How can we measure the strength of acids and alkalis?

How can we measure the strength of acids and alkalis?

    1. An acid produces hydrogen ions in aqueous solution. The acidity of a solution is a measure of the concentration of the hydrogen ions in the solution.
  1. A base produces hydroxide ions in aqueous solution. The alkalinity of a solution is a measure of the concentration of the hydroxide ions in the solution.
  2. In 1909, Soren Sorensen proposed the pH scale for measuring acidity or alkalinity of an aqueous solution.

The pH scale

How can we measure the strength of acids and alkalis 1

  1. The pH scale normally has a range of values from 0 to 14 to indicate how acidic or alkaline an aqueous solution is.
  2. The pH value measures the concentration of hydrogen ions or hydroxide ions.
    pH < 7 → acidi solution
    pH = 7 → neutral solution
    pH > 7 → alkaline solution
  3. The lower the pH value, the higher the concentration of hydrogen ions.
    The higher the pH value, the higher the concentration of hydroxide ions.
  4. The pH of values some common solutions used in daily life are shown below.
    How can we measure the strength of acids and alkalis 2
  5. The pH value of an aqueous solution can be measured by using
    (a) Universal Indicator
    (b) pH meter
    (c) acid-base indicators
    How can we measure the strength of acids and alkalis 3

Table shows examples of acid-base indicators.

IndicatorpH rangeColour change
Acid         NeutralAlkali
Methyl orange3.0 – 5.0RedOrangeYellow
Bromothymol blue6.0 – 8.0YellowGreenBlue
Phenolphthalein8.0-10.0ColourlessColourlessPink

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Strong and weak acids

  • The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water.
  • The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water.
  • A strong acid is an acid which ionises or dissociates completely in water to produce a high concentration of hydrogen ions.
  • Hydrochloric acid is a strong acid. It is 100% ionised in water. All the hydrogen chloride molecules that dissolve in the water ionise completely into hydrogen ions and chloride ions.
    How can we measure the strength of acids and alkalis 4
  • A weak acid is an acid which ionises partially in water to produce a low concentration of hydrogen ions.
  • Ethanoic acid is a weak acid. Dilute ethanoic acid is about 0.4% ionised, only 4 out of every 1000 ethanoic acid molecules ionise. As fast as the acid molecules ionise to produce ions, the ions combine back again to give the original acid molecules in a reversible reaction.
    How can we measure the strength of acids and alkalis 5
    How can we measure the strength of acids and alkalis 6
  • Examples of strong and weak acids are shown in Table.
    AcidNameParticles
    Strong acidHydrochloric acid, HClH+, Cl
    Nitric acid, HNO3H+, NO3
    Sulphuric acid, H2S04H+, HSO4, SO42-
    Weak acidCarbonic acid, H2C03H+, HCO3, CO32-, H2CO3
    Ethanoic acid, CH3COOHH+, CH3COO, CH3COOH
    Sulphurous acid, H2SO3H+, HSO3, SO32-, H2SO3
  • Strong acid and weak acid are defined as follows.
    A strong acid is completely ionised in water to produce a high concentration of hydrogen ions.
    A weak acid is partially ionised in water to produce a low concentration of hydrogen ions.
  • The pH of an acid solution changes with concentration of the acid. The concentration of the acid affects the concentration of hydrogen ions produced. For example:
    10 mol dm-3 hydrochloric acid: pH = 1
    01 mol dm-3 hydrochloric acid: pH = 2
  • pH values can be used to compare the acid strength of different acids. All the acids to be compared must be of the same concentration.
    For example:
    10 mol dm-3hydrochloric acid: pH = 1
    10 mol dm-3 ethanoic acid: pH = 3
  • For two different acids of the same concentration, the acid with the lower pH value is the stronger acid, i.e. higher degree of ionisation in water.

Strong and weak alkalis

  • Sodium hydroxide is a strong alkali. It ionises fully when dissolved in water.
    How can we measure the strength of acids and alkalis 7
  • A strong alkali is an alkali which is fully ionised in water to produce a high concentration of hydroxide ions.
  • A weak alkali is an alkali which ionises partially in water to produce a low concentration of hydroxide ions.
  • Ammonia is an example of a weak alkali. It is only partly ionised in water, which means the ionisation of ammonia in water is incomplete. Only a small amount of ammonia molecules are ionised in water to produce ammonium ions and hydroxide ions.
    How can we measure the strength of acids and alkalis 8
  • Examples of strong and weak alkalis are shown in Table.
    AlkaliNameParticles
    Strong alkaliSodium hydroxide, NaOHNa+, OH
    Potassium hydroxide, KOHK+, OH
    Barium hydroxide, Ba(OH)2Ba2+, OH
    Weak alkaliAmmonia, NH3NH3, NH4+, OH
    Methylamine, CH3NH2CH3NH2, CH3NH3+, OH
    Hydrazine, N2H4N2H4, N2H5+, OH
  • Strong alkali and weak alkali are defined as follows.
    A strong alkali is completely ionised in water to produce a high concentration of hydroxide ions.
    A weak alkali is partially ionisesed in water to produce a low concentration of hydroxide ions.
  • The pH of an alkali solution changes with concentration of the alkali. The concentration of the alkali affects the concentration of hydroxide ions produced.
    For example:
    10 mol dm-3m-3 sodium hydroxide solution pH = 13
    01 mol dm-3 sodium hydroxide solution pH = 12
  • pH values can be used to compare the strength of different alkalis. All the alkalis to be compared must be of the same concentration. For example:
    10 mol dm-3 sodium hydroxide: pH = 13
    10 mol dm-3 ammonia solution: pH = 11
  • For two different alkalis of the same concentration, the alkali with the higher pH value is the stronger alkali, i.e. higher degree of ionisation in water.

 

Action of Heat on Salts

Action of Heat on Salts

  1. Heating a salt may cause it to decompose. The decomposition may result in
    (a) a colour change
    (b) evolution of a gas
    (c) liberation of water vapour
  2. Gases such as carbon dioxide, sulphur dioxide, nitrogen dioxide, ammonia or oxygen can be evolved. By identifying the gas or gases liberated, it is possible to pinpoint the anion present in the salt.
  3. Examination of the residue can provide information to identify the cation in the salt.
    Action of Heat on Salts 1

Action of Heat on Carbonate Salts

  1. Most metal carbonates decompose on heating to produce metal oxides and carbon dioxide gas.
    Metal carbonate → metal oxide + carbon dioxide
    Action of Heat on Salts 2
  2. When the carbon dioxide gas is bubbled through limewater, it will turn the limewater milky.

Table shows the action of heat on carbonate salts.

Carbonate saltAction of heat
Potassium carbonate
Sodium carbonate
Do not decompose
Calcium carbonate
Magnesium carbonate
Aluminium carbonate
Zinc carbonate
Iron(III) carbonate
Lead(II) carbonate
Copper(II) carbonate
Decompose to produce metal oxide and carbon dioxide
Metal carbonate → metal oxide + carbon dioxide
For example,
CaCO3(s) → CaO(s) + CO2(g)
Silver carbonate

Decomposes to produce metal, oxygen and carbon dioxide
2Ag2CO3(s) → 4Ag(s) + O2(g) + 2CO2(g)

Ammonium carbonate

Decomposes to produce ammonia, water and carbon dioxide
(NH4)2CO3(s) → 2NH3(g) + H2O(l) + CO2(g)

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Action of Heat on Carbonate Salts Experiment

Aim: To investigate the action of heat on carbonate salts.
Materials: Limewater, sodium carbonate, magnesium carbonate, calcium carbonate, zinc carbonate, lead(II) carbonate, copper(II) carbonate and potassium carbonate.
Apparatus: Test tubes, tongs, spatula, Bunsen burner and stopper with delivery tube.
Procedure:

  1. About two spatulaful of copper(II) carbonate are placed in a test tube.
  2. The colour of the carbonate salt is noted.
  3. The test tube is stoppered with a delivery tube dipping into limewater as shown in Figure.
    Action of Heat on Salts 3
  4. The carbonate salt is then heated strongly.
  5. Any changes that occur to the limewater and also the colour of the residue when it is hot and when it is cold are recorded.
  6. Steps 1 to 5 are repeated using each of the carbonate salts listed in Table to replace the copper(II) carbonate.

Observations:

Carbonate saltColour of salt before heatingColour of residueEffect on limewater
HotCold
Copper(II) carbonateGreenBlackBlackLimewater turns milky.
Sodium carbonateWhiteNo change.
Potassium carbonateWhiteNo change.
Calcium carbonateWhiteWhiteWhiteLimewater turns milky.
Magnesium carbonateWhiteWhiteWhiteLimewater turns milky.
Zinc carbonateWhiteYellowWhiteLimewater turns milky.
Lead(II) carbonateWhiteBrownYellowLimewater turns milky.

Discussion:

  1. Alkali metal carbonates such as sodium carbonate and potassium carbonate are stable to heat.
  2. Most metal carbonates decompose on heating to produce metal oxides and liberate carbon dioxide gas.
  3. The carbon dioxide gas forms a white precipitate with limewater, making the limewater milky.

Conclusion:
Heating a metal carbonate will decompose it into a metal oxide and liberate carbon dioxide. Group 1 metal carbonates are not decomposed by heat.

Action of Heat on Nitrate Salts

  1. Nitrate salts also undergo decomposition on heating.
  2. Most metal nitrates decompose to produce a metal oxide, nitrogen dioxide and oxygen.
    Action of Heat on Salts 4
  3. Sodium nitrate and potassium nitrate decompose to produce nitrite salts and oxygen.
  4. Nitrogen dioxide is a brown gas. It is an acidic gas that turns moist blue litmus paper red. Hence, dissolving it in water produces a colourless acidic solution.
    2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)
  5. The colourless oxygen gas rekindles a glowing wooden splint.

Table: Action of heat on nitrate salts

Nitrate saltAction of heat
Potassium nitrate
Sodium nitrate
Decompose to produce a nitrite salt and oxygen
2KNO3(s) → 2KNO2(s) + O2(g)
2NaNO3(s) → 2NaNO2(s) + O2(g)
Calcium nitrate
Magnesium nitrate
Aluminium nitrate
Zinc nitrate
Iron(II) nitrate
Iron(III) nitrate
Lead(II) nitrate
Copper(II) nitrate

Decompose to produce metal oxide, nitrogen dioxide and oxygen
Metal nitrate → metal oxide + nitrogen dioxide + oxygen
For example,
2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g)

Silver nitrateDecomposes to produce metal, nitrogen dioxide and oxygen
2AgNO3(s) → 2Ag(s) + 2NO2 (g) + O2(g)
Ammonium nitrateDecomposes to produce nitrous oxide and water
NH4NO3(s) → N2O(g) + 2H2O(l)

Action of Heat on Nitrate Salts Experiment

Aim: To investigate the action of heat on nitrate salts.
Materials: Sodium nitrate, magnesium nitrate, calcium nitrate, zinc nitrate, lead(II) nitrate, copper(II) nitrate, potassium nitrate, iron(III) nitrate, iron(II) nitrate, blue litmus paper and wooden splint.
Apparatus: Test tubes, tongs, spatula and Bunsen burner.
Procedure:

  1. About two spatulaful of copper(II) nitrate are placed in a test tube.
  2. The colour of the nitrate salt is noted.
  3. The nitrate salt is then heated strongly as shown in Figure.
    Action of Heat on Salts 5
  4. The gases liberated are tested by
    (a) lowering a glowing wooden splint into the test tube.
    (b) bringing a piece of moist blue litmus paper to the mouth of the test tube.
  5. The colour of the residue when it is hot and when it is cold are recorded.
  6. Steps 1 to 5 are repeated using each of the nitrate salts listed in Table to replace the copper(II) nitrate.

Observations:

Nitrate saltColour of salt before heatingColour of residueTests for gases evolved
HotColdColour of gasGlowing splintBlue litmus paper
Copper(II) nitrateBlueBlackBlackBrown gas and colourless gasRekindlesTurns red
Sodium nitrateWhiteWhiteWhiteColourlessRekindlesNo change
Potassium nitrateWhiteWhiteWhiteColourlessRekindlesNo change
Calcium nitrateWhiteWhiteWhiteBrown gas and colourless gasRekindlesTurns red
Magnesium nitrateWhiteWhiteWhiteBrown gas and colourless gasRekindlesTurns red
Zinc nitrateWhiteYellowWhiteBrown gas and colourless gasRekindlesTurns red
Iron(II) nitrateGreenBlackBlackBrown gas and colourless gasRekindlesTurns red
Iron(III) nitrateBrownBrownBrownBrown gas and colourless gasRekindlesTurns red
Lead(II) nitrateWhiteBrownYellowBrown gas and colourless gasRekindlesTurns red

Discussion:

  1. When nitrate salts are heated, they decompose to liberate nitrogen dioxide and oxygen.
  2. Only sodium nitrate and potassium nitrate decompose to liberate oxygen.
  3. Nitrogen dioxide is a brown gas that turns moist blue litmus paper red.
  4. Oxygen is a colourless gas that relights a glowing wooden splint.

Conclusion:
Most metal nitrates decompose to produce a metal oxide, nitrogen dioxide and oxygen except sodium nitrate and potassium nitrate which decompose to produce nitrite salts and oxygen.

Action of Heat on Sulphate Salts

  1. The normal sulphate salts are more stable to heat compared to the carbonates and nitrates.
  2. Group 1 metal sulphates such as sodium sulphate and potassium sulphate do not decompose on heating. Group 2 metal sulphates such as calcium sulphate also do not decompose when heated.
  3. The sulphates of heavy metals are decomposed into metal oxides and sulphur trioxide when heated.
    Action of Heat on Salts 6
  4. Sulphur trioxide is a typical acidic oxide and dissolves in water to form sulphuric acid.
    SO3(g) + H2O(l) → H2SO4(aq)
  5. An exceptional case is iron(II) sulphate because it also forms sulphur dioxide gas.
    2FeSO4(s) → Fe2O3(s) + SO3(g) + SO2(g)
    The green crystals of iron(II) sulphate turn into a brown solid of iron(III) oxide.
  6. Ammonium sulphate sublimes when first heated. Further heating decomposes the salt into ammonia and hydrogen sulphate.
    (NH4)2SO4(S) – 2NH3(g) + H2SO4(g)

Action of Heat on Chloride Salts

  1. Chloride salts are stable to heat except ammonium chloride.
  2. Initial heating of ammonium chloride causes the salt to sublime.
    NH4Cl(S) → NH4Cl(g)
  3. On further heating, decomposition takes place to produce ammonia and hydrogen chloride.
    NH4Cl(g) → NH3(g) + HCl(g)
  4. When ammonium chloride is heated in a test tube, the lighter ammonia gas will emerge first and turn a piece of moist red litmus paper blue. Hydrogen chloride, coming up next, will change the litmus paper from blue back to red.
    Action of Heat on Salts 7

Identification of salts by action of heat

  1. When a salt is heated strongly, it may decompose. One or more gases may be liberated.
  2. Each gas can be identified by
    • noting its colour.
    • testing it with moist blue or red litmus paper.
    • testing it with limewater.
    • testing it with glowing wooden splint.
    • testing it with acidified potassium dichromate(VI) solution or acidified potassium manganate(VII) solution.
  3. The colour of the residue when hot and cold must be noted to help in the identification of the salt.
  4. The following table shows how to identify salts P, Q, R, S and T through gases liberated by the action of heat.
TestObservationInference
Heat P in a test tube. Identify the gas/gases given off.
  • A colourless gas is liberated.
  • It forms a white precipitate with limewater, i.e. limewater turns milky.
  • Carbon dioxide gas, CO2, is produced.
  • Carbonate ion, CO22-, is present.
  • P is a carbonate salt.
Heat Q in a test tube. Identify the gas/gases given off.
  • A brown gas is liberated which turns damp blue litmus paper red.
  • A colourless gas is evolved which relights a glowing wooden splint.
  • Solid residue is yellow when hot and white when cold.
  • Brown gas is nitrogen dioxide, NO2.
  • Colourless gas is oxygen, O2.
  • Nitrate ion, NO3, is present.
  • Residue is zinc oxide.
  • Q is zinc nitrate.
Heat R in a test tube. Identify the gas/gases given off.
  • A colourless gas is given out.
  • It turns orange acidified potassium dichromate(VI) green.
  • Sulphur dioxide gas, SO2 is produced.
  • Sulphate ion, SO42-,is present.
  • R is a sulphate salt.
Heat S in a test tube. Identify the gas/gases given off.
  • A colourless pungent gas is evolved.
  • It turns damp red litmus paper blue.
  • It forms dense white fumes with concentrated hydrochloric acid.
  • Ammonia gas, NH3, is produced.
  • Ammonium ion, NH4+, is present.
  • S is an ammonium salt.
Heat Tin a test tube. Identify the gas/gases given off.
  • A yellow-green gas is liberated.
  • It turns damp blue litmus paper red and then white.
  • Chlorine gas, Cl2, is produced.
  • Chloride ion, Cl, is present.
  • T is a chloride salt.

 

What is the most common type of plastic?

What is the most common type of plastic?

Plastics:

  1. Plastics are made from the products of cracking of petroleum fractions such as alkenes molecules through addition polymerisation.
  2. Plastics are the largest group of synthetic polymers with the following properties:
    • Can be easily moulded and coloured
    • Low density
    • Strong
    • Inert to chemicals
    • Insulator of heat and electricity
    • Impermeable
    • Non-biodegradable
  3. Table shows some examples of commonly used plastics, how they are formed, their properties and uses.
Name of polymerEquation for polymerisationPropertiesUses

Polyethylene
(polythene)

What is the most common type of plastic 1Durable, light, impermeable, inert to chemicals, easily melted, insulatorShopping bags, plastic cups and plates, toys
Polypropylene
(polypropene)
What is the most common type of plastic 2Durable, light, impermeable, inert to chemicals, easily melted, insulator, can be moulded and colouredBottles, furniture, battery casing, pipes, toys
Polyvinyl chloride (polychloroethene)What is the most common type of plastic 3Low softening temperature, durable, elastic, can be colouredPipes, pipe fittings, wire and cable casing, raincoats, footwear, bags
PolystyreneWhat is the most common type of plastic 4Heat insulator, light, can be moulded, impermeableDisposable cups and plates, packaging materials, toys, heat insulators.
PerspexWhat is the most common type of plastic 5Transparent, strong, lightReplacement for glass, lenses and optical fibres
TeflonWhat is the most common type of plastic 6Durable, non-stick, chemically inert, strong, impermeableCoatings for non-stick pans, electrical insulators

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Synthetic rubbers

  1. Synthetic rubbers are elastomers that are elastic and can regain their original length and shape after being stretched and pressed.
  2. The structure of synthetic rubbers is similar to that of natural rubber.
  3. Synthetic rubbers are produced by addition polymerisation.
  4. Neoprene and styrene-butadiene rubber (SBR) are examples of synthetic rubber.
  5. Neoprene is elastic and resistant to oil and is suitable to make rubber host and toys.
  6. Styrene-butadiene rubber, SBR rubber is very suitable for making car tires and shoes because it is elastic and hard.

Constructing ionic equations using the continuous variation method

Constructing ionic equations using the continuous variation method

  1. The ionic equation for the formation of an insoluble salt can be constructed if we know the number of moles of cation and anion reacted together to form 1 mole of the insoluble salt.
  2. For example:
    (a) 1 mole of silver chromate(VI) is formed from 2 moles of Ag+ ions and 1 mole of CrO42- ions.
    Ionic equation:
    2Ag+(aq) + CrO42- (aq) → Ag2CrO(s)
    (b) 1 mole of lead(II) bromide is formed from 1 mole of Pb2+ ions and 2 moles of Br ions. Ionic equation:
    Pb2+(aq) + 2Br(aq) → PbBr2(s)
  3. The number of moles of cation and anion which combine to form 1 mole of the insoluble salt can be determined from experiment by a continuous variation method.
  4. The method involves the following steps.
    1. Carry out a reaction between a fixed volume of reactant A with varying volumes of a second reactant B.
    2. Determine the volume of reactant B required to react completely with the fixed volume of reactant A.
    3. Use the results of the experiment to calculate the number of moles of reactant A and number of moles of reactant B which reacted completely with each other.
    4. Determine the simplest mole ratio of reactant A to reactant B which combine to form one mole of the insoluble salt.
    5. Use the ratio to construct the ionic equation.

Constructing ionic equations example

1. 10 cm3 of 0.25 mol dm-3 lead(II) nitrate solution reacts completely with 5 cm3 of 1.0 mol dm-3 potassium iodide solution. A yellow precipitate of lead(II) iodide is formed. Construct the ionic equation for the formation of lead(II) iodide.
Solution:
Constructing ionic equations using the continuous variation method 1

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Constructing ionic equations experiment

Aim: To construct an ionic equation for the formation of lead(II) chromate(VI).
Problem statement: How to construct an ionic equation for the formation of lead(II) chromate(VI)?
Hypothesis: As the volume of potassium chromate(VI) solution increases, the height of the yellow precipitate increases until all the lead(II) nitrate has reacted.
Variables:
(a) Manipulated variable : Volume of potassium chromate(VI) solution
(b) Responding variable : Height of yellow lead(II) chromate(VI) precipitate
(c) Controlled variables : Volume and concentration of lead(II) nitrate solution, concentration of potassium chromate(VI) solution, size of test tubes
Materials: 0.5 mol dm-3 lead(II) nitrate solution and 0.5 mol dm-3 potassium chromate(VI) solution.
Apparatus: 7 test tubes, test tube rack, burettes, metre rule, stopper, dropper, retort stand and clamp.
Procedure:

  1. Seven test tubes of the same size are labelled from 1 to 7 and are placed in a test tube rack.
  2. A burette is filled with 0.5 mol dm-3 lead(II) nitrate solution. 5.00 cm3 of the lead(II) nitrate solution is added to each of the seven test tubes.
  3. A second burette is filled with 0.5 mol dm-3 potassium chromate(VI) solution. The potassium chromate(VI) solution is added to each of the seven test tubes according to the volumes specified in Table.
  4. Each test tube is stoppered and shaken well. The test tubes are left aside for about half an hour to allow the precipitate to settle.
  5. The height of precipitate in each test tube is measured. The colour of the solution above the precipitate in each test tube is noted.
  6. All readings and observations are recorded in Table.

Results:

Test tube1234567
Volume of 0.5 mol dm-3 lead(II) nitrate solution5555555
Volume of 0.5 mol dm-3 potassium chromate(VI) solution (cm3)1234567
Height of precipitate (cm)0.61.21.82.43.03.03.0
Colour of solution above precipitateColourlessColourlessColourlessColourlessColourlessYellowYellow

Graph:
A graph of the height of precipitate against the volume of potassium chromate(VI) solution added is plotted.
Constructing ionic equations using the continuous variation method 2Interpreting data:
From the graph:
Volume of potassium chromate(VI) solution required to completely react with 5.0 cm3 of lead(II) nitrate solution = 5.00 cm3
Constructing ionic equations using the continuous variation method 3
Number of moles of Pb2+ ions in 5.00 cm3 of 0.5 mol dm-3 lead(II) nitrate solution in each of the test tubes
Constructing ionic equations using the continuous variation method 4
Discussion:

  1. A yellow precipitate of lead(II) chromate(VI) is formed in each of the seven test tubes.
  2. The height of the precipitate increases gradually from test tubes 1 to 5 because more and more lead(II) chromate(VI) is formed due to the increasing amount of potassium chromate(VI) added to the test tubes.
  3. The first test tube to achieve maximum constant height of precipitate indicates a complete reaction has taken place. The volume of potassium chromate(VI) solution (5.00 cm3) added is just sufficient to react completely with the lead(II) nitrate in the test tube.
  4. The colour of the solution above the precipitate in test tubes 1 to 4 are colourless due to the excess lead(II) nitrate.
  5. In test tube 5, a complete reaction has taken place. No reactants are present in excess.
  6. The colour of the solution above the precipitate in test tubes 6 and 7 are yellow due to the excess potassium chromate(VI).

Conclusion:
The ionic equation for the formation of lead(II) chromate(VI) can be obtained from a precipitation reaction. The hypothesis is accepted.

Relationship between pH values and molarity of acids and alkalis

Relationship between pH values and molarity of acids and alkalis

 

The relationship between pH values and concentration of hydrogen ions is given below:

  • Concentration of hydrogen ions increases → pH value decreases
  • In an acidic solution, the concentration of hydrogen ions depends on the concentration or molarity of the acidic solution. An acid with a higher molarity will have a higher concentration of hydrogen ions and hence a lower pH value.
    Higher molarity of acid → lower pH value
  • Figure shows the pH values of acidic solutions of different molarity.
    Relationship between pH values and molarity of acids and alkalis 1
  • The relationship between pH values and concentration of hydroxide ions is given below:
    Concentration of hydroxide ions increases → pH value increases
  • In an alkaline solution, the concentration of hydroxide ions depends on the concentration or molarity of the alkaline solution.
  • By increasing the molarity of an alkaline solution, concentration of hydroxide ions increases and its pH value increases.
    Higher molarity of alkali → higher pH value
  • Figure shows how the pH values vary for different molarity of sodium hydroxide solutions.
    Relationship between pH values and molarity of acids and alkalis 2

The pH value of an acid or an alkali depends on:

  1. the strength (degree of ionisation) of acid/alkali
    (i) stronger acid (higher degree of ionisation) → lower pH
    (ii) stronger alkali (higher degree of ionisation) → higher pH
  2. the concentration (molarity) of acid/alkali
    (i) higher concentration (molarity) of acid → lower pH
    (ii) higher concentraton (molarity) of alkali → higher pH
  3. the basicity of acid/alkali
    (i) higher basicity of acid → lower pH
    (ii) higher basicity of alkali → higher pH

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Relationship between pH values and molarity experiment

Aim: To investigate whether:
(a) an increase in the molarity of an acid will decrease its pH value and
(b) an increase in the molarity of an alkali will increase its pH value
Problem statement: (a) Does an increase in the molarity of an acid cause its pH value to decrease?
(b) Does an increase in the molarity of an alkali cause its pH value to increase?
Hypothesis: Increasing the molarity of an acid will decrease its pH value, whereas increasing the molarity of an alkali will increase its pH value.
Variables:
(a) Manipulated variable: Molarity of acid or alkali
(b) Responding variable: pH values
(c) Controlled variable: Types of acid or alkali
Materials: Hydrochloric acid with concentrations of 1.0 mol dm-3, 0.1 mol dm-3, 0.01 mol dm-3, 0.001 mol dm-3, and 0.0001 mol dm-3; sodium hydroxide solution with concentrations of 1.0 mol dm-3, 0.1 mol dm-3, 0.01 mol dm-3, 0.001 mol dm-3 and 0.0001 mol dm-3.
Apparatus: Boiling tubes, test tube rack and pH meter.
Procedure:

  1. Five boiling tubes are arranged in a test tube rack and labelled 1, 2, 3, 4 and 5.
  2. Each boiling tube is filled with about 10 cm3 of acid as follows.
    • Boiling tube 1 – 1.0 mol dm-3 hydrochloric acid
    • Boiling tube 2 – 0.1 mol dm-3 hydrochloric acid
    • Boiling tube 3 – 0.01 mol dm-3 hydrochloric acid
    • Boiling tube 4 – 0.001 mol dm-3 hydrochloric acid
    • Boiling tube 5 – 0.0001 mol dm-3 hydrochloric acid
  3. A clean, dry pH meter is dipped info each acid to measure and record its pH value.
  4. The boiling tubes are cleaned and steps 1 to 3 are repeated using sodium hydroxide solution to replace the hydrochloric acid.
  5. The readings are recorded in Table.

Results:

Boiling tube numberHydrochloric acidSodium hydroxide solution
MolaritypHMolaritypH
11.001.014
20.110.113
30.0120.0112
40.00130.00111
50.000140.000110

Discussion:

  1. Molarity measures the number of moles of acid or alkali in 1 dm3 of solution.
  2. Based on Table increasing the molarity of an acid decreases its pH value.
    Reason:
    Hydrochloric acid is a strong acid. When the molarity of hydrochloric acid increases,

    • the number of moles of hydrochloric acid dissolved in 1 dm3 of solution increases
    • more hydrochloric acid molecules ionise to hydrogen ions
    • the concentration of hydrogen ions also increases
    • the pH value decreases
      Hence, increasing the molarity of an acid will decrease its pH value.
  3. Based on Table, increasing the molarity of an alkali increases its pH value.
    Reason:
    Sodium hydroxide solution is a strong alkali. When the molarity of sodium hydroxide solution increases,

    • the number of moles of sodium hydroxide dissolved in 1 dm3 of solution increases
    • more sodium hydroxide dissociates to produce hydroxide ions
    • the concentration of hydroxide ions also increases
    • the pH value increases
      Hence, increasing the molarity of an alkali will increase its pH value.

Conclusion:
Increasing the molarity of an acid will decrease its pH value. Increasing the molarity of an alkali will increase its pH value. The hypothesis can be accepted.