally do to simplify calculations.

Why does a strong acid dissociate?

Gas phase

Liquid phase

Rank the following acids, from strongest to weakest: HClO3 pKa = - 3 HSO31Ka = 1 x 10-7 HF pKa = 3.1 HC2H3O2 Ka = 2 x 10-5

Rank the following acids, from strongest to weakest: HClO3 pKa = - 3 HSO31Ka = 1 x 10-7 HF pKa = 3.1 HC2H3O2 Ka = 2 x 10-5

HClO3 > HF > HC2H3O2 > HSO31-

Sample pH problem

What is the pH of 3.5 x 10-3 M HCl? (Ka > 1)

Will 7.5 x 10-4 M HCl have a higher or lower pH?

Sample pH problem

What is the pH of 3.5 x 10-3 M HCl? (Ka > 1) pH = - log(3.5 x 10-3) = 2.46 Will 7.5 x 10-4 M HCl have a higher or lower pH? higher

Sample pH problem (in reverse)

What is [HCl] in a pH 2.30 solution?

Sample pH problem (in reverse)

What is [HCl] in a pH 2.30 solution? - log[H3O1+] = 2.30 log[H3O1+] = - 2.30 [H3O1+] = 10-2.30 = 5.0 x 10-3 M

Therefore, the initial [HCl] = 5.0 x 10-3 M

Sample pH problem considering base

What is the pH of 1.2 x 10-2 M Ca(OH)2?

Will 7.0 x 10-4 M Ca(OH)2 have a higher or lower pH?

Sample pH problem considering base

What is the pH of 1.2 x 10-2 M Ca(OH)2? pOH = - log(2.4 x 10-2) = 1.62

pH = 14.00 - 1.62 = 12.38

12 mol OH 2 -2 1.2 x 10 M cpd x  2.4 x 10 M 1 mol cpd

Will 7.0 x 10-4 M Ca(OH)2 have a higher or lower pH? lower

Measurement of pH A pH meter measures the electrochemical potential between a known liquid inside the glass electrode (membrane) and an unknown liquid outside. Because the thin glass bulb allows mainly the hydrogen ions to interact with the glass, the glass electrode measures the electrochemical potential of hydrogen ions. To complete the electrical circuit, also a reference electrode is needed. Note that the instrument measures a voltage difference.

Acid rain SO2 is the most important gas which leads to acidification of lakes and soil. Emissions of nitrogen oxides which are oxidized to form nitric acid are of increasing importance. 70 Tg(S) per year in the form of SO2 comes from fossil fuel combustion and industry, 2.8 Tg(S) from wildfires and 7-8 Tg(S) per year from volcanoes. SO2 + H2O  H2SO3

SO3 + H2O  H2SO4 2NO2 + H2O  HNO2 + HNO3

Acid rain SO2 is the most important gas which leads to acidification of lakes and soil. Emissions of nitrogen oxides which are oxidized to form nitric acid are of increasing importance. 70 Tg(S) per year in the form of SO2 comes from fossil fuel combustion and industry, 2.8 Tg(S) from wildfires and 7-8 Tg(S) per year from volcanoes. SO2 + 2H2O  HSO3- + H3O+

SO3 + H2O  HSO4- + H3O+ 2NO2 + 3H2O  NO2- + NO3- + 2H3O+

Ocean acidification CO2 that is emitted into the atmosphere is taken up by the oceans. There is an equilibrium between the CO2 in the atmosphere and the water. Once the CO2 is dissolved it can dissociate. Thus, increasing CO2 levels mean increased acidity in the ocean. CO2 + 2H2O  HCO3- + H3O+ HCO3- + H2O  CO32- + H3O+

Goals • Write net equations for acid/base reactions • Understand the dissociation of water [H3O1+] < --- > [OH1-] • Rank acids in strength • Calculate pH of strong acids and strong bases

Strong Acid

Strong Base

Weak Acid

Weak Base

know [H3O1+] calc pH Strong Acid Weak Acid

Strong Base Weak Base

know [OH1-] calc pOH

O1+]

know [H3 calc pH

Strong Acid

Strong Base

Weak Acid

Weak Base

pH of Weak Acids & Bases • How do we find [H3O1+]? • How do acid and base strengths relate? Text : Sections 6.4, 6.7

pH of Weak Acids What is the pH of a 0.20 M solution of HC2H3O3? (pKa = 4.74)

pH of Weak Acids What is the pH of a 0.20 M solution of HC2H3O3? (pKa = 4.74) Solution: Set A = HC2H3O3 Step 1: Solve for Ka = 1.82 x 10-5 Step 2: Make a reaction table Molecule Initial Difference Equilibrium

HA 0.2 -x 0.2-x

A0 x x

H+ 0 x x

pH of Weak Acids What is the pH of a 0.20 M solution of HC2H3O3? (pKa = 4.74) Step 3: Solve for x

pH of Weak Acids What is the pH of a 0.20 M solution of HC2H3O3? (pKa = 4.74) Step 3. Calculate pH

Note that if peroxy acetic acid were a strong acid the pH would have been

pH of Weak Acids What is the pH and % ionization of a 0.0084 M solution of HClO? (pKa = 7.46) Solution: Write down the reaction Step 1: Solve for Ka Step 2: Make a reaction table

pH of Weak Acids HClO problem (contd) (pKa = 7.46) Molecule Initial Difference Equilibrium

Step 3: Solve for x

Step 4: Solve for pH

HClO 0.0084 -x 0.0084-x

ClO0 x x

H+ 0 x x

pH of Weak Acids HClO problem (contd) % ionization Step 5: Calculate percent ionization

Goals • Convert between Ka, pKa • Calculate the pH of solutions of weak acids and weak bases