Unit 11

Acids and Bases

Name the following Acids H2CrO4

HF H3PO4

HClO4

Acid vs Base All aqueous solutions contain H+

and OH- ions. Relative amounts determine

whether the solution is acid, base, or neutral Acid soln – more H+ (Hydronium ion) Basic soln- more OH-

Neutral- equal amounts of each

Basic Acidic Neutral

H+

H+

H+OH-

OH-

OH-

Solution Solution Solution

Which solution is Acidic? Basic? Neutral???

Arrhenius Model Acid is a substance that contains

hydrogen and ionizes to produce hydrogen ions in aqueous solns

Base is a substance that contains a hydroxide group and dissociates to produce a hydroxide ion in aqueous solns

Exception-NH3

Acids and Bases

Produce H+ ions in water

Have a sour taste Break down metals Formula starts with

H Poisonous and

corrosive to skin pH less than 7

Produce OH- ions in water

Have a bitter taste and a slippery feel

Break down fats and oils

Formula ends with OH

Poisonous and corrosive to skin

pH greater than 7

Brønsted-Lowry Model Acid- hydrogen ion donor Base- hydrogen ion acceptor

HX (aq) + H20 (l) H30+ (aq) + X- (aq)Base Conjugate AcidAcid Conjugate Base

Conjugates Conjugate Acid

Species produced when a base accepts a hydrogen ion from an acid

Conjugate Base Species that results when an acid donates a

hydrogen ion to a base Conjugate acid- base pair

Consists of 2 substances related to each other by donating and accepting of a single H+

Conjugates HF + H2O H3O+ + F- (H3O+ Conjugate acid)

(F- Conjugate base)

NH3 + H20 NH4+

+ OH- (NH4+ Conjugate acid)

(OH- Conjugate base) Amphoteric- substances that can act as both

acids and bases Monoprotic- HCl, HF Polyprotic- H2SO4, H3PO4

Warm up H3PO4 + H2O H3O+ + H2PO4

-

HClO4 + H2O -ClO4 + H3O+

H2CrO4 + NH3 +NH4 + HCrO4-

Acid Strength Strong acids – ionize completely Weak acids- do not ionize

completely

Ka =

HCN + H2O H3O+ + CN-

Ka =

][.]..][.[

AcidBCAC

][]][[ 3

HCNCNOH

Practice Problems Write an ionization equation and

acid ionization constant expression for Nitrous Acid.

HNO2

HNO2 + H2O H3O+ + NO2-

Ka = ][

]][[2

23

HNONOOH

One More Practice Problem Write an ionization equation and

acid ionization constant expression for Chlorous Acid.

HClO2

HClO2 + H2O H3O+ + ClO2-

Ka = ][]][[

2

23

HClOClOOH

Base Strength Strong Bases- completely

dissociate into metal ions and hydroxide ions

Weak bases- partially dissociateBase ionization constant

Kb = ][.]..][.[

BaseBCAC

Write ionization equations and base ionization constant expressions for the carbonate ion.

CO32-

CO32- + H2O HCO3

- + OH-

Kb =

Practice Problems

][

]][[2

3

3

CO

OHHCO

One More Practice Problem Write ionization equations and base

ionization constant expressions for the hydrogen sulfite ion.

HSO3-

HSO3- + H2O H2SO3 + OH-

Kb = ][

]][[

3

32

HSO

OHSOH

Warm up Write a Disassociation equation to

describe the following reactions. CO3

2- + H2O HCO3- + OH-

HClO2 + H2O H3O+ + ClO2-

pH Measure of H+ ions in soln pH = -log[H+] Acidic solutions have a pH below 7 Basic solutions have a pH above 7 pH 7 is neutral Change of 1 pH unit represents a

tenfold change. (exponential)

pOH Measures concentration of OH- ion

pOH = - log [OH-]

pH + pOH = 14.00

Practice Problems Calculate the pH and pOH of

aqueous solutions having the following ion concentrations.

[OH-] = 6.5 x 10-6

pOH = -log[OH-] pH = 14.00 – pOHpOH = -log[6.5 x 10-6] pH = 14.00 –

5.19pOH = -[log 6.5 + log 10-6] pH = 8.81pOH = -[0.81 + (-6)]pOH = 5.19

One more, one more time! Calculate the pH and pOH of aqueous

solutions having the following ion concentrations.

[H+] = 3.6 x 10-9

pH = -log[H+] pOH=14.00 – pHpH = -log[3.6 x 10-9] pOH=14.00-8.44pH = -[log 3.6 + log 10-9] pOH=5.56pH = -[0.56 + (-9)]pH = 8.44

What is the pH of the following concentrations?

1.0 E -9 2.4 E -3 6.7 E -7 4.4 E -5

Classify each as Acid or Base

Buffers A buffer is a mixture of a weak acid

and its conjugate base OR, a weak base and its conjugate acid.

This mixture resists changes in pH. The amount of acid or base a

buffer can absorb without significant change in pH is called the buffer capacity.

Neutralization Reactions

When an acid is added to a base, the end products are always salt and water. (neutral)

A salt is defined as the neutral end product of an acid/base reaction.

ACID + BASE SALT + WATER H2S + Ca(OH)2 CaS + H2O What is wrong with this equation???

Balance the final equation!

H2S + Ca(OH)2 CaS + H2O

1 Ca 11 S 14 H 22 O 1

H2S + Ca(OH)2 CaS + 2 H2O

Neutralization Reactions

Try another example:Acid + Base Salt + Water

H2SO4 + NaOH Na2SO4 + H2O1 Na 21 SO4 13 H 21 O 1

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

Take it one step further…

Sulfurous acid and sodium hydroxide yields sodium sulfite and water.

H2SO3 + NaOH Na2SO3 + H2O

1 Na 21 SO3 1

3 H 21 O 1

H2SO3 + 2NaOH Na2SO3 + 2H2O

One Last Step

Hydrosulfuric acid and calcium hydroxide yields what???

H2S + Ca(OH)2

1) One product will always be water. H2S + Ca(OH)2 H2O +

2) The other product will be the + ion of the base bonded with the – ion of the acid.

H2S + Ca(OH)2 2H2O + CaS

pH Indicators A chemical substance that changes color in

the presence of an acid and/or a base.1) pH paper – Dip the paper, match color to

scale on vial to determine numeric pH.pH<7 = acid, pH>7 = base, pH = 7 neutral

2) Litmus – Dip one red and one blue paper.Red stays red, blue turns red AcidBlue stays blue, red turns blue BaseRed stays red, blue stays blue Neutral

pH Indicators

3) Bromthymol Blue – Add a few drops of bromthymol blue to the substance.

If the blue color turns to yellow AcidIf the blue color stays blue Base

4) Phenolphthalein – Add a few drops of phenolphthalein to the substance.

If the clear liquid turns to pink BaseIf the clear liquid remains clear Acid

Concentration

Strength of an acid or base is determined by the amount of ionization. Concentration is determined by the amount of water added to the substance.

Molarity (M)

The number of moles of solute dissolved in each liter of solution.

Molarity = moles of solute

liters of solution

Example Problem #1

If 1.00 liter of sugar water contains exactly 1.00 mole of sugar, what is its molarity?Molarity = 1.00 mol

1.00 LMolarity = 1.00 M

Example Problem #2

If 1.00 liter of sugar water contains exactly 2.00 mole of sugar, what is its molarity?Molarity = 2.00 mol

1.00 LMolarity = 2.00 M or 2.00 mol/L

(Twice as concentrated…)

Example Problem #3

What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?

Molarity = 0.75 mol = 0.30 mol/L or 0.30M

2.50 L

In Lab, grams are typically used in place of moles.

If you wanted to make 2.00L of a 6M HCl solution, how much HCl would you need?

First, calculate the molar mass of the acid.H 1 x 1.00795 = 1.00795Cl 1 x 35.453 = 35.453

36.46095 = 36.461

If you wanted to make 2.00L of a 6M HCl solution, how much HCl would you need?

First, calculate the molar mass of the acid.

HCl contains 36.461 g/mol

It would take 36.461 g of HCl to make 1 liter of a 1M HCl solution. How many grams would it take to make 2L of a 1M solution?

2 x 36.461g = 72.922g

If you wanted to make 2.00L of a 6M HCl solution, how much HCl would you need?

It takes 72.922g of HCl to make 2 liters of a 1M solution. How much would it take to make 2 liters of a 6M solution?

6 x 72.922g = 437.532 g

Try One More Suppose you wanted to make 2 liters of

a 0.5 M solution of HCl. How much HCl would you need?

Each mole of HCl is equal to 36.461g For a 0.5 M solution, you would need

half that much. 36.461 x 0.5 = 18.2305g.

However, you want to make 2 liters, so double that amount. 18.2305 x 2 = 36.461g.