Physical Sciences Gr12, Cheat Sheet of Physics

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Physical Sciences

GRADE

SELF STUDY GUIDE

ACIDS AND BASES

TABLE OF CONTENTS Page

• 1. Introduction
• 1. How to use this self-study guide
  • 3.1 Extract from the examination guidelines 4 – 3. Acid and basis
  • 3.2 Mind map
  • 3.3 Definitions
  • 3.4 Arrhenius and Lowrey-Brønsted models
  • 3.5 Strong acids/bases and weak acid/bases 9 –
  • 3.6 Concentrated/Dilute acids/bases
  • 3.7 Conjugate acid-base pairs 12 –
  • 3.8 Hydrolysis 13 –
  • 3.9 pH and the pH scale and pH calculations 15 –
  • 3.10 Titrations and stoichiometric calculations 21 –
  • 3.11 Ka and Kb values 28 –
  • 4.1 Multiple choice questions 30 - 4. Exercises
  • 4.2 Structured questions
  • 4.3 Typical examination questions 34 -
• 1. Solutions to exercises 37 –
• 1. Examination guidance
• 1. Study and examination tips
• 1. References
• 1. Acknowledgement

2 How to use the booklet

This booklet is meant to help you improve your understanding of the subject Physical Sciences. It

summarises the work that must be studied for examination purposes. This guide does not give full

explanations of all the concepts. Its aim is to assist you in the understanding of the important facts as

highlighted in the examination guidelines, it gives tips and suggested methods of solving problems,

and on how to answer certain questions, your textbook will explain the work in depth, thus it does not

replace your textbook, it should be used in conjunction with your preferred textbook. The authors

wrote this booklet using their experience in a classroom situation.

This guide aims to help you improve your performance, or to help you score marks in Acids and

Bases. Examples are given with solutions/answers, some explanations are provided next to the

solution/answer to enhance your understanding. After studying a certain example, using a blank

paper/cardboard shield the solution/answer for that example and try to solve it on your own without

looking at the answer. Use the provided solutions to mark your own work. If you fail to get it right on

the first attempt, do not give up, keep on trying until you can solve it successfully.

Solutions to the exercises are provided in this booklet. Attempt the exercises without looking at the

solutions. After doing an exercise compare your solution to the one provided and go through the

provided solutions carefully and make sure you understand the steps taken to solve the

problem/question. If you cannot do a certain exercise, go back to the relevant section/theory and

study it again.

͵Ǥ Acids and Bases 

3.1 Extracts from the Examination Guidelines

Acid-base reactions

x Define acids and bases according to Arrhenius and Lowry-Brønsted:

Arrhenius theory: Acids produce hydrogen ions (H +) in solution. Bases produce

hydroxide ions (OH-) in solution.

Lowry-Brønsted theory: An acid is a proton (H +^ ion) donor. A base is a proton (H +

ion) acceptor.

x Distinguish between strong acids/bases and weak acids/bases with examples.

Strong acids ionise completely in water to form a high concentration of H 3 O +^ ions.

Examples of strong acids are hydrochloric acid; sulphuric acid and nitric acid.

Weak acids ionise incompletely in water to form a low concentration of H 3 O +^ ions.

Examples of weak acids are ethanoic acid and oxalic acid.

Strong bases dissociate completely in water to form a high concentration of OH -

ions. Examples of strong bases are sodium hydroxide and potassium hydroxide.

Weak bases dissociate/ionise incompletely in water to form a low concentration of

OH-^ ions. Examples of weak bases are ammonia, calcium carbonate, potassium

carbonate, calcium carbonate, sodium hydrogen carbonate.

x Distinguish between concentrated and dilute acids/bases.

Concentrated acids/bases contain a large amount (number of moles) of acid/base

in proportion to volume of water.

Dilute acids/bases contain a small amount (number of moles) of acid/base in

proportion to volume of water.

x Write down the reaction equations of aqueous solutions of acids and bases.

Examples: HCℓ(g) + H 2 O(ℓ) → H 3 O + (aq) + Cℓ - (aq) (HCℓ is a monoprotic acid.)

NH 3 (g) + H 2 O(ℓ) → NH  4 (aq) + OH-(aq)

H 2 SO 4 (aq) + 2H 2 O(ℓ) → 2H 3 O +^ (aq) + SO 24 (aq) (H 2 SO 4 is a diprotic

acid.)

x Identify conjugate acid-base pairs for given compounds. When the acid, HA, loses

a proton, its conjugate base, A - , is formed. When the base, A - , accepts a proton,

its conjugate acid, HA, is formed. These two are a conjugate acid-base pair.

x Describe a substance that can act as either acid or base as ampholyte. Water is a

good example of an ampholyte substance. Write equations to show how an

ampholyte substance can act as acid or base.

o Describe measures that need to be in place to ensure reliable results.

o Interpret given results to determine the unknown concentration.

pH and pH scale

x Explain the pH scale as a scale of numbers from 0 to 14 used to express the acidity

or alkalinity of a solution.

x Calculate pH values of strong acids and strong bases using pH = -log[H 3 O +].

x Define Kw as the equilibrium constant for the ionisation of water or the ionic product

of water, i.e. Kw = [H 3 O +][OH-]=1 × 10^14 at 298 K.

x Explain the auto-ionisation of water i.e. the reaction of water with itself to form H 3 O +

ions and OH-^ ions.

x Interpret Ka values of acids to determine the relative strength of given acids.

Interpret Kb values of bases to determine the relative strength of given bases.

x Compare strong and weak acids by looking at:

o pH (monoprotic and diprotic acids)

o Conductivity

o Reaction rate

3.2 Mind Map

Conjugate Acid-Base Pairs

Lowry-Brønsted Models x^

Lowry-Brønsted acid

is a proton donor. x^

Lowry-Brønsted base

is a proton acceptor. Strong Acid

-ionises

completely in water. Strong base

-dissociates

completely in water. Weak Acid

• ionises

incompletely in water. Weak Base

dissociates/ionises incompletely in water. Concentrated Acids/Bases

Dilute Acids/Bases

Neutralisation Reactions ofCommon laboratory Acids

and Bases

Hydrolysis

• the reaction of

pH Scale and pH a salt with water

calculations

Titration Calculations-

ACIDS AND BASES stoichiometric calculations

Arrhenius Models Acids

produce hydrogen

ions (H

+^ ) in solution.

Bases

produce hydroxide

ions (OH

• ) in solution.

3.4 Arrhenius and Lowry-Brønsted Models

Examination Guidelines

Arrhenius theory: Acids produce hydrogen ions (H +^ ) in solution. Bases produce hydroxide ions (OH -^ ) in solution.

Lowry-Brønsted theory: An acid is a proton (H +^ ion) donor. A base is a proton (H +^ ion) acceptor.

According to the Arrhenius theory Acids produce hydrogen ions (H +^ ) in solution.

e.g. HNO 3 (aq) → H+^ (aq) + NO 3 - (aq), H+^ was produced thus it’s an Arrhenius acid,

Bases produce hydroxide ions (OH - ) in solution.

e.g. KOH(aq) → K+(aq) + OH-(aq), OH-^ was produced, thus it’s an Arrhenius base.

According to the Lowry-Brønsted theory, an acid is a proton (H +^ ion) donor, and a base is

a proton (H+^ ion) acceptor.

e.g HNO 3 (aq) + H 2 O(ℓ) ֖ H 3 O +^ (aq) + NO 3 - (aq)

HNO 3 donated a proton to H 2 O, it is a Lowry-Brønsted acid, and H 2 O accepted the proton,

so it is a Lowry-Brønsted base.

3.5 Strong acids/bases and weak acids/bases

Examination Guidelines

Strong acids ionise completely in water to form a high concentration of H 3 O +^ ions. Weak acids ionise incompletely in water to form a low concentration of H 3 O +^ ions. Strong bases dissociate completely in water to form a high concentration of OH -^ ions. Weak bases dissociate/ionise incompletely in water to form a low concentration of OH -^ ions.

H +

Acids

Strong acids ionise completely in water to form a high concentration of H 3 O +^ ions.

Weak acids ionise incompletely in water to form a low concentration of H 3 O +^ ions.

Bases

Strong bases dissociate completely in water to form a high concentration of OH -^ ions.

(In a NaOH solution, all the base dissociate to form ions, there are now Na+^ and OH-^ ions

in the container)

Weak bases dissociate/ionise incompletely in water to form a low concentration of OH -

ions. (There are fewer ions in the solution compared to the base itself, e.g. in a Zn(OH) 2

solution there is a low amount of Zn2+^ and OH-^ ions compared to the Zn(OH) 2 molecules)

Please familiarise yourself with the list/table given below.

STRONG ACIDS STRONG BASES

HCℓ - Hydrochloric acid (monoprotic) HNO3 – Nitric acid (monoprotic) H2SO4 – Sulphuric acid (diprotic) H3PO4 – Phosphoric acid (triprotic)

NaOH – Sodium hydroxide KOH – Potassium hydroxide LiOH – Lithium hydroxide Ba(OH)2 – Barium hydroxide WEAK ACIDS WEAK BASES CH3COOH – Acetic acid (COOH)2 – Oxalic acid

NH3 - Ammonia Zn(OH)2 – Zinc hydroxide Note:

A monoprotic acid is an acid that can donate one proton only.

e.g. H Cℓ

A diprotic acid is an acid that can donate a maximum of two protons.

e.g. H 2 SO 4

HA

A-

H +

HA

A-

H +

The acid HA ionised completely. There are no HA molecules present. Only ions are present.

The acid HA ionised incompletely. HA molecules are still present. Very few ions are present.

Only ONE proton is available/can be donated.

TWO protons are available/can be donated.

3.7 Conjugate acid-base pairs

Examination Guidelines

Identify conjugate acid-base pairs for given compounds. When the acid, HA, loses a proton, its conjugate base, A -^ , is formed. When the base, A -^ , accepts a proton, its conjugate acid, HA, is formed. These two are a conjugate acid-base pair.

Describe a substance that can act as either acid or base as ampholyte. Water is a good example of an ampholyte substance. Write equations to show how an ampholyte substance can act as acid or base.

You must be able to write balanced equations of aqueous solutions of acids and bases and be able to identify conjugate acid-base pairs. Conjugate acid-base pairs are compounds that differ from each other by a proton (H +^ ion)

To form a conjugate acid, add a proton.

Example

The conjugate acid of:

(a) NH 3 is NH 4 +^ (A proton (H +^ ) was added to NH 3 )

(b) HCO 3 -^ is H 2 CO 3 (A proton (H +^ ) was added to HCO 3 -^ )

To form a conjugate base, remove a proton.

Example

The conjugate base of:

(a) HCO 3 -^ is CO 3 ଶି^ (A proton was removed from HCO 3 )

(b) H 2 O is OH -^ (A proton was removed from H 2 O)

Example Identify conjugate acid base pairs.

The conjugate acid-base pairs are:

NH 3 and N 4 +^ and, H 2 O and H 3 O +^.

Forward reaction

NH 3 is a base because it accepts a proton from H 2 O, and H 2 O is an acid because it donates a proton to NH 3.

Reverse reaction

NH 4 +^ is an acid because it donates a proton to OH -^ , and OH -^ is a base because it accepts a proton from NH 4 +^.

This may assist especially when answering multiple choice questions.

Base 1 Acid 1

Acid 2 Base 2

NH 3 (aq) + H 2 O ֖  NH (aq) + OH -^ (aq)

Exercise

Identify conjugate acid-base pairs in the following balanced equations.

(a) HCO 3 -^ (aq) + H2O(ℓ) ֖ H 2 CO 3 (aq) + OH - (aq)

(b) HCO 3 -^ ( aq) + H2O(ℓ) ֖  CO 3 ଶ-^ ሺaq) + H 3 O +(aq)

Solution

(a) H 2 CO 3 is the conjugate acid of HCO 3 -^ (base)

OH-^ is the conjugate base of H 2 O (acid)

(b) CO 3 -^ is the conjugate base of HCO 3 -^ -^ (acid)

H 3 O +^ is the conjugate acid of H 2 O (base)

3.8 Hydrolysis

Examination Guidelines

Define hydrolysis as the reaction of a salt with water. Determine the approximate pH (equal to, smaller than or larger than 7) of salts in salt hydrolysis.

o Hydrolysis of the salt of a weak acid and a strong base results in an alkaline

solution i.e. the pH > 7. Examples of such salts are sodium ethanoate, sodium

oxalate and sodium carbonate.

o Hydrolysis of the salt of a strong acid and a weak base results in an acidic solution

i.e. the pH < 7. An example of such a salt is ammonium chloride.

o The salt of a strong acid and a strong bases does not undergo hydrolysis and the

solution of the salt will be neutral i.e. pH = 7.

Hydrolysis is the reaction of a salt with water.

The salt of a strong acid and a weak base is acidic, pH < 7.

The salt of a weak acid and a strong base is basic, pH > 7.

The salt of a strong acid and a strong base does not undergo hydrolysis.

In both equations, it can be noted that in (a) HCO 3 -^ is a base and in (b) it is an acid, thus HCO 3 -^ is an ampholyte , i.e., it can act as either an acid or a base. H 2 O is also an ampholyte as can be seen in both (a) and (b).

Hydroxide ion

Hydronium ion

Is the salt KHCO 3 acidic or basic? Explain with the aid of a balanced chemical equation.

Solution

HCO 3 - (aq) + H 2 O(ℓ) ֖ H 2 CO 3 (aq) + OH -^ (aq)

The salt is basic , OH -^ is formed which is a base (A strong base reacted with a weak acid)

3.9 pH and the pH scale and pH calculations

Examination Guidelines

x Explain the pH scale as a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution. x Calculate pH values of strong acids and strong bases using pH = -log[H 3 O +^ ]. x Define the concept of K w as the equilibrium constant for the ionisation of water – the ionic product of water (ionisation constant of water). x Explain the auto-ionisation of water i.e. the reaction of water with itself to form H 3 O +^ ions and OH -^ ions.

The pH scale is a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution.

Acids have a pH less than 7 (pH<7), a neutral solution has a pH equal to 7 (pH=7), bases have a pH greater than 7 (pH>7).

KHCO 3

K+^ HCO^3 -

KOH H^2 CO^3

Strong base Weak acid

Acid-base indicators are used to test the pH of solutions, the common indicators are:

x Methyl orange (red in an acidic solution and yellow in a basic solution)

x Bromothymol blue (yellow in an acidic solution and blue in a basic solution)

x Phenolphthalein (colourless in an acidic solution and pink in a basic solution)

Calculation of pH

To calculate the pH of a solution the concentration of H 3 O +^ must be known or be determined. The reaction of water with itself produces H 3 O +^ and OH -^ ions, it is called the auto-ionisation of water. The equation is given below:

H 2 O(ℓ) + H 2 O(ℓ) ֖ H 3 O +^ (aq) + OH -^ (aq)

The expression for the equilibrium constant of the chemical equation given above is:

k (^) w = ൣH^3 O

• (^) ൧[OH - (^) ] [H 2 O]^2

the concentration of water is very large (constant) compared to the concentrations of H 3 O +^ and OH - /water is a pure liquid so its concentration will be 1 mol.dm-3, it can therefore be omitted from the

equilibrium expression, thus k (^) w = ൣH^3 O

• (^) ൧[OH - (^) ] 1 , this expression is called the^ ionic product^ of water, the symbol Kw is used, Kw = [H 3 O +^ ][OH -^ ]. Kw =1×10-14^ at 25°C.

For a neutral solution [H 3 O +^ ] = [OH -^ ] = 1×10-7^ mol·dm-^.

For an acidic solution [H 3 O +^ ] > 1×10-7^ mol·dm -3^ and [H 3 O +^ ] > [OH -^ ].

For a basic solution [H 3 O +^ ] < 1×10-7and [H 3 O +^ ] < [OH -^ ].

Copy and complete the following table, use Kw = [H 3 O +^ ][OH -^ ] = 1×10-

pH 2 3 5 6 7 9 10 12 13 [H 3 O +^ ] mol·dm -3^ 1×10-2^ 1×10-3^ 1×10-6^ 1×10-9^ 1×10-12^ 1×10- [OH -^ ] mol·dm -3^ 1×10-9^ 1×10-8^ 1×10-7^ 1×10-5^ 1×10 -

Solution

Use the expression [H 3 O +^ ][OH -^ ]=1×10-14^ to calculate the unknown value,

e.g. [H 3 O +^ ][OH -^ ]=1×10 -

H 3 O +^ =1×10-

Therefore, [H 3 O +^ ]=1×10-1^ mol·dm -

pH 2 3 5 6 7 9 10 12 13 [H 3 O +^ ] mol·dm -3^ 1×10-2^ 1×10-3^ 1×10-5^ 1×10-6^ 1×10-7^ 1×10-9^ 1×10 -10^ 1×10-12^ 1×10- [OH -^ ] mol·dm -3^ 1×10-12^ 1×10-11^ 1×10-9^ 1×10-8^ 1×10-7^ 1×10-5^ 1×10 -4^ 1×10-2^ 1×10-

Examples

1. Calculate the pH of a 0,4 mol·dm -3^ NaOH solution.

NaOH(aq) ֖ Na +^ (aq) + OH -^ (aq)

0,4 mol·dm -3^ 0,4 mol·dm-3^ ( 1:1 ratio )

׵ [OH -^ ] = 0,4 mol·dm -

. [H 3 O +^ ][OH -^ ]=1×10 -

H 3 O +^ =1×10 -

׵ [H 3 O +^ ] = 2,5 ×10-14^ mol·dm -

pH = -log [H 3 O +^ ]

= -log (2,5 ×10-14^ )

= 13,

2. Calculate the pH of a 0,01 mol·dm -3^ Ba(OH) 2 solution.

Ba(OH) 2 (aq) ֖ Ba2+^ (aq) + 2OH -^ (aq)

0,01 mol·dm-3^ 2(0,01) mol·dm -3^ ( 1:2 ratio )

׵ [OH -^ ] = 0,02 mol·dm -

[H 3 O +^ ][OH -^ ]=1×10 -

H 3 O +^ =1×10-

׵ [H 3 O +^ ] = 5 ×10-13^ mol·dm-

pH = -log [H 3 O +^ ]

= -log (5 ×10 -13)

= 12,

Calculating the concentration of an acid or base if given the pH

1. Calculate the concentration of the acid HCℓ with a pH = 4,

Solution

x Write down a balanced equation for the ionisation reaction of the acid (reaction with

water)

x Substitute the pH value in the formula pH = -log[H 3 O +^ ] and calculate the concentration of

[H 3 O +^ ] using antilog, use the button on your calculator, i.e., press 2 nd^ F/SHIFT

then

x Use ratios to determine the concentration of the acid.

HCℓ(aq) + H 2 O(ℓ) ֖ H 3 O +^ (aq) + Cℓ -^ (aq)

pH = -log [H 3 O +]

4,5 = -log [H 3 O +^ ]

[H 3 O +^ ] = 10 -4,

= 3,162 × 10 -5^ mol·dm -

Therefore [HCℓ] = 3,162 × 10 -5^ mol·dm -3^ ( 1:1 ratio )

log