M6 ABR2

IQ2 What is the role of water in solutions of acids and bases?

IQ2 SYLLABUS STATEMENTS

2.1 conduct a practical investigation to measure the pH of a range of acids and bases

2.2 calculate pH, pOH, hydrogen ion concentration ([H+]) and hydroxide ion concentration ([OH–]) for a range of solutions

2.3 conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids and bases

2.4 construct models and/or animations to communicate the differences between strong, weak, concentrated and dilute acids and bases

2.5 calculate the pH of the resultant solution when solutions of acids and/or bases are diluted or mixed

2.6 write ionic equations to represent the dissociation of acids and bases in water, conjugate acid/base pairs in solution and amphiprotic nature of some salts, for example:

– sodium hydrogen carbonate

– potassium dihydrogen phosphate

Using the Bronsted-Lowry Theory

NOTE: You will see variations of the hydronium ion throughout chemistry: H3O+ (complex of H2O plus H+) is preferred in equations to H+, while H+ is simpler to use in pH calculations.

2.1 conduct a practical investigation to measure the pH of a range of acids and bases

Introduction

Previously we looked at indicators, especially universal indicator, as a method for identifying the acidity of a solution. pH meters and/or probes can also be used to identify the acidity or basicity of a solution.

In 1909, the Danish chemist Søren Sørensen published a technical paper on the effect of hydrogen ion concentration on enzyme activity. Later, he devised (created) the pH scale, which is of enormous practical use in the life sciences. In 1935, sensitive milli-voltmeters became available to provide convenient electrochemical measurement of pH. (Pearson Education Australia, 2019)

Practical 2.1.1 Measuring pH

Method

1 Warning: Wear safety glasses!

2 Design a table to record the data obtained in steps 4, 5 and 6 below.

3 Familiarise yourself with the correct procedure for using a pH meter or probe (PASCO pH Logger).

4 Transfer 5 mL of each test sample into a test tube. Select 10 different substances for testing purposes.

5 Measure and record the pH of each test sample using a pH meter.

6 Add a few drops of universal indicator and record the resultant colour. Identify the pH using a universal indicator colour chart.

Practical 2.1.2

Procedure

Step 1: Attach the pH probe and adaptor to the pH meter.

NOTE: On the pH meter, there is a electronic display screen that shows the pH value which you can read off from and determine the pH of the substance.

Step 2: Attach the adaptor to the electrical power supply

Step 3: Wash the lower part of the probe with distilled water.*

Step 4: Pour 25ml of 0.1M HCl, 0.1 NaOH, 0.1M ammonia, orange juice, vinegar, red wine, distilled water, soapy water, oven cleaner into separate test tubes and label as test tubes 1-9 respectively along with the name of solution it contains.

Step 5: Submerge the pH probe into test tube 1 and record the pH shown on the pH meter.

Step 6: Wash the pH thoroughly with distilled water.*

Step 7: Repeat steps 5-6 to measure the pH of a different substance in another test tube.

*NOTE: You can also insert the pH probe in a buffer solution with a pH of 7 (i.e. neutral buffer) in a separate beaker to reset the pH reference point to 7 rather than rinsing the probe with distilled water. Note that distilled water have a pH of 7 which we used to rinse the pH probe.

Results

You should have obtained a pH value of less than 7 for all the acids that you measured using your pH probe. Vice versa, you should have obtained a pH value of more than 7 for all the bases that you have measured using your pH probe. For neutral substances, such as distilled water, they should have a pH of 7.

The key objective of this practical, apart from learning how to measure pH of substances, is for you to identify that:

Acids have a pH of less than 7.
Bases have a pH of more than 7.
Neutral substances have a pH equal to 7.

VIEW Video:

ABR#10 https://www.youtube.com/watch?v=ACdnX4nuXc8&index=5&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS [6.39 mins]

TASK 2.1.1

1. (2009, Q3)

Which of the following groups contains ONLY acidic substances?

(A) Antacid tablets, baking soda, laundry detergents

(B) Blood, oven cleaner, seawater

(D) Vinegar, wine, aspirin

2. Evaluate the accuracy of universal indicator as a tool for determining the pH of a solution. (3 marks)

3. What information can be inferred from a pH probe which has a reading of 9.3? (2 marks)

2.2 calculate pH, pOH, hydrogen ion concentration ([H+]) and hydroxide ion concentration ([OH–]) for a range of solutions

The pH scale is the most convenient and comprehensive tool for identifying the relative acidity or basicity of solutions. The table below shows the colours and their corresponding pH values, and some examples of substances which have the corresponding pH. We can use the pH scale not only to compare acids and bases, but also to compare relative strengths of different acids.

Look at the table on the following page and discuss the following questions:

a) What could you say about the acid strength in black coffee compared to ‘pure’ water?

b) Why do you think toothpaste has such a high pH?

Research has found that the purest form of water weakly conducts electricity (is a weak electrolyte), which indicates the presence of small quantities of ions. These ions are hydrogen and hydroxide ions, formed when water molecules (H2O) dissociate.

The interaction between two water molecules leads one molecule to act as an Bronsted-Lowry acid and the other water molecule to act as a Bronsted-Lowry base.

H₂O_(l)+ H₂O_(l) <-> H₃O⁺_(aq) + OH^-_(aq)

H₂O ↔ H⁺ + OH⁺ (written with the hydrogen ion)

It is this reaction that leads to the auto-ionisation of water, in which the interaction between the two water molecules brings about their dissociation .

NOTE: A closed system of water reaches equilibrium: there will be water molecules, hydrogen ions and hydroxide ions in the system. This will therefore have an equilibrium constant. Since water molecules are the only reactants, we call this the auto-ionisation of water constant, Kw.

Kw = [H₃O⁺] x [OH^-]

Recall that pure liquid water does not change its concentration, so it is not be included in the equilibrium constant expression.

At 25^oC: Kw = 1.0 x 10^-14

Recall that equilibrium constant values, such as Kw, will change depending on temperature

Kw = 1.0 x 10^-14 = [H₃O⁺] x [OH^-]

There is a mathematical way to determine the acidic, basic or neutral nature of a solution, that is to use a function called pH:

pH, a measure of acidity, means the [H⁺] is being multipled with -log₁₀.

pH = -log₁₀[H⁺].

At 25^oC, [H⁺] = 10^-7 mol.L^-1. This corresponds to a pH of 7.

Similarly, pOH, a measure of basicity, means the [OH^-] is being multipled with -log₁₀.

pOH = log₁₀[OH^-].

At 25^oC, [OH^-] = 10^-7 mol.L^-1. This corresponds to a pH of 7

And therefore: pH + pOH = 14

What is the significance of this?

This means that if we know pH, we can determine the pOH using pH + pOH = 14.
This also means that if we know the [H⁺] or [OH^-] we can calculate both pH and pOH. Or, you can calculate it through using using the ionisation of water constant: Kw = 1.0 x 10^-14 = [H₃O⁺] x [OH^-]

If the pH of the solution is 7 at 25^oC, then [H⁺] and [OH^-] are both equal to 1.0 x 10^-7

Neutral solutions: pH = 7 [H⁺] = 10^-7 mol L^-1

If the pH of the solution is less than 7 (acidic) at 25^oC, the [H₃O⁺] must be greater than 1.0 x 10^-7.

Acid solutions: pH < 7 [H⁺] > 10^-7 mol L^-1

If the pH of the solution is greater than 7 (basic) at 25^oC, the [OH^-] must be greater than 1.0 x 10-^-7

Basic solutions: pH > 7 [H⁺] < 10^-7 mol L^-1

There are several important mathematical relationships shown below.

pH = - log₁₀ [H⁺]

pOH = - log₁₀ [OH^-]

pH + pOH = 14

[H⁺] x [OH^-] = 10^-14

Review:

Q1. 0.005 mol of sulfuric acid is dissolved in sufficient water to make 500 mL of solution. Calculate the concentration of hydroxide and hydrogen ions in this solution.

If the hydrogen ion concentration of a strong acid solution is known, the pH can be calculated by substituting the hydrogen ion concentration into the pH formula:

pH = - log₁₀ [H⁺]

If the concentration of a strong acid solution is known, then the hydrogen ion concentration can be determined based on whether the acid involved is monoprotic, diprotic or triprotic:
- The hydrogen ion concentration of a monoprotic strong acid solution is equal to the concentration of the acid itself.
- The hydrogen ion concentration of a diprotic strong acid solution is equal to twice the concentration of the acid itself.
- The hydrogen ion concentration of a triprotic strong acid solution is equal to three times the concentration of the acid itself ie [H^{+ ]}= 3 x [H₃PO₄]

H₃PO₄ → 3H⁺ + PO₄^3-

Q: Why is a single arrow used, not an equilibrium arrow?

Type 1: Calculating pH

To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per litre (molarity). The pH is then calculated using the expression:

pH = - log [H₃O⁺]

Example: Find the pH of a 0.0025 M HCl solution. HCl is a strong acid and is 100% ionised in water. The hydronium ion concentration is 0.0025 M. Thus:

pH = - log (0.0025) = - ( - 2.60) = 2.60

Type 2: Calculating the [H3O⁺] (or [H⁺]) given pH

The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH.

[H₃O⁺] = 10^-pH or [H₃O⁺]= antilog (- pH)

Example: What is the hydronium ion concentration in a solution that has a pH of 8.34?

8.34 = - log [H₃O⁺]

- 8.34 = log[H₃O⁺]

[H₃O⁺] = 10^-8.34 = 4.57 x 10^-9 M

On a calculator, calculate 10^-8.34, or "inverse" log ( - 8.34).

Type 3: Calculating pOH

To calculate the pOH of a solution you need to know the concentration of the hydroxide ion in moles per litre (molarity). The pOH is then calculated using the expression:

pOH = - log [OH^-]

Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M?

pOH = - log (4.82 x 10^-5) = - ( - 4.32) = 4.32

Type 4: Calculating the [OH^-] given pOH

The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.

[OH-] = 10^-pOH or [OH^-] = antilog ( - pOH)

Example: What is the hydroxide ion concentration in a solution that has a pOH of 5.70?

5.70 = - log [OH^-]

-5.70 = log[OH^-]

[OH-] = 10^-5.70 = 2.00 x 10^-6 M

On a calculator calculate 10^-5.70, or "inverse" log (- 5.70).

Relationship between pH and pOH

The pH and pOH of a water solution at 25^oC are related by the following equation.

pH + pOH = 14

If either the pH or the pOH of a solution is known, the other can be quickly calculated.

Example: A solution has a pOH of 11.76. What is the pH of this solution?

pH = 14 - pOH

pH = 14 - 11.76

pH = 2.24

Note in the following example how polyprotic acids/bases are treated: the concentration of the hydronium or hydroxide ion is doubled (diprotic) or tripled (triprotic) from the original acid molecule concentration, but the terms in the Kw expression are not squared or cubed.

Kw is a specific type of equilibrium expression and is always just [H+] x [OH-].

Carry out simulation 1: https://phet.colorado.edu/en/simulation/ph-scale-basics

Click on the Play button.

1. Choose micro from menu.

2. Switch between linear and log scale

3. Select battery acid to start. Investigate relationship between [H⁺], [OH^-] and pH, and molecule concentrations.

5. Click the orange arrow (bottom) to reset. Change substance.

Carry out simulation 2: https://phet.colorado.edu/en/simulation/ph-scale

VIEW Videos:

ABR#11 https://www.youtube.com/watch?v=e1_bMCXrMJU&index=4&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS [7.07]
Calculating pH etc (shows calculator usage) https://www.youtube.com/watch?v=UiK37I159fc [10.51 mins]
What is pH?: https://www.youtube.com/watch?v=iHxAlR7IwDM&list=PLB755F0C836855EAB&index=39 [14.06 mins]
pH and pOH Crash Course Chemistry https://www.youtube.com/watch?v=LS67vS10O5Y [11.23]
Calculating pH https://youtu.be/-xVR8Sl01F0 [9 mins]
Ka Kb Kw pH etc https://www.youtube.com/watch?v=VbubK1HhWTA [69.54 mins]

TASK 2.2.1

1. (2007, Q10)

A 0.1 mol /L HCl solution has a pH of 1.0. What volume of water must be added to 90 mL of this solution to obtain a final pH of 2.0?

(A) 10 mL (B) 180 mL (C) 810 mL (D) 900 mL.

2. (2006, Q17, 4 marks)

a) Calculate the pH of a 0.2 mol/ L solution of hydrochloric acid. (1 mark)

b) Calculate the pH after 20 mL of 0.01 mol /L sodium hydroxide is added to 50 mL of 0.2 mol /L hydrochloric acid. Include a balanced chemical equation in your answer. (3 marks)

3. (2005, Q8)

What would be the pH of a 0.1 mol L^-1 solution of sulfuric acid?

(A) Less than 1.0

(B) Exactly 1.0

(D) Greater than 7.0

4. A small quantity of tartaric acid (a diprotic acid) was dissolved in water to give a final hydrogen ion concentration of 1.3 x 10-6 mol.L^-1. Calculate the pH of this solution.

5. (2003, Q8)

A sulfuric acid solution has a concentration of 5 × 10−4 mol L^-1.

What is the pH of this solution, assuming the acid is completely ionised?

(A) 3.0

(B) 3.3

(D) 4.0

6. (2004, Q7)

The figure shows the pH values of some substances. Based on the pH values shown in the figure, which of the following statements about the concentration of hydrogen ions is correct?

(A) It is twice as great in milk as that in lemon juice.

(B) It is 1 000 000 times greater in soap than in wine.

(D) It is 1 000 times greater in distilled water than in soap.

TASK 2.2.2

Complete WS (Answers included at back)

12C6 2.2c pH WS w Ans.docx

REVIEW 2.2

Complete Text WS 6.2b P64 Calculating pH
Complete Text WS 6.3 p65 Calculating pH and pOH

PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)

2017 Q20

2016 Q12; Q18

2015 Q13

2014 Q12; Q14

2013 Q14

2012 Q28 3

2008 Q14

2007 Q8; Q10

2005 Q8

2004 Q7

2.3 conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids and bases

Introduction

Acids are substances that have one or more hydrogen atoms covalently bonded to one or more non-metal atoms with a higher electronegativity than hydrogen. In water the hydrogen "loses" its electron to the non-metal atom(s), and becomes a positive ion, a proton (the hydrogen ionises). The non-metal atom or group gains the electron and becomes a negative ion (ionises).

Acids are of different strengths. The strength of an acid relates to its degree of ionisation. The stronger the acid, the more readily it is ionised (considered to be COMPLETELY IONISED). The weaker the acid, the less it is ionised (PARTIALLY IONISED).

All the molecules of a strong acid dissociate completely into the constituent ions (H⁺ ions and the anion) in water. This is called complete ionisation, or 100% degree of ionisation / dissociation.

All the molecules of a weak acid do not dissociate completely into constituent ions in water. Some of the acid molecules dissociate, some do not, i.e. partial dissociation.

Those acid molecules that do not dissociate remain as acid molecules dissolved in water, rather than ions.

It is the difference in the degree of ionisation of acid molecules in water that classifies an acid as a strong acid or a weak acid.

An equilibrium exists between the acid molecules that have not ionised, and the ions formed from molecules that have ionised and dissociated:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq)

CH₃COOH _(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)

In reality, the proton attaches to a water molecule, forming the hydronium ion, but this is often just written in short form as the hydrogen ion.

H⁺ + H₂O _(l) ⇌ H₃O⁺

The extent to which acids are ionised can be determined using a special equilibrium constant known as the acid ionisation constant, Ka. We will look at this in detail in 3.4, so it is just an introduction here.

For weak monoprotic acids which undergo partial ionisation:

[H⁺] [A^-]

Ka = ---------------

[HA]

Dissociation of acetic acid, a weak acid. (See model below.)

[H⁺] [CH₃COO^-]

Ka = --------------------

[CH₃COOH]

The smaller the value of Ka the weaker the acid (the lower the numerator and the higher the value of the denominator.)

In the same way:

The same is true for bases. Bases may already be ions, in which case they are not actually ionised, but the ions dissociate (separate) in water.

Strong bases have 100% degree of ionisation/dissociation where all of the base molecules are ionised in water.

NaOH_(aq) ⇌ Na⁺_(aq) + OH^-_(aq)

For weak bases, not all of the base molecules ionise in water, only a portion do, i.e. partial ionisation/dissociation. Those base molecules that do not dissociate in water remain as base molecules dissolved in solution rather than ions (e.g. OH– and its cation/conjugate acid).

Na₂CO_3(aq) + H₂O_(l) ⇌ 2Na⁺_(aq) + HCO₃^-_(aq) + OH^-_(aq)

[The equilibrium constant in this case is Kb but we will look at these in greater detail in 3.4.]

VIEW Videos:

ABR#12 https://www.youtube.com/watch?v=W9GW8dZssWc&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS&index=12 [6.22 mins]

PRACTICAL 2.3.1

Practical Activity 6.1 Comparing the strengths of acids and of bases Text PP 75-77

PRACTICAL 2.3.2

Introduction

The strength of an acid is dependent on its ability to ionise in water. A strong acid ionises completely; a weak acid ionises poorly. These concepts are easily illustrated by comparing 0.10 mol.L^-1 solutions of hydrochloric, citric and acetic acids:

HCl_(g) + H₂O_(l)→ H₃O⁺_(aq) + Cl^-_(aq) (100% ionised)

C₆H₈O_7(aq) + H₂O_(l) → H₃O⁺_(aq)+ C₆H₇O₇^-_(aq) (8.6% ionised)

CH₃COOH_(aq) + H₂O_(l) → H₃O⁺_(aq) + CH₃COO^-_(aq) (1.3% ionised)

In this experiment you will determine the pH of acetic, citric and hydrochloric acids, and assess their acid strength using thymol blue indicator and the pH probe.

Below pH 1.2 thymol blue is coloured red/pink; in the transition range of pH 1.2–2.8 it is orange; and above pH 2.8 it is yellow.

We will also look at three other solutions: NaOH(aq), NaHCO₃(aq) and Na2CO₃(aq) (all solutions will have a molarity of 0.1 M)

Method

1 Warning: Wear safety glasses!

2 Warning: Use a pipette filler.

3 Transfer 10 mL of 0.10 M acetic acid, 10 mL of 0.10 M citric acid, 10 mL of 0.10 M HCl, 10 mL of 0.10 M sodium hydroxide solution, 10 mL of 0.10 M sodium bicarbonate solution and 10 mL of 0.10 M sodium carbonate solution into separate labelled test tubes.

4 Use the 0.10 M HCl to prepare 10.0 mL of 0.010 M HCl by dilution.

5 Use the formula: C1V1 = C2V2.

6 Store the solution in a labelled test tube.

7 Use the 0.010 M HCl to prepare 10 mL of 0.0010 M HCl by dilution.

8 Use the formula: C1V1 = C2V2.

9 Store the solution in a labelled test tube.

10 You should now have eight test tubes filled with 0.10 M acetic acid, 0.10 M citric acid, 0.10 M HCl, 0.010 M HCl, 0.0010 M HCl, 0.10 M sodium hydroxide, 0.10 M sodium bicarbonate and 0.10 M sodium carbonate.

11 Use the pH probe to record the pH of each solution.

12 Add six drops of thymol blue indicator to each test tube. Observe and record the resultant colours.

Note: Below pH 1.2 thymol blue is red in colour; above pH 2.8 it is yellow.

PRACTICAL 2.3.3

Step 1: Attach the pH probe and adaptor to the pH meter.

NOTE: On the pH meter, there is a electronic display screen that shows the pH value which you read off from and determine the pH of the substance.

Step 2: Attach the adaptor to the electrical power supply

Step 3: Wash the lower part of the probe with distilled water.*

Step 4: Pour 25ml of 0.1M HCl, HNO3, acetic acid (CH3COOH), H3PO4, KOH, NaOH, ammonia (NH3) and NaF into separate test tubes and label as test tubes 1-8 respectively along with the name of solution it contains.

Step 5: Submerge the pH probe into test tube 1 and record the pH shown on the pH meter.

Step 6: Wash the pH thoroughly with distilled water.*

Step 7: Repeat steps 5-6 to measure the pH of a different substance in another test tube.

Expected Results

In the condition of equal acid molecules concentrations, compared to strong acids, weak acids will have a higher pH.

This is because strong acids have more moles of hydrogen ions in solution due to complete dissociation as we have explored in the previous learning objective. This would mean that strong acids have higher concentration of hydrogen ions in solution than weak acids provided the strong and weak acids have equal concentration to start with.

In the condition of equal base molecule concentrations, compared to strong bases, weak bases will have a lower pH.

This is because strong bases have more moles of hydroxide ions in solution if both the strong and weak bases due to 100% ionisation into hydroxide ions and its conjugate acid. This would mean that strong bases have higher concentration of hydroxide ions in solution than weak bases provided than the strong and weak bases have equal concentration.

Acids experiment results:

HCl and HNO3 are strong acids.
CH3COOH and H3PO4 are weak acids. You will expect these to have higher pH (less acidic) than HCl and HNO3 since the concentration of the strong and weak acids are the same.

Bases experiment results:

KOH, NaOH are strong bases
NH3 and NaF are weak bases. You will expect these weak bases to have lower pH (less basic) than KOH and NaOH since the concentration of the strong and weak bases are the same.

TASK 2.3.1

1. (2010, Q21, 3 marks)

A 0.001 M solution of hydrochloric acid and a 0.056 M solution of ethanoic acid both have a pH of 3.0.

Why do both solutions have the same pH?

2. (2007, Q21, 5 marks)

a) State what colour the red cabbage indicator would be in a 0.005 M solution of H₂SO₄. Show your working. (1 mark)

b) Using the red cabbage indicator, what colour would the solution be if 10 mL of 0.005 M H₂SO₄ was diluted to 100 mL? (1 mark)

3. (2004, Q24, 5 marks)

The diagram shows three reagent bottles containing acids.

b) Explain the difference in pH between the three acids in the diagram. (2 marks)

HINT: You are expected to know that citric acid is triprotic (like H3PO4) and weak, and acetic acid is monoprotic and weak.

2.4 construct models and/or animations to communicate the differences between strong, weak, concentrated and dilute acids and bases

Strong acid: An acid that completely ionises in water to form hydrogen ions, leaving no neutral acid molecules remaining.
Weak acid: An acid that only partially ionises in water to from hydrogen ions, leaving neutral molecules remaining.
Concentrated solution: A solution in which the total concentration of acid is high (a large amount of solute is contained in a small amount of solution). The majority of chemists agree that a solution that has a molarity of 3M or higher is considered concentrated.
Dilute solution: A solution in which the total concentration of acid is low (a small amount of solute is contained in a large amount of solution).

Acids may also be classified on the basis of the number of protons they are capable of donating in an acid-base reaction.

• Monoprotic: Donate one proton per molecule eg. HCl

• Diprotic: Donate two protons per molecule eg. H₂SO₄

• Triprotic: Donate three protons per molecule eg. H₃PO₄

PRACTICAL 2.4.1

a) Strong and weak acids

The complete ionisation of a strong acid can be modelled using the following procedure:
- Use molecular model kits to construct four hydrogen chloride molecules and four water molecules, ensuring that the oxygen atom in water has three attachment points.
- Remove a hydrogen atom from each hydrogen chloride molecule and attach each one to a water molecule.
- This forms four chloride ions and four hydronium ions, leaving no neutral acid molecules.
The partial ionisation of a weak acid can be modelled using the following procedure:
- Use molecular model kits to construct four acetic acid molecules and four water molecules.
- Remove a hydrogen atom from one acetic acid molecules and attach it to one water molecule.
- This forms one acetic acid ion and one hydronium ion, leaving three neutral acid molecules.

c) Strong and weak bases

Use the model kits to demonstrate strong and weak bases.

c) Concentrated and dilute acids

Use the model kits, or construct your own, to develop a model to show the difference between concentrated and dilute acids/bases.

VIEW PPTS:

12C6 2.3a Strong Weak 1 PPT.pptx

12C6 2.3a Strong Weak 2 PPT.pptx

TASK 2.4.1

Complete Worksheet

12C6 2.4a AcidBase Model WS for MOV.pdf

VIEW Videos:

ABR#16 https://www.youtube.com/watch?v=fYSl8On-V-A&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS&index=16 [7.13 mins]
Model: https://www.youtube.com/watch?v=6W_Et8I3Mtk&list=PLB755F0C836855EAB&index=37 [9.27 mins]
Strong/weak, dilute/concentrated: https://youtu.be/Y3hW3PolRHw [6.41 mins]

Ka and acid strengthHow to write an equilibrium expression for an acid-base reaction and how to evaluate the strength of an acid using Ka

TASK 2.4.2

a) Explain how models can be used to show the difference between a strong and a weak acid.

b) What are the drawbacks to using a model?

2. (2007, Q8)

Acid X and acid Y are both monoprotic weak acids of equal concentration. Acid X is a stronger acid than acid Y.

Which statement about acid X and acid Y is correct?

(A) Acid Y is completely ionised in solution.

(B) The solution of acid X is less ionised than the solution of acid Y.

(D) 1 mole of acid Y requires a greater volume of 1.0 M NaOH for neutralisation than 1 mole of acid X.

3. (2015, Q13)

Which of the following solutions has the highest pH?

(A) 1.0 M acetic acid (B) 0.10 M acetic acid

4. Why is it incorrect to identify a strong acid solution, such as hydrochloric acid in water, as an equilibrium?

REVIEW 2.4a

VIEW videos

Task 2.4.2

Complete Text WS 6.2a Modelling acids and bases P63

REVIEW 2.4b

PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)

2014 Q8 B
2012 Q18 C
2011 Q15 A
2010 Q21 3

Hydrochloric acid is a strong acid and therefore fully ionises in aqueous solution. Ethanoic acid is a weak acid and does not fully ionise in aqueous solution. The total concentration of ethanoic acid must be higher than that of HCl, due to incomplete ionisation of CH3COOH, to give an equivalent [H+ ] and therefore pH, as pH is equal to –log10 [H+ ].

2009 Q21A 3, 21B 1

Question 21 (a) Sample answer: Acid 1 is a strong acid, equivalence point is at pH = 7 and initial pH is at 1. Acid 2 is a weak acid as pH at equivalence point is >7 and initial pH is >1. Acid 2 has a higher concentration than Acid 1 as it takes more KOH to neutralise it. Answers could include: 38 mL KOH for Acid 2 and 25 mL KOH for Acid 1. Acid 2 is more concentrated since more base is used. Question 21 (b) Answers could include: Potassium acetate CH3COOK Potassium ethanoate.

2007 Q21A 1

Question 21 (a) Better responses included clear working, showing the correct calculation of pH with the correct link to colour and recognition of successive ionisation of diprotic acids, with suggestions of pH changes over a small range corresponding to ionization levels. Weaker responses provided intuitive guesses of colour without reference to pH and confused representations of log formula. Weaker responses included a simple statement of pH, without any working. They also ignored the need to double the acid concentration and double the pH value (based on acid concentration) to find the pH of H+ in H2SO4.

(b) Better responses included clear recognition of the effect of dilution on pH. Weaker responses included a simple pH statement with no colour or colour with no link to pH.

(c) Better responses included balanced equations, a clear statement of formulae and well setout working. Weaker responses showed a lack of understanding of the mole relationship and a poor ability to balance equations. Mid-range responses included an understanding of the mole relationship, but lack of recognition of 2:1 mole ratio, along with limited working.

2006 Q8 B

2.5 a) calculate the pH of the resultant solution when solutions of acids and/or bases are

diluted

The pH of solutions can change as a result of the addition of water, called dilution

2.5 b) calculate the pH of the resultant solution when solutions of acids and/or bases are

mixed

The pH of solutions can change as a result of neutralisation with either an acid or base.

PRACTICAL 2.5.1 Neutralisation Challenge

VIEW PPT

NOTE: In reality, H+ does not exist, it is H3O+. All other ions would be surrounded by water molecules (hydrated) as well, eg copper sulfate is really CuSO4.5H2O (copper sulfate pentahydrate). H+ is often used for simplicity, as long as we remember we are doing that.

12C6 2.5b Neutralisation Prac PPT.pptx

TASK 2.5.1

1. (2008, Q14)

20 mL of 0.08 M HCl is mixed with 30 mL of 0.05 M NaOH. What is the pH of the resultant solution?

(A) 1.1 (B) 2.7 (C) 4.0 (D) 7.0

2. (2007, Q10)

A 0.1 M HCl solution has a pH of 1.0. What volume of water must be added to 90 mL of this solution to obtain a final pH of 2.0?

(A) 10 mL (B) 180 mL (C) 810 mL (D) 900 mL.

3. (2006, Q10)

Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation.

P₂O_5(s) + 3H₂O_(l) → 2H₃PO_4(aq)

Phosphoric acid reacts with sodium hydroxide according to the following equation.

H₃PO_4(aq) + 3NaOH_(aq) → Na₃PO_4(aq) + 3H₂O_(l)

A student reacted 1.42 g of phosphorus pentoxide with excess water.

What volume of 0.30 M sodium hydroxide would be required to neutralise all the phosphoric acid produced?

(A) 0.067 L (B) 0.10 L (C) 0.20 L (D) 5.0 L

4. (2016, Q12)

Which of the following could be added to 100 mL of 0.01 M hydrochloric acid solution to change its pH to 4?

A) 900 mL of water B) 900 mL of 0.01 M hydrochloric acid

C) 9900 mL of water D) 9900 mL of 0.01 M hydrochloric acid

5. (2004, Q24, 5 marks)

The diagram shows three reagent bottles containing acids.

a) Calculate the pH after 10.0 mL of 0.01 M hydrochloric acid solution is diluted by the addition of 90.0 mL of distilled water. (1 mark)

Better responses set out the calculation for both dilution and pH very clearly. Most candidates knew the equation for calculating pH. A significant proportion of responses did not arrive at the correct dilution. Several candidates correctly calculated dilution, but did not subsequently calculate pH.

20.0 mL of a 1.00 mol.L-1 LiOH solution is added to 30.0 mL of a 0.500 mol.L-1 HNO₃ solution.

a) which reactant is the limiting agent?

b) what mass of LiNO₃ will the reaction mixture contain when the reaction is complete?

c) what is the pH of the final solution?

Task 2.5.2

Text WS 6.5 Calculating pH of mixtures and dilutions P67

REVIEW 2.5

PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)

2015 Q13 B
2014 Q12 C
2010 Q21 3

Question 21 Sample answer: Hydrochloric acid is a strong acid and therefore fully ionises in aqueous solution. Ethanoic acid is a weak acid and does not fully ionise in aqueous solution. The total concentration of ethanoic acid must be higher than that of HCl, due to incomplete ionisation of CH3COOH, to give an equivalent [H+ ] and therefore pH, as pH is equal to –log10 [H+ ].

2005 Q9 D

2.6 a) write ionic equations to represent

the dissociation of acids and bases in water

You have covered much of this, and will continue to do so through the rest of the module.

VIEW Videos:

ABR#13 https://www.youtube.com/watch?v=0AKA0-3SbC8&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS&index=13 [8.34 mins]
Ionic equations: https://www.youtube.com/watch?v=NZ934P6KjqI&list=PLB755F0C836855EAB&index=36 [10.21 mins]

TASK 2.6.1

Below is a model of the dissociation of acetic acid. See if you can work out what each species is.

TASK 2.6.2

1. Explain the meaning of the following statement: “When a weak acid dissolves in water an equilibrium is established between the acid molecule and its constituent ions.”

2. Write an ionisation equation for the following acids in water. (Davis, Disney, & Smith, 2018, p. 158)

a) Nitric acid

b) Hydrofluoric acid

3. Write the dissociation reaction for the following bases in water. (Davis, Disney, & Smith, 2018, p. 158)

a) Barium hydroxide

b) Calcium oxide

2.6 b) write ionic equations to represent

conjugate acid/base pairs in solution

Conjugate pairs are species involved in the transfer of protons.

Generally:

Acid 1 + Base 2 ⇌ Conjugate Base 1 + Conjugate Acid 2

If the difference between species is H⁺, then they are a conjugate acid-base pair.

The equilibrium position of an acid base system is determined by the relative strengths of the bases involved.

If the equilibrium lies to the right, Acid 1 is stronger and its conjugate Base 1 is weaker. Strong bases have weak conjugate acids.

The strength of an acid is a measure of the amount of dissociation to its component ions.

HCl is a strong acid, as it dissociates almost completely.

CH₃COOH (acetic acid) is a weak acid, as it does not dissociate completely; it forms an equilibrium in solution.

Weak Acid ⇌ Hydrogen ion + Conjugate Base

HA ⇌ H⁺ + A^-

The strength of a base is a measure of the amount of dissociation to its component ions.

NaOH is a strong base, as it dissociates almost completely.

NH₃ is a weak base, as it does not completely react; it forms an equilibrium in solution.

Weak Base + H⁺ ⇌ Conjugate Acid

A^- + H⁺ ⇌ HA

eg NH_{3 (aq)} + H⁺ ⇌ NH₄⁺_(aq)

VIEW Videos:

ABR#14 https://www.youtube.com/watch?v=YWjzjXIRLiw&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS&index=14 [8.05 mins]
Conjugate acid: https://www.youtube.com/watch?v=xJ3RG8Y6m1w&list=PLB755F0C836855EAB&index=51 [12.13 mins]

TASK 2.6.3

1. Complete the equations below by identifying the product and one conjugate pair.

a) HCl → H⁺+

b) HNO₃ → H⁺ +

c) H₃O+ → H⁺ +

d) HSO₄- → H⁺ +

2. Complete the equations below by identifying the products and two conjugate pairs

a) HF + H₂O →

b) HCOOH + H₂O →

c) CH₃COOH + H₂O →

d) NH₄⁺ + H₂O →

e) H₂O + H₂O →

3. Identify the conjugate pairs in the following reaction:

HCl + NH₃ → NH₂⁺ + Cl^-

4. (2009, Q7) What is the conjugate base of HSO₂^- ?

(A) SO₃^2-

(B) SO₂^2-

(D) HSO₃^-

TASK 2.6.4

Complete Worksheet: (Pages 3 and 4 are answers)

12C6 2.6b Conjugate Pairs WS.doc

REVIEW 2.6b

PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)

2017 Q14
2016 Q10
2014 Q10
2013 Q25 4
2011 Q25 3
2009 Q7
2008 Q26 4; Q27B 2; Q28B 1

2.6 c) write ionic equations to represent

amphiprotic nature of some salts, for example:

– sodium hydrogen carbonate

– potassium dihydrogen phosphate

There are some species which are capable of behaving as either an acid or a base, proton donor or proton acceptor according to the Bronsted-Lowry definition, given certain circumstances. We say these species are amphiprotic (you may sometimes also see them referred to as amphoteric).

Eg. HCO₃^-, HSO₄^- and H₂O are good examples.

These often form from a polyprotic acid (Remember: a polyprotic species is an acid which can donate more than one proton.)

The term ‘amphiprotic‘ means capable of acting both as a proton donor and proton acceptor. The word ‘amphi’ means ‘both’ or ‘two’

A Bronsted-Lowry acid is a proton donor and a Bronsted-Lowry base is a proton acceptor.

Therefore, if a substance is amphiprotic, it is capable of acting as both a Bronsted-Lowry acid or a Bronsted-Lowry base, depending on the substance with which the amphiprotic substance is reacting.

Although not assessable in HSC Chemistry, you can determine if an amphiprotic substance will act as an acid or base by comparing the equilibrium constant of the two possible reactions to determine which reaction is more favoured.

NOTE: For a substance to be amphiprotic, it MUST have at least one hydrogen atom. This is because if it does not have an hydrogen atom, it cannot act as a Bronsted-Lowry acid. Hence, it cannot be amphiprotic.

Potassium dihydrogen phosphate

dissociating

KH₂PO_{4 (aq)} <-> HPO₄^2-_(aq) + K⁺_(aq)

acting as a Bronsted-Lowry Acid (proton donor):

K⁺_(aq) + H₂PO₄^-_(aq)+ H₂O _(l) <-> HPO₄^2-_(aq) + H₃O⁺_(aq)+ K⁺_(aq)

Note: K+ is a spectator ion so we can remove it from both sides of the equation to get net ionic equation.

H₂PO₄^-_(aq)+ H₂O _(l) <-> HPO₄^2-_(aq) + H₃O⁺_(aq)

Potassium dihydrogen phosphate

dissociating

KH₂PO_{4 (aq)} <-> HPO₄^2-_(aq) + K⁺_(aq)

acting as a Bronsted-Lowry Base (proton acceptor):

K⁺_(aq)+H₂PO₄^-_(aq) + H₂O _(l) <-> H₃PO_4(aq) + OH^- (_(aq) + K⁺ _(aq)

Full ionic: K⁺_(aq)+H₂PO₄^-_(aq) + H₂O _(l) <-> 3H⁺_(aq) + PO₄^3-_(aq) + OH^- _(aq) + K⁺ _(aq)

Net ionic: H₂PO₄^-_(aq) + H₂O _(l) <-> 3H⁺_(aq) + PO₄^3-_(aq) + OH^- _(aq)

When potassium dihydrogen phosphate was dissolved in water, it reacted with water and accepted a proton (H+)

Note: H₃PO₄ would actually be ionised

Sodium hydrogen carbonate: dissociates in water

NaHCO_{3 (aq)} <-> HCO₃^-_(aq) + Na⁺_(aq)

_{The sodium ion is a spectator ion, so can be removed for the net ionic equations.}

VIEW Videos:

ABR#15 https://www.youtube.com/watch?v=3j1GnzHZZ90&list=PLeFSFSJ9WqSB27YUnCfMsSpmNluCyhZbS&index=15 [8.01 mins]
Amphiprotic: https://www.youtube.com/watch?v=QAd6Lw1GXVc&list=PLB755F0C836855EAB&index=53 [7.53 mins]

PRACTICAL

Practical Activity 6.2 Amphiprotic substances in water PP78-9

Place 20 drops of each solution into separate test tubes: 0.1M NaHCO3; NaHSO4; K2HPO3; KH2PO4
Add 1 drop universal indicator to each OR use pH meter
Use universal indicator chart to determine pH
Write equations for the acid and base hydrolysis reaction undergone by each amphiportic anion tested. Indicate from the pH which reaction predominates.

TASK 2.6.4

1. Write equations to show that water is an amphiprotic substance

2. Complete the following sentences:

a) Polyprotic acids are acids which...

b) An example of a diprotic acid is...

c) An example of a triprotic acid is...

3. (04, Q22, 3 marks)

(a) Define the term amphiprotic. (1 mark)

(b) Write TWO chemical equations to show that the dihydrogen phosphate ion (H₂PO₄^- ) is amphiprotic. (2 marks)

4. Use the dihydrogen phosphate ion as a starting point and write a sequence of steps to show how this ion can change depending on whether it is acting as a Brønsted-Lowry acid or base.

eg H₂PO₄^-_(aq) + H₂O_(l) ⇌

REVIEW 2.6c

PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)

2017 Q5
2011 Q25 3
2007 Q25 5
2005 Q10
2004 Q22A 1; 22B 2

REVIEW MDQ2

VIEW videos

12C6 2.7c Review WS1.doc

12C6 2.7c Review WS2.pdf

12C6 1.6 Definitions PPT.ppt

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