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Unit 4: ACIDS, BASES AND SALTS 4.1 Arrhenius Acids and Bases Acids release H+ in water Bases release OH- in water Salts are products of an acid-base neutralization reaction. The salt is an ionic compound, which is neither an acid nor a base. Example 1: Balance the following and identify the acid, the base and the salt. HCl + NaOH H2SO4 + KOH H3PO4 + Sr(OH)2 HNO3 + Al(OH)3 Properties of Acids: React with bases Electrolytes React with some metals to form H2 (g) Turn litmus paper red Taste sour

Properties of Bases: React with acids Electrolytes Feel slippery (saponification – makes soap) Turn litmus paper blue Taste bitter

H+

(aq) ions in water will combine with the lone pairs in the water molecule to form H3O+(aq)

H2O(l) + H+

(aq) H3O+(aq)

Understand that even though H+

(aq) is written, it does not exist in solution.

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4.2 Brønsted-Lowry Acids and Bases Acids are proton (H+) donors Bases are proton (H+) acceptors Therefore, Brønsted-Lowry acids and bases must come in pairs. Example 2: Identify the acids and bases in the reactants for the following reactions: (a.) HCl + H2O H3O+ + Cl- (b.) H2SO4 + S2- HS- + HSO4

- (c.) NH3 + H2O OH- + NH4

+ Acids can donate one or more protons Monoprotic acids: have one proton to donate Diprotic: have two protons to donate Triprotic: have three protons to donate Polyprotic: more than one proton to donate Look at Example 2 again. What do you notice about water in (a) and (c)? A substance that can act as an acid or a base is called amphiprotic. H3O+ H2O OH- H2SO3 HSO3

- SO3-

An amphiprotic substance can be identified by the presence of:

A negative charge (base)

An easily removable proton (acid) Water is not negatively charged, but it has a lone pair of electrons that allows it to act as a base.

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4.3 Strengths of Acids and Bases In an equilibrium, an acid (HA) will donate a proton and become a base (A-).

HA + H2O ⇄ H3O+ + A-

HA and A- are a “Conjugate Acid-Base Pair” A- is the conjugate base of HA HA is the conjugate acid of A- Example 1: Write the reaction for the following and identify the conjugate acid base pairs. CH3COOH + F- HPO4

2- + SO32-

Strong Acids are acids that will dissociate completely in a reaction with water. HCl + H2O H3O+ + Cl- Single arrow = NO EQUILIBRIUM Strong Electrolyte See the top left of the Acid-Base Table (p6 of the Data Booklet) for a list of the strong acids. Note the single arrow means that a reverse reaction does NOT occur. Strong Bases will also completely dissociate in water. Most of these contain the OH- ion. O2- and NH2

- are also strong bases. These bases are listed on the bottom right on the Acid-Base table. Again, note the single arrow, which means no equilibrium. Forward reaction does not occur!

Weak Acids are acids that do not dissociate completely in water, but form equilibrium with water. Note “⇄”

They are listed in order of decreasing strength on the left side of the Acid-Base table. The two “acids” at the bottom of the table are so weak that they never act as acids with water.

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Weak Bases are bases that do not dissociate completely with water, and also form an equilibrium. Note “⇄”

They are listed in order of increasing strength on the right side of the Acid-Base table. The six “bases” at the top of the table are so weak that they never act as bases with water. Example 2: Write the reaction of the following acids in water: H2SO3 HIO3 HF H2S Which will produce the highest [H+]? Example 3: Write the reaction of the following bases in water: SO3

2- IO3

- NH3 H2BO3

- Which will produce the highest [OH-]?

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Other Relationships:

1. For polyprotic acids, the acid with the most H is the strongest. H3PO4 > H2PO4

- > HPO42-

2. The stronger the acid, the weaker its conjugate base.

3. Concentrated and dilute refer to molarity

Strong acids H3O+ + conjugate base 1.0 M 1.0 M 1.0 M

Weak acids ⇄ H3O+ + conjugate base

1.0 M < 1.0 M < 1.0 M

Levelling Effect

H3O+ is the strongest acid that can exist in water. An acid that is stronger than H3O+ will give its proton to water to form H3O+. HClO4 + H2O H3O+ + ClO4

- OH- is the strongest base that can exist in water. A stronger base will form OH-.

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4.4 Kw and pH

All of the acid-base chemistry we will be exploring takes place in water. Recall that water is an amphiprotic substance. The first equilibrium we will examine is that of water reacting with itself:

2 H2O(l) ⇄ H3O+(aq) + OH-

(aq)

The equilibrium expression, therefore, is: Kw = [H3O+][OH-] This is called the self-ionization, or auto-ionization, of water. In an acidic solution:

[H3O+] > [OH-]

In a basic solution:

[H3O+] < [OH-]

In a neutral solution:

[H3O+] = [OH-]

Example 1: Using Le Châtelier’s Principle, describe what happens to the equilibrium when an acid (H+) or base (OH-) is added to neutral water. At 25oC, Kw = 1.0 X 10-14

Since Kw = [H3O+][OH-] = 1.0 X 10-14, [H3O+] or [OH-] can be calculated if one of them is known.

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Example 2: Find [H3O+] and [OH-] for:

Pure water

0.20 M HCl

0.0030 M NaOH The auto-ionization of water is an endothermic reaction:

2 H2O(l) + 59 kJ ⇄ H3O+(aq) + OH-

(aq)

What happens to Kw, [H3O+], and [OH-] in pure water as it is heated? when it is cooled? If the [H3O+] and [OH-] are changing, is the water still neutral?

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Acidity and basicity is measured with a logarithmic scale called pH or pOH.

pH = -log [H3O+] [H3O+] = 10-pH pOH = -log [OH-] [OH-] = 10-pOH Example 3: Convert the following: Note: for pH and pOH, only digits after the decimal are significant.

[H3O+] = 2.1 X 10-2 M, pH = ?

[OH-] = 7.18 X 10-5 M, pOH = ?

pH = 4.2, [H3O+] = ?

pOH = 6.34, [OH-] = ? Combining auto-ionization of water with pH we find a relationship between pH and pOH.

1.0 X 10-14 = [H3O+][OH-]

Taking the –log of both sides:

14.00 = pH + pOH (at 25oC)

This shows the relationship between acidity and basicity. If one of the values is known, all others can be found:

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Example 4: Is the solution acidic or basic? Find the values for pH, pOH, [H3O+], and [OH-] that are not given.

pH = 3.36

pOH = 8.841

[H+] = 3.7 X 10-11 M

[OH-] = 0.003 M Example 5: At some temperature, Kw = 1 X 10-13. Is this temperature above or below 25oC? Find the pH of neutral water at this temperature.

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Example 6: At some temperature, the pOH of neutral water is 7.30. What is the value of Kw at this temperature? Is this temperature above or below 25oC? Final notes: A pH change of 1 unit is a tenfold change in concentration. At 25oC: pH < 7 is acidic pH = 7 is neutral pH > 7 is basic Always assume 25oC, unless specified otherwise.

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4.5 Strong Acid-Base Mixtures

A strong acid and strong base mixture will produce neutral water and a salt. If the acid and base are not mixed in a stoichiometric ratio, one will be in excess, and consequently the solution will either be acidic or basic. This will be a limiting reagent question. Example 1: 20.0 mL of 0.50 M HCl is mixed with 40.0 mL of 0.20 M NaOH. What is the pH of the resulting solution? Example 2: 1.5 L of 0.510 M hydrobromic acid is mixed with 0.80 L of 0.480 M strontium hydroxide. What is the pH of the resulting solution?

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Example 3: 100.0 mL of 0.100 M NaOH solution is treated with hydrogen iodide gas. If the volume of solution does not change, how much gas must be added to change the pH to 9.50? Sometimes you can have a mixture of two strong acids or of two strong bases. Example 4: If 15.0 mL of 0.250 M HCl is added to 75.0 mL of 0.100 M HNO3, determine the resulting pH. Example 5: 45.0 mL of 0.0500 M NaOH is mixed with 25.0 mL of 0.133 M Sr(OH)2. What is the resulting pH?

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4.6 Ka and Kb A weak acid dissociates partially in water:

HA(aq) + H2O(l) ⇄ H3O+(aq) + A-

(aq)

The equilibrium expression for this reaction is designated Ka A weak base dissociates partially in water:

B-(aq) + H2O(l) ⇄ HB(aq) + OH-

(aq)

The equilibrium expression for this reaction is designated Kb The acid-base data table lists the weak acids with their Ka values. As the acids get weaker, they dissociate less; therefore, they have a smaller equilibrium constant. Kb values can be calculated from the Ka values of a conjugate acid. For a conjugate acid-base pair,

Kw = Ka Kb

For example, for NH4

+ in water:

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Example 1: For each of the following, give the reaction with water, the equilibrium expression, and the equilibrium value:

HIO3

F-

NH3

CH3COOH

HCO3

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The relative strengths of acids and bases can be used to determine the position of equilibrium. In an acid-base equilibrium, the strength of the acid in the products can be compared to the strength of the acid in the reactants. The stronger acid will give away its protons. Example 2: Complete the following reactions and indicate if the resulting equilibrium favours reactants or products.

HF + NO2-

H2S + HCO3-

HSO3- + HCO3

-

NH4HCO3 (aq)

HCOOH + H2O

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4.7 Hydrolysis

The hydrolysis of a salt is the reaction of a cation and/or an anion with water to produce an acidic or basic solution. We will only consider the reactions of soluble ions with water. Cations: Alkali metals (Group 1) and alkaline earth metals (Group 2) will not react with water to make an acidic or basic solution. Fe3+, Cr3+, and Al3+ can react with water. They are highly charged and small. They can weaken the O-H bonds in water molecules near them.

Fe(H2O)63+ ⇄ H+ + Fe(H2O)5OH2+

Cr(H2O)63+ ⇄ H+ + Cr(H2O)5OH2+

Al(H2O)63+ ⇄ H+ + Al(H2O)5OH2+

Fe3+, Cr3+, and Al3+ will make an acidic solution in water. Another acidic cation is ammonium:

NH4+ ⇄ H+ + NH3

Anions: Anions that are conjugate bases of strong acids will not react with water to make a basic solution. (Note the direction of the arrow in the acid-base data table). Conjugate bases of weak acids will act as weak bases if dissolved in water.

F- + H2O ⇄ HF + OH-

Anions can also be amphiprotic. These anions can act as weak bases and weak acids. (Note HSO4

-)

HCO3- + H2O ⇄ H2CO3 + OH-

HCO3- + H2O ⇄ CO3

2- + H3O+

If there is more than one possible hydrolysis reaction we must determine which equilibrium is predominant.

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The predominant equilibrium will have the highest equilibrium constant. This is the reaction that will be most important. The other reactions still occur, but their contribution of H+ or OH- is negligible. Example 1: Will the following salts produce an acidic, basic, or neutral solution?

a.) NaCl

b.) KF

c.) AlCl3

d.) CaSO4

e.) NaHSO3

f.) FePO4

g.) NH4CN

h.) LiCH3COO

i.) NH4CH3COO

j.) NH4HCO3

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4.8 Ka

Equilibrium calculations involving acids have three variables:

Initial concentration of the acid

Ka value

[H3O+] or pH Problems will provide two of these variables and ask you to find the third. Notes: Use of quadratic solutions are not required. Equilibrium will only involve the donation of one proton for polyprotic acids All steps and assumptions must be shown Example 1: Find the pH of a 0.10 M solution of acetic acid. Example 2: A 0.50 M solution of an unknown diprotic acid has a pH of 5.30. What is the value of the equilibrium constant?

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Example 3: A solution of ammonium chloride has a pH of 3.850. What was the initial concentration of the solution?

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4.9 Kb

Equilibrium calculations of weak bases are similar to weak acids. The variables are: initial concentration of weak base, Kb, [OH-] or pH. Remember, Kw = Ka X Kb Take care in converting from [OH-] to pH. The solutions are basic. Example 1: Find the pH of a 0.300 M solution of sodium nitrite. Example 2: A 2.0 M solution of an unknown base has a pH of 12.80. What is the value of the equilibrium constant for this base?

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Example 3: A 2.0 L solution of potassium fluoride has a pH of 9.208. How many grams of potassium fluoride were dissolved into the initial solution?

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4.10 Metal and Non-metal Oxides

Oxides that dissolve in water produce acidic or basic solutions. Metal oxides form basic solutions. Na2O(s) 2 Na+

(aq) + O2-(aq)

Due to the leveling effect: O2-

(aq) + H2O(l) 2OH-(aq)

Non-metal oxides form acidic solutions: CO2 (g) + H2O(l) H2CO3 (aq)

SO2 (g) + H2O(l) H2SO3 (aq)

SO3 (g) + H2O(l) H2SO4 (aq)

2NO2 (g) + H2O(l) + ½ O2 (g) 2HNO3 (aq)

Non-metal oxides are responsible for acid rain. Rain with a pH of < 5.6 is called acid rain. CO2 naturally makes rain acidic with a pH of ~5.6. Acid rain is caused by nitrogen oxides, NOx, and sulphur oxides, SOx. Nitrogen oxides are mostly produced by internal combustion engines. Sulphur oxides are mostly produced in industrial processes:

steel making (coal)

coal combustion (sulphur in coal)

burning heavy fuels that have sulphur in them (ships, industries, … ) Effects of acid rain:

Damages plant growth

Kills lakes – eventually almost all living organisms in lakes will die

Leaches toxic minerals from earth

Corrodes iron, marble and concrete

Water contamination Solutions for acid rain:

Produce fewer pollutants: o More efficient, properly tuned engines o Burn cleaner fuels

Remove the pollutants after they are made: o Catalytic converters in cars o Scrubbers in coal exhausts

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4.11 Titrations

In this unit, you may be asked to solve three types of problems with a titration question:

Concentration of an acid or base solution

Molar mass of an acid or base

Percentage purity of an acid or base sample Definitions: Equivalence point or Stoichiometric point: the point at which the acid and the base are neutralized. Endpoint: the point at which titration is stopped after the indicator has declared that the equivalence point has been reached. Ideally, the endpoint and the equivalence point of the titration will be very close. Solve titration questions using the following steps:

Write the reaction

Indicate the information given in the question

Perform the stoichiometric calculation

Check to make sure the question has been answered. Example 1: A 35.00 mL solution of sulphuric acid is titrated to its equivalence point with 0.120 M sodium hydroxide solution. Given the following results, what is the initial concentration of the acid solution? Trial 1 17.80 mL Trial 2 17.62 mL Trial 3 17.60 mL

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Example 2: 0.250 g of an unknown solid base is titrated with 23.78 mL of 0.110 M hydrochloric acid solution. What is the molar mass of the base? (assume 1:1 ratio between acid and base) Example 3: A 0.700 g sample of acetic acid is titrated with 0.200 M strontium hydroxide. The titration uses 25.88 mL of base. What is the purity of the acetic acid?

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4.12 Indicators

An indicator is a weak acid or base that shows a different colour in its acidic and basic forms. We use indicators to determine approximate pH values for a solution based on the colour we observe. As a weak acid or base, an indicator will form equilibrium with water. For the indicator HIn:

HIn + H2O ⇄ H3O+ + In-

(yellow) (blue) How will this indicator react to addition of an acid? A base? Example 1: A solution with bromcresol green indicator is blue. What can you conclude about the solution? An indicator will change colour at different pH values, depending on its strength as a weak acid. We can determine the strength of the indicator by examining it at its transition colour (a mixture of its acid and base colours)

HIn + H2O ⇄ H3O+ + In-

(yellow) (blue) The transition colour for this indicator is green. If it is green, then [HIn] = [In-] At the transition point, Ka = [H3O+], and pH = pKa

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Example 2: Find the Ka values of the following and rank them in order of decreasing acid strength: orange IV, phenolphthalein, methyl red, phenol red, thymol blue. Example 3: An indicator mixture consists of methyl orange, bromothymol blue, and phenolphthalein. What colours will it indicate at what pH ranges?

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4.13 Buffers

A buffer is a solution with appreciable amounts of a weak acid and its conjugate weak base. A buffer will minimize the change in pH of the solution if a strong acid or base is added to the solution. A buffer minimizes the change in pH due to Le Châtelier’s Principle. Consider an acetic acid buffer:

CH3COOH + H2O ⇄ H3O+ + CH3COO-

If there are equal amounts of the weak acid and the conjugate base, the pH = pKa. Example 1: Show why this is true. Example 2: Using Le Châtelier’s Principle, show what happens to this equilibrium, and the pH, if a strong acid or base is added.

CH3COOH + H2O ⇄ H3O+ + CH3COO-

The key is that the equilibrium tries to minimize the stress added to the equilibrium. So pH changes slightly.

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Example 3: Show what happens to the pH of a buffer solution if water is added to the solution.

CH3COOH + H2O ⇄ H3O+ + CH3COO-

In a buffer with equal amounts of a weak acid and a conjugate weak base, pKa = pH To make a buffer with a specific pH, you must choose a weak acid with an appropriate Ka value. Example 4: How would you make a buffer with pH ~ 3, or pH ~ 10?

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A buffer will only function if both the weak acid and the conjugate base are present. If the buffer has shifted too far, and one of these is used up, the solution will no longer be a buffer. Many biological reactions need to be at specific pH values to occur. Buffers are important in biological systems. The optimum pH of blood is 7.35. Haemoglobin in blood binds with oxygen through the following equilibrium:

HHb + O2 + H2O ⇄ H3O+ + HbO2-

What happens if blood pH gets too high or low? One of the main blood buffers is the H2CO3, HCO3

- buffer. If you hyperventilate, you are breathing out too much CO2. What does that do to blood pH, and to the haemoglobin equilibrium? Why does breathing into a bag help?

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4.14 Titration pH Curves In an acid-base titration, an indicator changes colour at the equivalence point of a titration. An indicator will change colour with the addition of one drop of an acid or a base. By graphing the change in pH during a titration, we can see why an indicator works, and what a titration curve looks like for different types of titrations. Graph the data given on p166 (Hebden) This data represents the titration of 1.0 L of 1.0 M HCl with 1.0 M NaOH. (The HCl is in the flask and the NaOH is in the burette) For a strong acid-strong base titration (SASB), the products at the equivalence point are water and a salt. The salt will be neutral, because they are conjugates of strong acids and strong bases. The pH at the equivalence point will be 7. The indicator must change colour in the pH 5-9 range (steep part of the graph) A strong base-strong acid curve will be the reverse:

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A weak acid titrated with a strong base will have three important differences:

Initial pH will be higher (partial dissociation)

Buffer region in the titration where the pH will not change greatly (a weak acid and its conjugate base)

Equivalence point pH will be higher, pH > 7. The neutralized weak acid will form a weak base. A weak base titrated with a strong acid will have a reverse curve. Three main differences compared to SBSA curve:

Lower initial pH

Buffer region

Final pH < 7 For titrations with weak acids or bases, the indicator must be chosen carefully. For a WASB titration, the indicator should change colour at pH ~ 9. For a WBSA titration, the indicator should change colour at pH ~ 5. An incorrect indicator can have the endpoint and the equivalence at different volumes. There are three main calculations with titrations:

The initial concentration of one of the reactants

The Ka or Kb given the initial pH and concentration

The Ka or Kb given the pH at the midpoint of the titration Given the initial pH and concentration, the calculation is the same as the calculations we did during the Ka calculations section. If the pH at the midpoint of the titration (half the equivalence point volume) is in the middle of the buffer region, the amount of weak acid and conjugate base are equal, and pH = pKa (from the Buffers section)

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Example 1: An unknown monoprotic acid is titrated to its endpoint with 35.00 mL of 0.1000 M NaOH. After 17.50 mL of NaOH is added, the pH is 5.30. What is the Ka of the unknown acid? Example 2: 17.25 mL of 0.150 M HCl is used to titrate 20.00 mL of an unknown weak base. The initial pH of the solution was 12.15. What indicator should be used for this titration? What is the Kb of this weak base? Keypoints of Titration Curves:

1. Shape 2. Features

a. Initial pH b. pH at equivalence point (with a suitable indicator) c. Volume at equivalence point

3. Calculations a. Volume at equivalence point b. Initial pH of WA/SB or WB/SA curves, given initial [ ] c. Ka or Kb given pH or [ ]