Unit 8 - Acids & Bases

Unit 8 - Acids & Bases

8.1 Introduction to Acids and Bases

8.2 pH and pOH of Strong Acids and Bases 8.3 Weak Acid and Base Equilibria

8.4 Acid-Base Reactions and Buffers 8.5 Acid-Base Titrations

8.6 Molecular Structure of Acids and Bases 8.7 pH and pKa

8.8 Properties of Buffers

8.9 Henderson-Hasselbalch Equation

8.10 Buffer Capacity

This logo shows it is a Topic Question - it should only require knowledge included in this Topic and it should be giving practice in the Science Practice associated with this Topic.

8.1 Introduction to Acids and Bases

Original Definitions of Acids & Bases - Arrhenius Definition

In earlier courses, you were probably introduced to the idea that Acids are substances that release H⁺ ions into water, for example,

HCl

_(g)

HCl

_(aq)

H

⁺_(aq)

+ Cl

^—_(aq)

hydrogen chloride gas hydrogen chloride solution hydrochloric acid

H

₂

SO

_4(l)

H

₂

SO

_4(aq)

2 H

⁺_(aq)

+ SO

₄^2—_(aq)

hydrogen sulfate liquid hydrogen sulfate solution sulfuric acid

CO

_2(g)

CO

_2(aq)

H

₂

CO

_3(aq)

2 H

⁺_(aq)

+ CO

₃^2—_(aq)

carbon dioxide gas carbon dioxide solution hydrogen carbonate solution carbonic acid

All of our acids start as covalent molecules that ionise when dissolved in water. The significance of is that we would consider the acid has ionised 100% and we would define the acid as a strong acid.

The significance of is that we would consider that the acid has only partially ionised and we would define the acid as a weak acid.

Again, our earliest definition was that a Base is a substance that releases OH^`— ions into water, for example,

NaOH

_(s)

Na

⁺_(aq)

+ OH

^—_(aq)

sodium hydroxide sodium hydroxide solution

Al(OH)

_3(s)

Al

³⁺_(aq)

+ 3 OH

^—_(aq)

K_sp = 1.8 x 10^-5

aluminium hydroxide aluminium hydroxide solution

NH

_3(g)

NH

_3(aq)

NH

₄⁺_(aq)

+ OH

^—_(aq)

ammonia gas ammonia solution ammonium hydroxide solution

Apart from ammonia, most bases are already ionic so the significance of is that we would consider the base has dissolved 100% and we would define the base as a strong base.

The significance of is that we would consider that the base has only partially

dissolved or that the base has only partially ionised and we would define the acid as a weak base.

Role of Water

Though water has a major role as the solvent, it plays even more crucial roles in acid-base chemistry.

Hydrogen ions are too small (basically they are the positive nucleus of the original hydrogen atom) to exist as stable particles.

Instead they form a coordinate covalent bond, using one of the lone pairs of electrons on a water molecule, and form a number of complex ions such as the hydroxonium or hydronium ion.

With strong bases, water is usually just a solvent but with weak bases, water gets involved in the ionisation and is the source of the OH^— ions.

This time a H⁺ ion forms a coordinate covalent bond with the base, using one of the lone pairs of electrons on the NH₃ molecule to form an ammonium ion, NH₄⁺.

However, water has an even more pervasive influence on acid-base chemistry due to the

autoionisation of water:

H

₂

O

_(l)

+ H

₂

O

_(l)

⇌ H

₃

O

⁺_(aq)

+ OH

^—_(aq)

Water Constant, K

_w

The significance of the autoionisation of water is that there will be H⁺ ions and OH ^— ions present in all solutions, and their relative concentrations will be linked by their equilibrium constant.

K = [H₃O⁺][OH^—] but measurements show that (at 25°C) [H⁺] = [OH^—] = 10^-7 mol L^-1, [H₂O]² whereas in a Litre of water, [H₂O] would be 1000/18 ≃ 55 mol L^-1 Therefore, any change in [H₂O] will be negligible and we can incorporate [H₂O] into our constant.

This is consistent with the rule met in Unit 7 - Equilibrium, where pure liquids and pure solids are treated as having an 'activity' = 1 which means 'effective concentration' or [H₂O] = 1

K [H₂O]² = [H₃O⁺][OH^—] becomes

K

_w

= [H

₃

O

⁺

][OH

^—

] or K

_w

= [H

⁺

][OH

^—

]

= 1.0 x 10

^-14

at 25° C

⇌

Effect of Temperature on K

_w

H

₂

O

_(l)

⇌ H

⁺_(aq)

+ OH

^—_(aq)

The ionisation of water is endothermic (bond breaking) so will be favoured by an increase in

temperature, as shown in the graph opposite.

At 25^° C, neutral solutions will have:

pK_w = 14.0 so K_w = 1.0 x 10^-14 mol² L^-2 K_w = [H⁺][OH^—] = 1.0 x 10^-14 mol² L^-2 [H⁺] = [OH^—] = 1.0 x 10^-7 mol L^-1

and pH = pOH = 7

However, neutral solutions will have different pH's at different temperatures.

At 250^° C, neutral solutions will have: pK_w = 11.0 so K_w = 1.0 x 10^-11 mol² L^-2

K_w = [H⁺][OH^—] = 1.0 x 10^-11 mol² L^-2 [H⁺] = [OH^—] = √1.0 x 10^-11 = 3.16 x 10^-6 mol L^-1 and pH = pOH = 5.5 At ~0^° C, neutral solutions will have: pK_w = 14.8 so K_w = 2.0 x 10^-15 mol² L^-2

K_w = [H⁺][OH^—] = 2.0 x 10^-15 mol² L^-2 [H⁺] = [OH^—] = √2.0 x 10^-15 = 3.98 x 10^-8 mol L^-1 and pH = pOH = 7.4

While the vast majority of our data will be measures at 25°C (s.t.p.), you need to be aware that values such as K_w and hence, pH and pOH , are temperature dependent.

But at 25°C

As acid is added: [H⁺]⬀ so [OH^—]⬂

[H⁺] > 10^-7 mol L^-1 pH < 7

In neutral: [H⁺] = [OH^—]

[H⁺] = 10^-7 mol L^-1 pH = 7

As base is added: [OH^—]⬀ so [H⁺]⬂

[H⁺] < 10^-7 mol L^-1 pH > 7

K = [H O

⁺

][OH

^—

] = 1.0 x 10

^-14

The 'potent' Scale

In an attempt to simplify working with very small numbers, Søren Sørensen, a Swedish chemist, suggested, in 1909, a scale based on the 'powers of ten' ('potent' is German for power).

This was first applied to [H⁺]'s so [H⁺] = 1.0 x 10^-7 became pH = 7 Mathematically,

pH = - log [H

⁺

]

and inversely

[H

⁺

] = 10

^-pH

similarly,

pOH = - log [OH

^—

]

and inversely

[OH

^—

] = 10

^-pOH

and,

pK

_a

= - log Ka

and inversely

K

_a

= 10

^-pKa

As well as :

K

_w

= [H

₃

O

⁺

][OH

^—

] = 1.0 x 10

^-14

Other useful relationships include:

pK

_w

= pH + pOH = 14

Due to the logarithmic nature of the pH scale:

a change of one pH unit represents a 10 fold increase or decrease in

ion concentration 1M HCl, [H⁺] = 10⁰, pH = 0 10M HCl, [H⁺] = 10¹, pH = -1 but 100M HCl not possible so pH scale has a 'minimum' between -1 and -2.

1M NaOH, [H⁺] = 10^-14, pH = 14 10M NaOH, [H⁺] = 10^-15, pH = 15 but 100M NaOH not possible so pH scale has a 'maximum' between 15 and 16.

Most strong acids and strong bases form saturated solutions in the range 20 - 30 M.

8.2 pH and pOH of Strong Acids and Bases

pH = - log [H

⁺

] pOH = - log [OH

^—

] K

_w

= [H

₃

O

⁺

][OH

^—

] = 10

^-14

pK

_w

= pH + pOH = 14

a change of one pH unit represents a

10 fold increase or decrease in ion concentration

Strong Acids

For strong acids, since we can assume 100% ionisation, there is a quick and easy route to calculating pH and, if necessary, pOH.

0.20 M HCl [HCl] = 0.20 M [H⁺] = 0.20 M

pH = - log [H⁺] = -log(0.20) = 0.70 pOH = 14 - pH = 14 - 0.70 = 13.30 0.20 M H₂SO₄ [H₂SO₄] = 0.20 M [H⁺] = 0.40 M

pH = - log [H⁺] = -log(0.40) = 0.40 pOH = 14 - pH = 14 - 0.40 = 13.60 0.02 M H₂SO₄ [H₂SO₄] = 0.02 M [H⁺] = 0.04 M

pH = - log [H⁺] = -log(0.04) = 1.40 pOH = 14 - pH = 14 - 1.40 = 12.62 Notice that diluting the H₂SO₄ by a factor of 10 (0.20M → 0.02M) causes pH and pOH to change by 1 unit.

Strong Bases

For strong bases, since we can again assume 100% ionisation, there is a similar quick and easy route to calculating pOH though we will normally want to convert this to pH.

0.015 M NaOH [NaOH] = 0.015 M [OH^—] = 0.015 M

pOH = - log [OH^—] = -log(0.015) = 1.82 pH = 14 - pOH = 14 - 1.82 = 12.18 0.015 M Ca(OH)₂ [Ca(OH)₂] = 0.015 M [OH^—] = 0.030 M

pOH = - log [OH^—] = -log(0.030) = 1.52 pH = 14 - pOH = 14 - 1.52 = 12.48 0.15 M Ca(OH)₂ [Ca(OH)₂] = 0.15 M [OH^—] = 0.30 M

pOH = - log [OH^—] = -log(0.30) = 0.52 pH = 14 - pOH = 14 - 0.52 = 13.48 Notice that increasing the [Ca(OH)₂] by a factor of 10 (0.015M → 0.15M) causes pH and pOH to change by 1 unit.

Effect of Temperature

In the previous lesson we saw that K_w is temperature dependent. At ~0^° C, pK_w = 14.8

Changing temperature, however, has no effect on stoichiometry and our formulae for calculating pH and pOH do not change. So,

0.20 M HCl [HCl] = 0.20 M [H⁺] = 0.20 M pH = - log [H⁺] = -log(0.20) = 0.70

However, pOH = 14.8 - pH = 14.8 - 0.70 = 14.1

8.3 Weak Acid and Base Equilibria

Brønsted–Lowry Definition of Acids and Bases

This concept is based on the fact that acid–base reactions involve the transfer of H⁺ ions from one substance to another.

Conjugate pairs, that differ by a H⁺ ion, will exist with one molecule being the acid while the other is the base.

One consequence of this definition is that water, H₂O, will often fulfill the role of an acid or base in many reactions.

An ACID is a substance (molecule or ion) that donates a proton (H⁺)

A BASE is a substance (molecule

or ion) that accepts a proton (H⁺)

Weak Acids

Most acids do not ionise 100%, they are described as weak acids and are represented by the following equation:

HA

_(aq)

+ H

₂

O

_(l)

⇋

H

₃

O

⁺_(aq)

+ A

^—_(aq)

The equilibrium constant is

K = [H₃O⁺] [A^—]

[HA] [H₂O]

but, as usual, [H₂O] can be assumed to stay

unchanged so, we re-define the equilibrium constant as the acid-dissociation constant, K_a

K

_a

= [H

⁺

] [A

^—

]

[HA]

HCl ⇋ Cl^— NH₃ ⇋ NH₄⁺

As always, an equilibrium constant provides useful information about the ratio of products and reactants and helps us compare the relative strengths of different weak acids.

Calculating pH of Weak Acids

We can use the acid-dissociation constant to calculate [H⁺] ( and then pH = -log[H⁺]) but only if we make an important assumption about the value of [HA]equilibrium.

In essence, we have to assume the acid is so weak and such a small amount of ionisation will take place that:

[HA]equilibrium = [HA]_initial = C (referred to as the analytical concentration) In other words, if we start with a 0.01 M ethanoic acid solution, we will assume [CH₃COOH]_equ is still 0.01 M.

So

K

_a

= [H

⁺

] [A

^—

]

becomes

K

_a

= [H

⁺

] [A

^—

]

We also know that [HA]equilibrium

[HA]_initial

[H

⁺

] = [A

^—

]

so

K

_a

= [H

⁺

]

² and

[H

⁺

] = √(K

_a[HA]_initial

)

[HA]_initial

then

pH = -log[H

⁺

]

Example 1: Calculate the pH of a 0.01 M solution of ethanoic (acetic) acid given that K_a = 1.8 x 10^-5

Method 1

step 1: [H⁺] = √(K_a [HA]_initial ) = √(1.8 x 10^-5 x 0.01)

= 4.2 x 10^-4 M

step 2: pH = -log [H⁺] = -log (4.2 x 10^-4)

= 3.37

A Second Method would be to use the ICE table, met in the Equilibrium Unit, to calculate [H⁺] at equilibrium and then use pH = -log[H⁺].

CH

₃

COOH

_(aq)

⇋ H

⁺(aq)

+ CH

₃

COO

^—_(aq)

Initial 0.01 0.00 0.00

Change - x + x + x

Equilibrium 0.01 - x x x

K_a = [H⁺] [CH₃COO^—] = x ^x x = 1.8 x 10^-5

[CH₃COOH] 0.01 - x

At this point, we are faced with using the quadratic expression to solve for x. Instead, we use our previous assumption, that [CH₃COOH]equilibrium ≃ [CH₃COOH]_initial = 0.01M,

to make life easier.

x² = 1.8 x 10^-5 x 0.01 so x = √(1.8 x 10^-7) = 4.2 x 10^-4 so [H⁺] = 4.2 x 10^-4 M and pH = -log(4.2 x 10^-4) = 3.37

Assumptions

There are two major assumptions made in these calculations:-

① that we can ignore the [H⁺] being produced by the auto-ionisation of water.

Given that this is normally ≃ 10^-7 M, this will only be an issue if our weak acid is at very low concentration.

② that we can assume [HA]equilibrium ≃ [HA]_initial

This is justifiable as long as the % of acid molecules that ionise is <5%. However, % ionisation increases at lower concentrations so ...

Example 2: Calculate the ionisation constant, K_a, of a 0.1 M solution of a weak monoprotic acid that has a pH equal to 4.0.

Method

step 1: [H⁺] = 10^-pH = 1.0 x 10^-4 M

step 2: K_a = [H⁺]² / [HA]_initial = (1.0 x 10^-4)² / 0.1 = 1 x 10^-7

Calculating Percent Ionisation of a Weak Acid

We have seen that the magnitude of K_a indicates the strength of an acid. Another measure of the strength of an acid is its percent ionisation, which is defined as

percent ionisation = ionised acid concentration at equilibrium x 100 % initial concentration of acid

The stronger the acid, the greater the percent ionisation. For a monoprotic acid HA, the

concentration of the acid that undergoes ionisation is equal to the concentration of the H⁺ ions or the concentration of the A^— ions at equilibrium.

Therefore, we can write the percent ionisation as percent ionisation = [H⁺] x 100 %

[HA]

Example: The pH of a 0.036 M nitrous acid (HNO₂) solution is measured as 2.4.

HNO_2(aq) ⇋ H⁺_(aq) + NO₂^—_(aq)

Calculate the percent ionisation for nitrous acid.

step 1: [H⁺] = 10^-pH = 10^-2.4 = 3.98 x 10^-3 M

step 2: percent ionisation = [H⁺] / [HA] x 100 % = 3.98 x 10^-3 / 0.036 x 100 %

= 11.1 %

So about 1 in 9 molecules of nitrous acid would ionise which is consistent with a weak acid.

Earlier it was mentioned that one of

our main assumptions for pH, pK_a

calculations relies on ionisation being

< 5%, so we would be unable to rely on

these calculations for 0.036 M HNO₂ .

Calculating pH of Weak Bases

The ionisation of weak bases is treated in the same way as the ionisation of weak acids. When ammonia dissolves in water, it undergoes the reaction

NH_3(aq) + H₂O_(l) ⇋ NH₄⁺_(aq) + OH^—_(aq) The equilibrium constant is given by

K = [NH₄⁺][OH^—]

[NH₃][H₂O]

Compared with the total concentration of water, very few water molecules are consumed by this reaction, so we can treat [H₂O] as a constant. Thus, we can write the base ionisation constant (K_b), which is the equilibrium constant for the ionisation reaction, as

K[H₂O] =

K

_b

= [NH

₄⁺

][OH

^—

] = [BH

⁺

][OH

^—

]

[NH

₃

] [B]

Similarly, the magnitude of K_b indicates the strength of a base. Notice also that the stronger the base, the weaker is the conjugate acid. (K_a x K_b = 1.0 x 10^-14 for a conjugate pair)

We can use the base-dissociation constant to calculate [OH^—] ( and then, if necessary, convert to [H⁺] to allow pH = -log[H⁺]). But again we need to make an important assumption about the value of [NH₃]equilibrium.

In essence, we have to assume the base is so weak and such a small amount of ionisation will take place that:

[NH₃]equilibrium = [NH₃]_initial

In other words, if we start with 0.01 M ammonia solution, we will assume [NH₃]_equ is still 0.01 M.

So

K

_b

= [NH

₄⁺

][OH

^—

]

becomes

K

_b

= [NH

₄⁺

][OH

^—

]

We also know that [NH₃]equilibrium

[NH₃]_initial

[NH

₄⁺

] = [OH

^—

]

so

K

_b

= [OH

^—

]

² from here there are a number of different routes that can be taken [NH₃]_initial

Example 1: What is the pH of a 0.40 M ammonia solution? K_b = 1.8 x 10^-5 step 1: [OH^—] = √(K_b C) = √(1.8 x 10^-5 x 0.4)

= 2.7 x 10^-3 M

step 2: [H⁺] = 1.0 x 10^-14 /[OH^—]

K

_w

= [H

⁺

][OH

^—

] = 1.0 x 10

^-14

= 1.0 x 10^-14 / 2.7 x 10^-3

= 3.7 x 10^-12

step 3: pH = -log [H⁺] = -log (3.7 x 10^-12) = 11.4

Example 1 again: What is the pH of a 0.40 M ammonia solution? K_b = 1.8 x 10^-5 step 1: [OH^—] = √(K_b C) = √(1.8 x 10^-5 x 0.4)

= 2.7 x 10^-3 M step 2: pOH = -log[OH^—]

= -log(2.7 x 10^-3)

= 2.6

step 3: pH = 14 - pOH

pK

_w

= pH + pOH = 14

= 14 - 2.6 = 11.4

Again, we are assuming that OH^— ions from auto-ionisation of water will be insignificant and that the weak base does not ionise enough (< 5%) to ensure [base]equilibrium = [base]_initial.

Formulae for Calculations

We have an almost bewildering number of formulae that we can either derive as needed or attempt to memorise.

Lets start, however, with the formulae provided in the 'AP® Chemistry Equations & Constants' sheet.

Acid Equations

① ② ③

Base Equations

④ ⑤ ⑥

General Equations

⑦

⑧

⑨ only applies to conjugate pairs

The following could be very useful and can then be derived or memorised.

From ① we can derive From ④ we can derive

From ⑨ we can derive

pK

_w

= 14 = pK

_a

+ pK

_b

only applies to conjugate pairs

You should also be able to do the inverse of ② , ③ , ⑤ and ⑥

[H

⁺

] = 10

^-pH

[OH

^—

] = 10

^-pOH

K

_a

= 10

^-pKa

K

_b

= 10

^-pKb

Over the rest of this unit you should try solving questions using the minimum number of

memorised formulae - wherever possible, use the formulae provided by 'AP® Chemistry Equations

& Constants' sheet and derive any others needed.

K_b = [OH^—]²

[B]_initial

8.4 Acid-Base Reactions and Buffers

Strong Acid–Strong Base Neutralisation Reaction

The reaction between a strong acid (say, HCl) and a strong base (say, NaOH) can be represented by: NaOH_(aq) + HCl_(aq) NaCl_(aq) + H₂O_(l)

or more generally:

OH^—_(aq) + H⁺_(aq) H₂O_(l) net ionic equation

Changes in pH can be monitored by setting the reaction up as a titration and using a pH meter to measure the pH throughout :-

A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M HCl solution in an Erlenmeyer flask. This curve is sometimes referred to as a titration curve.

The equivalence point represents the stoichiometric balance when [H⁺] = [OH^-]. This can be different from the 'end-point' of a titration - the point at which an indicator changes colour as a slight excess of H⁺ ions or OH^— ions may be needed to complete a colour change.

Only for strong acid / strong base titrations will pH = 7 at equivalence point.

Reactions involving weak acids or weak bases will produce acidic salts or basic salts , so pH ≠ 7 Changes in pH can be calculated using our 'traditional' and 'new' formulae:

C = n / L n = m / M pH = -log [H

⁺

] [H

⁺

] = 10

^-pH

K

_w

= 10

^-14

= [H

⁺

] [OH

^—

]

or

14 = pH + pOH pOH = -log [OH

^—

]

Example 1: A 25.0 mL sample of H₂SO₄ requires 46.23 mL of a standard 0.203 M NaOH solution to reach the equivalence point. Calculate the [H₂SO₄].

step1: calculate number of moles of OH^— ions added,

n = C x L

n = 0.203 x 0.04623 = 0.00938 moles of OH^— ions step2: use the stoichiometry of the balanced equation to convert to moles of H₂SO₄ ,

2 NaOH_(aq) + H₂SO_4(aq) Na₂SO_4(aq) + 2 H₂O_(l)

2 moles ≣ 1 mole

0.0938 moles of OH^— ions ≣ 0.00938/2 = 0.00469 moles of H₂SO₄

step3: calculate [H₂SO₄],

C = n /L

C = 0.00469 / 0.0250 = 0.188 M

Example 2: A 25.0 mL sample of 0.188 M H₂SO₄ has 50 mL of a standard 0.203 M NaOH solution added. Calculate the pH of the resulting solution.

step 1: calculate number of moles of H⁺ ionspresent at start,

n = C x L

n = 0.188 x 0.025 = 0.00470 moles of H₂SO₄ = 0.00940 moles of H⁺ step 2: calculate number of moles of OH^— ions added,

n = C x L

n = 0.203 x 0.050 = 0.01015 moles of OH^— ions step 3: subtract to determine moles of excess ions,

0.01015 moles of OH^— ions

-

0.00940 moles of H⁺ = 0.00075 moles of excess OH^— step 4: calculate [OH^—], (don't forget to combine volumes for overall volume of solution)

C = n /L

C = 0.00075 / 0.075 = 0.01 M [OH^—] = 10^-2 step 5: convert to [H⁺],

K

_w

= 10

^-14

= [H

⁺

] [OH

^—

]

^{[H+] = 10-14 / 10-2 = 10-12 M}

step 6: calculate pH,

pH = -log[H

⁺

]

pH = -log(10^-12) = 12

Weak Acid–Strong Base Neutralisation Reaction

The calculations in the previous sections are made 'easier' by the fact that everything - acids, bases and salts - ionise 100%. Next 'easiest' is where we have one weak / one strong.

The reaction between a weak acid (say, CH₃COOH) and a strong base (say, NaOH) can be represented by:

NaOH_(aq) + CH₃COOH_(aq)

⇋

^NaCH₃^COO_(aq) + H₂O_(l) or more generally:

OH^—_(aq) + HA_(aq)

⇋

^A—_(aq) ^{+ H}₂^O_(l)

Changes in pH can be monitored by setting the reaction up as a titration and using a pH meter to measure the pH throughout :-

A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M CH₃COOH solution in an Erlenmeyer flask.

The equivalence point represents the stoichiometric balance when moles of CH₃COOH present at start = moles of NaOH added.

Only for strong acid / strong base titrations will pH = 7 at equivalence point. When a weak acid / strong base is used the salt formed will be basic so pH > 7.

Changes in pH can still be calculated using our 'traditional' and 'new' formulae but we may need to allow for the partial ionisation of the acid so extra equations needed:

C = n / L n = m / M pH = -log [H

⁺

] [H

⁺

] = 10

^-pH

K

_w

= 10

^-14

= [H

⁺

] [OH

^—

]

or

14 = pH + pOH pOH = -log [OH

^—

]

K

_a

= [H

⁺

] [A

^—

] / HA

or

K

_a

= [H

⁺

]

²

/ C etc.

NB. When excess base is used, the weak acid molecule can be treated as ionising 100% as the equilibrium is driven over to the right until all the molecules are consumed. So,

NaOH_(aq) + CH₃COOH_(aq) NaCH₃COO_(aq) + H₂O_(l)

Similarly, we can assume 100% removal of acid molecules by OH^— ions, but any remaining acid molecules will be only partially ionising depending on

K

_a

= [H

⁺

] [A

^—

] / HA

^but [A^—] and [HA] will have changed as will [H⁺] and pH Example 1: Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium

hydroxide after the addition to the acid solution of a) 10.0 mL of 0.100 M NaOH, b) 25.0 mL of 0.100 M NaOH, c) 35.0 mL of 0.100 M NaOH. K_a = 1.8 x 10^-5. a) step 1: calculate number of moles of OH^— ions added,

n = C x L

n = 0.100 x 0.010 = 0.0010 moles of OH^— ions step 2: calculate the number of moles of CH₃COOH present at start

n = C x L

n = 0.100 x 0.025 = 0.0025 moles of CH₃COOH step 3: calculate the number of moles of CH₃COOH remaining and the number of moles of

CH₃COO^— formed in the new equilibrium mixture,

NaOH_(aq) + CH₃COOH_(aq) NaCH₃COO_(aq) + H₂O_(l) CH₃COOH = 0.0025 - 0.0010 = 0.0015 moles CH₃COO^— = 0.0010 moles

step 4: calculate [H⁺],

K

_a

= [H

⁺

] [A

^—

] / HA

so [H⁺] = K_a [HA] / [A^—]

we are about to substitute values in moles for what should be Molar concentrations. We could calculate concentrations using new overall volumes but this would not change the ratio [HA] / [A^—] so we can get the [H⁺] quicker and easier.

[H⁺] = K_a [HA] / [A^—] = (1.8 x 10^-5) (0.0015)/(0.0010) = 2.7 x 10^-5 M step 5: calculate pH,

pH = -log[H

⁺

]

so pH = -log (2.7 x 10^-5) = 4.57

A solution containing a reasonably high concentration of acid molecules (CH₃COOH) and high concentration of salt ions (CH₃COO^—) will produce a buffering effect. This explains why pH changes 'slow down' in the previous pH curve, between about 5 and 20 ml added.

Buffers will be covered in more detail later.

Example 1: Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid by sodium hydroxide after the addition to the acid solution of a) 10.0 mL of 0.100 M NaOH, b) 25.0 mL of 0.100 M NaOH, c) 35.0 mL of 0.100 M NaOH. K_a = 1.8 x 10^-5.

b) This would be the equivalence point - all CH₃COOH molecules would have reacted and there would be no excess OH^— ions.

The pH would be determined by the salt ions (CH₃COO^—). Later in this unit, you will learn how to calculate the pH of salt solutions.

c) step 1: calculate number of moles of OH^— ions added,

n = C x L

n = 0.100 x 0.035 = 0.0035 moles of OH^— ions step 2: calculate the number of moles of CH₃COOH present at start

n = C x L

n = 0.100 x 0.025 = 0.0025 moles of CH₃COOH step 3: calculate the number of moles excess OH^— ions

NaOH_(aq) + CH₃COOH_(aq) NaCH₃COO_(aq) + H₂O_(l) OH^— = 0.0035 - 0.0025 = 0.0010 moles

we are going to assume that the contribution due to the salt ions (CH₃COO^—) will be negligible compared to the excess OH^— ions.

step 4: calculate [OH^—],(don't forget to combine volumes for overall volume of solution)

C = n /L

C = 0.0010 / 0.060 = 0.0167 M [OH^—] = 1.67 x 10^-2 step 5: convert to [H⁺],

K

_w

= 10

^-14

= [H

⁺

] [OH

^—

]

[H⁺] = 10^-14 / (1.67 x 10^-2) = 5.99 x 10^-13 M step 6: calculate pH,

pH = -log[H

⁺

]

pH = -log(5.99 x 10^-13) = 12.22

In the previous example, we had a significant amount of our original acid molecule (HA) and the conjugate base (A^-) so we had to use our equilibrium constant, K_a , to calculate pH - in effect we were factoring in the buffering effect of the A ^- ion.

[H⁺] = K_a [HA] / [A^—]

Later you will be introduced to the Henderson- Hasselbalch equation that provides a direct route from relative amounts of salt (A^-) and acid (HA) to pH:

pH = pK_a + log( [A^—] / [HA])

8.5 Acid-Base Titrations

pH Titrations - Strong Acid / Strong Base

Though titrations are usually done with indicators, we learn much more by monitoring pH throughout the course of a titration as shown in the arrangement below.

The features of these Titration Curves are as follows:

Volume added (mL)

Adding NaOH to 0.1 M HCl Adding HCl to 0.1 M NaOH HCl_(aq) + NaOH_(aq) ⇨ NaCl_(aq) + H₂O_(l) NaOH_(aq) + HCl_(aq) ⇨ NaCl_(aq) + H₂O_(l) 0

Acid is 100% ionised so H⁺_(aq) and Cl^—_(aq) ions present in large quantities. [H⁺]=[Cl^—]

pH = -log (0.1) = 1.0

Small number of OH^—_(aq) ions (10^-13) present due to autoionisation of water.

Base is 100% ionised so Na⁺_(aq) and OH^—_(aq) ions present in large quantities. [OH^—]=[Na⁺]

pOH = -log (0.1) = 1, pH = 13.0 Small number of H⁺_(aq) ions (10^-13) present due to autoionisation of water.

0 - 20

[H⁺] drops to half - 0.05M (replaced by Na⁺_(aq))

pH = -log (0.05) = 1.30 pH changes very little.

[OH^—] drops to half - 0.05M(replaced by Cl^—_(aq))

pOH = -log (0.05) = 1.30, pH = 12.70 pH changes very little.

(90% completed)36

[H⁺] drops to 10% - 0.01M (replaced by Na⁺_(aq))

pH = -log (0.01) = 2.0

Tenfold dilution cause pH to change by one.

[OH^—] drops to 10% - 0.01M(replaced by Cl^—_(aq))

pOH = -log (0.01) = 2.0, pH = 12.0 Tenfold dilution cause pH to change by one.

(99% completed)39.6

[H⁺] drops to 1% - 0.001M (replaced by Na⁺_(aq))

pH = -log (0.001) = 3.0

Tenfold dilution cause pH to change by one.

[OH^—] drops to 1% - 0.001M (replaced by Cl^—_(aq))

pOH = -log (0.001) = 3.0, pH = 11.0 Tenfold dilution cause pH to change by one.

39.96

(99.9% completed, 1 drop (0.05 mL)

away from completion

[H⁺] drops to 0.1% - 0.0001M pH = -log (0.0001) = 4.0

Tenfold dilution cause pH to change by one.

[OH^—] drops to 0.1% - 0.0001M

pOH = -log (10^-4) = 4.0, pH = 10.0 Tenfold dilution cause pH to change by one.

Volume added (mL)

Adding NaOH to 0.1 M HCl Adding HCl to 0.1 M NaOH HCl_(aq) + NaOH_(aq) ⇨ NaCl_(aq) + H₂O_(l) NaOH_(aq) + HCl_(aq) ⇨ NaCl_(aq) + H₂O_(l) 39.96 - 40.0

(100% completed)

[H⁺] drops to 10^-7 M (same as water)

pH = -log (10^-7) = 7.0

During last 0.04 mL (about 1 drop) the pH will change by at least 3 units - sometimes more, as slight excess OH^— in last drop.

[OH^—] drops to 10^-7 M (same as water)

pOH = -log (10^-7) = 7.0, pH = 7.0 During last 0.04 mL (about 1 drop) the pH will change by at least 3 units - sometimes more, as slight excess H⁺ in last drop.

(equivalence 40 point)

moles NaOH added = moles HCl originally Concentration of acid solution could be calculated (if unknown) or checked (if known)

moles HCl added = moles NaOH originally Concentration of base solution could be calculated (if unknown) or checked (if known)

> 40

Effectively adding NaOH to water so [OH^—] will increase and pH will continue to rise.

Each addition of NaOH is having smaller effect than at start of titration due to larger volume of solution.

Tenfold increase in [OH^—] cause pH to change by one.

Effectively adding HCl to water so [H⁺] will increase and pH will continue to fall.

Each addition of HCl is having smaller effect than at start of titration due to larger volume of solution.

Tenfold increase in [H⁺] cause pH to change by one.

Understanding the shape and behaviour of a titration curve is crucial to selecting the correct indicator for a particular titration.

The large pH changes around

the equivalence point in a strong

acid/strong base titration make

it relatively easy to find suitable

indicators.

You will learn about indicators

in more detail in a later section,

but the crucial thing is that it

takes a pH change of 2 to ensure

a complete change in colour.

For example, phenolpthalein is

colourless at all pH's <8, but will be strong pink at pH > 10.

Phenolpthalein could be successfully used in a titration as described above but, to ensure a strong pink colour (pH 10), a slight excess of NaOH would be needed so the end-point detected could be slightly different from the true equivalence point. The difference, however, should be no more than a single drop from the burette, so may be acceptable.

Similarly, methyl red would complete it's colour change (4.5 - 6.5, yellow - red) slightly before the equivalence point. However, as previously shown, the last drop added from the burette will cause a massive pH change so end-point and equivalence point will probably still coincide.

Bromothymol blue would be ideal indicator as end-point and equivalence point should be same.

Rule of thumb: the pH of mid-point of the colour change and the pH of the equivalence point should be close as possible.

Remember that it is only for strong acid / strong base titrations that pH of equivalence point = 7.

pH Titrations involving Weak Acids and Weak Bases

There are a number of general differences between the curves involving a weak acid compared to a strong acid, and between a weak base compared to a strong base.

Strong Acid Weak Acid Weak Base Strong Base

e.g 0.1M HCl_(aq) e.g 0.1M CH₃COOH_(aq) eg. 0.1M NH_3(aq) e.g 0.1M NaOH_(aq) starts at low pH = 1, due to

100% ionisation starts at higher pH ≃ 3, due

to lower ionisation <1% starts at lower pH ≃ 11, due

to lower ionisation <1% starts at high pH = 13, due to 100% ionisation

pH changes slowly until very close to equivalence

pH changes rapidly at first but then more slowly until very close to equivalence

pH changes rapidly at first but then more slowly until very close to equivalence

pH changes slowly until very close to equivalence

large region of pH change around equivalence - about 8 pH units

smaller region of pH change around equivalence - about 4 pH units

smaller region of pH change around equivalence - about 4 pH units

large region of pH change around equivalence - about 8 pH units

At equivalence, pH = 7 At equivalence, pH > 7 At equivalence, pH < 7 At equivalence, pH = 7