The Chemistry Of Aqueous Solutions

Published Date: 02 Nov 2017

Please read the separate assignment sheet provided and answer all questions to obtain the minimum past grade for this unit. Each assignment question is directly related to the unit assessment criteria. Give clear annotated diagrams where applicable and the use of chemical equations when asked. Use clear and concise language for explanation and try not to make it sound ‘wordy’. Also, remember to include a bibliography, as assignment will not be marked

Date set:

Date for draft submission

(if applicable):

Date for final

submission: 14/3/12

Final Overall Grade

(to be inserted only once a grade has been awarded)

Final Grade:

Date:

Lecturer Signature:

Learner declaration: I certify that the attached assignment is all my work. I have provided references of the sources from which I have reproduced any figures, diagrams or tables of data and any texts from which I have extracted substantial amounts of information for this assignment.

Signature: Date:

TO THE LEARNER: Please attach this assignment brief to any written work you are handing in for assessment, or submit the brief as instructed. YOUR WORK CANNOT BE ASSESSED UNLESS YOU HAVE SIGNED AND SUBMITTED THIS FORM.

AC Level Three

Apply the units mol dm-3, g dm-3 and g/100 g solvent (% w/w) as used for concentration of chemicals in solution, and convert between these units.

Calculate concentration of substance in solution, and amount (mol) and mass (g) of substance in a given volume of solution.

Calculate concentration (mol dm-3) of ions in a solution containing one or more electrolytes.

Calculate number of particles (molecules or ions) in a given volume of solution of known concentration.

Calculate changes in concentration of chemicals in solution resulting from dilution.

Use data from volumetric analysis (titrations) to calculate the concentration (mol dm-3 and g dm-3) of chemicals in solution.

Explain the terms solute, solvent, solution, soluble, insoluble, miscible (liquids) and immiscible (liquids).

Explain the factors which affect solubility and miscibility.

Explain the structure and formation of hydrated ions.

Evaluate, interpret and perform a range of calculations from solubility data and solubility curves for solids and gases.

Define the meaning of acid, base and alkali (using Bronsted Lowry Theory).

Interpret common acid/base reactions (in aqueous solution and gas phase) as represented in chemical equations.

Define the meaning of neutral solution, acidic solution and alkaline solution.

Explain the chemical properties of acidic solutions and alkaline solutions and represent these with chemical equations.

Explain the meaning of pH solution, and interpret pH values.

Grade

If you have achieved all Level 3 criteria you will receive a grade (Pass, Merit or Distinction) against the following Grade Descriptors. (There are no descriptors for Pass; learners achieve a Pass by meeting all ACs for the unit at Level 3.)

Grade Descriptor

To achieve a Merit :

The learner, learner’s work or performance:

To achieve a Distinction :

The learner, learner’s work or performance:

GD1: Understanding of the subject

Demonstrates a very good grasp of the relevant knowledge base

Demonstrates an excellent grasp of the relevant knowledge base

GD3:

Application of skills

Applies appropriate skills and techniques with very good levels of confidence and accuracy

Applies appropriate skills and techniques with excellent levels of confidence and accuracy

GD7:

Quality

Taken as a whole demonstrates a very good response to the demands of the assignment

Taken as a whole demonstrates an excellent response to the demands of the assignment

What this means for this assignment

To be considered for a Merit, you must show all calculations and working where relevant, submit any drawings accurately annotated as relevant and produce accurately worked solutions and balanced equations where relevant. For a Distinction, you must consistently produce the above and demonstrate that you have fully explained your results in a manner that demonstrates that you have read beyond your class notes and can indicate the application of the results in a wider context

Assignment Feedback

Title of Access to HE Diploma: Science

Unit title(s):

Chemistry

Unit code(s):

RD1/2/LN/060

RD1/3/LN/059

Learner:

Tutor/Assessor:

PB

Title of Assignment:

The Chemistry of Aqueous Solutions

Assignment

Number:

of

for this Unit

Internally moderated?

Yes / No

IM’s signature:

KB

Part A: Feedback on credit level

AC no

Credit achieved ( L3 )

Location of evidence

Tutor/Assessor comments on assessment criteria

(you could also indicate on the work itself where each AC is met)

1.1

1.2

1.3

1.4

1.5

1.6

2.1

2.2

2.3

2.4

3.1

3.2

3.3

3.4

3.5

Additional feedback

Level

achieved:

Tutor/Assessor’s

signature:

Date:

In order to achieve credit at Level 3, learners must meet all the assessment criteria for that unit. If there is more than one assignment for a unit, the credit level for the unit cannot be awarded until all assignments have been completed.

Resubmission (if applicable)

If any of the assessment criteria for this assignment have not been met at Level 3, a resubmission may be permitted. Only one resubmission is allowed for each assignment.

Requirements for resubmission / Task set:

Date set:

Date due:

05/03/12

Date submitted:

Feedback on resubmission:

Level achieved

after resubmission:

Tutor/Assessor’s signature:

Date:

Part B: Feedback on grading

Applicable only if all assessment criteria achieved at level 3

Grade Descriptor

Tutor/Assessor comments against grade descriptors

Grade

indicator

(P/M/D)

GD1: Understanding of the subject

GD3: Application of skills

GD7:

Quality

Grade indicators for

assignment

Final unit grade, if applicable *

* If there is more than one assignment for a unit, the final grade for the unit cannot be awarded

until all assignments have been completed, and this space should be left blank.

Learner’s comments following return of assignment:

Learner signature

Access to HE Diploma: Science

Unit 2 Assignment – The Chemistry of Aqueous Solutions

Assessment Criteria (AC)

1.4 CALCULATE NUMBER OF PARTICLES (MOLECULES OR IONS) IN A GIVEN VOLUME OF SOLUTION OF KNOWN CONCENTRATION.

1. If a solution of calcium chloride contains fifty grams of the substance dissolved in 0.50 dm‾³ of solution

a. Calculate the number of Ca ions per dm‾³ of solution

Calcium ions in a solution Ca = Ca2+ therefore:

Ca2+- 2e+

2e- + Cl Cl2-

RMM of CaCl2

Ca = 40g

Cl2= 2 X 35.5g

Total RMM = 111g

We use 50g of CaCl2 therefore the number of moles using the equation Mass

RMM 50/111 = 0.45 mol CaCl2

This is the moles for 0.50dm-3 therefore the moles for 1dm3 will be double. 0.45 X 2 = 0.9 mol dm-3

To determine the Avogadro constant:

[Ca2+] 0.9 mol dm-3 X 6.022x1023 (Avogadro constant)

0.9

6.022x1023

5.42x1023

b. Calculate the number of Cl ions per dm‾³ of solution

[2Cl] 2 X 6.022x1023 (Avogadro constant)

5.42x1023 X 2 = 1.08x1024

1.5 CALCULATE CHANGES IN CONCENTRATION OF CHEMICALS IN SOLUTION RESULTING FROM DILUTION.

2. The concentrated hydrochloric acid in the store has a concentration of 12 mol dm‾³. The college technician has to fill all the reagent bottles in the lab with approximately 2 mol dm‾³hydrochloric acid. There are 24 bottles, each of which holds 250 cm‾³.

Describe showing all calculations how the technician should prepare the solution of dilute hydrochloric acid.

The first step is to find the number of moles HCl in one bottle of a solution of 2 mol dm-3 (250 ml)

No of mol = 250

1000 = 0.25 mol. This is the concentration for 1 mol dm-3. We need 2 mol dm-3 therefore the figure is doubled:

250

1000 = X 2(mol) = 0.5 mol

0.5 mol/12 = 0.04166 dm-3 this is for dm-3 where as we need the volume in cm-3 therefore the volume is multiplied by 1000:

0.04166 X 1000 = 41.67cm-3

The second step, the technician needs to make up 250cm-3 bottles and needs to dilute the solution with H2O. The exact amount is determined by subtracting the volume of H2O is added to the 12 mol-3 sol-n (41.67) to make 250cm-3. (250 – 41.67 = 208.33). 208.33 H2O is added.

This is the total for ONE 250cm-3 bottle and the technician needs to make up 24 X 250cm-3 bottles therefore the volumes are multiplied by 24 as thus:

HCl – 24 X 41.67 = 1000.08 cm-3

H2O – 24 X 208.34 = 49993.92 cm-3

Therefore 4999.92 cm-3 of H2O in a 1cm-3 of HCl concentration of 12 mol dm-3 are mixed to a volume of 6dm-3 of: H2O and 1 dm-3 HCl solutions. 6 dm-3 converts to cm-3 (6000/250 = 24 bottles of 250 cm-3)

1.6 USE DATA FROM VOLUMETRIC ANALYSIS (TITRATIONS) TO CALCULATE THE CONCENTRATION (MOL DM-3 AND G DM-3) OF CHEMICALS IN SOLUTION.

3. Use the data you obtained from the volumetric analysis (titration) you carried out in the laboratory between standard sodium carbonate solution and hydrochloric acid to calculate the concentration of the hydrochloric acid. You must show clearly all calculations carried out.

Na2CO3 + 2HCl 2NaCl +CO2 + H2O

1 mole of Na2CO3 neutralises 2 moles of HCl

Mol of Na2CO3 = Volume (dm3) X Concentration (mol dm-3)

25.0 X 10-3 X 0.100

25.0 X 10-3 mol

No of moles HCl = 2 x No of mol Na2CO3 = 5.00 X 10-3 mol

5.00 X 10-3 = 35.0 X 10-3 X C

C = (5.00 X 10-3) ÷ (35.0 X10-3) – C = 0.143

Concentration of HCl = 0.143 mol dm-3

2.1 EXPLAIN THE TERMS SOLUTE, SOLVENT, SOLUTION, SOLUBLE, INSOLUBLE, MISCIBLE (LIQUIDS) AND IMMISCIBLE (LIQUIDS).

4. From your understanding briefly explain the following chemical terms giving an example for each case.

Solute, Solvent, Solution, soluble, Insoluble, Miscible liquids, Immiscible liquids

Chemical Term

Definition

Example

Solute

A solute is a substance that is dissolved in a "solvent" to make a "solution. The solute is the substance to dissolve, the solvent is the substance that makes it dissolve and the solution is the end product of the two mixed. This could be a liquid, gas or solid. The solubility of a solute is directly related to the conditions it is in. Temperature, pressure and concentration levels all have a bearing.

An example would be a sugar lump (Solute) that when heated is dissolved in water, (Solvent). The sugar/solute is dissolved by the water/solvent and the two mixes to become the Solution.

Solvent

A solvent is the component in a solution that dissolves the solute to create the solution. This is often a liquid such as water, Aqueous Solution, but this can also be other liquids such as ethanol, this would be classed as a Non-Aqueous solution. Solvents can also be a solid or a gas. All substances, solvents and solutions react differently depending on conditions such as concentration, temperature and pressure. There would generally be more of a solvent than a solute.

The H2O solvent starts to undo or dissolve the sugar molecules from their crystal form. This is speeded up by heat and the sugar dissolves completely in the solution. As the solution cools, the sugar will return to a crystallised form as the concentration that is able to be dissolved by the solvent has limitations

Chemical Term

Definition

Example

Solution

When a solute is dissolved by a solvent, this then becomes a solution. This can be a liquid, solid or gas. As with Solvents and Solutes, different conditions affect the composition on a solution. Heat, concentration and pressure. This is also directly related to the "Solubility" of a solute.

Once the sugar as been dissolved by the solvent (H2O), the mixture of both substances becomes one solution.

Insoluble

Insoluble is a substance that will be dissolved in a solvent to make a solution. Although most substances can be dissolved in the right environment such as temperature, given a suitable solvent.

An example of insoluble would be sand as the solute and water as the solvent. It doesn’t mater what the volume of water; the grains of sand will not dissolve.

Miscible Liquids

Miscible liquids are two or more liquids that can be mixed together homogenously (having the same composition)

Ethanol is a miscible liquid with water. Commercial companies mix Ethanol with water to make commercial products such as alcoholic drinks.

Immiscible Liquids

Immiscible Liquids are the opposite of miscible liquids where they are unable to be mixed together.

Oil and water is a good example of Immiscible liquids. Non-polar Hydrophobic molecules within the oil repel the water molecules and as such are unable to form a homogenous solution. When shaken, the oil and water appear to mix into an emulsion but will quickly separate back to oil and water.

2.2 EXPLAIN THE FACTORS WHICH AFFECT SOLUBILITY AND MISCIBILITY.

5. Explain the factors that affect solubility and miscibility

There are a number of things that will affect the solubility and miscibility of a solute, the most common being temperature. Solubility and miscibility increases in most solvents when the temperature is increased.

Another affecting factor is polarity. Generally, solutes are dissolved by solvents that have a similar polarity. Polar solvents do not dissolve nonpolar solutes as nonpolar solvents do not dissolve polar solutes. "Like dissolves like" (Hill & Holmon, 1995)

Although for most solid and liquid solutes pressure doesn’t affect solubility, solubility and miscibility of gasses are affected by pressure. This is noted by "Henry’s Law" which states that "at constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid". This is demonstrated by a fizzy drinks bottle where CO2 is dissolved in the liquid and is kept under pressure. Once the bottle is opened, the pressure equalises the CO2 rushes out of the solution in the form of bubbles.

Finally molecular size or "mass" is a factor in miscibility and solubility. It is much more difficult for the solvent molecules to surround large molecules of solute to dissolve. The smaller the molecules, the easier it is to dissolve.

There are rules associated with solubility. The chart below indicates which solute is soluble in which solvent:

2.3 EXPLAIN THE STRUCTURE AND FORMATION OF HYDRATED IONS.

6. Using three relevant examples explain the structure and formation of hydrated ions. What happens when ionic solid dissolves in water)

When an ion is inserted into a water configuration, it changes the structure of the hydrogen bond network. A water molecule tends to rotate so that its polarized charge concentration faces the opposite charge of the ion.

As the water molecules orient themselves towards the ion, they break the hydrogen bonds to their nearest neighbours. The group of water molecules oriented around an ion is called a hydration shell.

The orientation of the molecules in the hydration shell results in a net charge on the outside of this shell, a charge of the same sign as that of the ion in the centre. The charge on the outside of the hydration shell tends to orient water molecules in the vicinity, leading to a second hydration shell.

The result of forming hydration shells is to weaken the structure of the hydrogen bond network. This explains the observation that salt water has a lower freezing point than pure water: Each ion in the liquid has a hydration shell of oriented water molecules around it, which prevents the water molecules from forming the hexagonal structure of ice.

"Enthalpy of hydration, Hhyd, of an ion is the amount of energy released when a mole of the ion dissolves in a large amount of water forming an infinite dilute solution in the process, Mz+(g) + mH2O ® Mz+(aq) where Mz+(aq) represents ions surrounded by water molecules and dispersed in the solution. The approximate hydration energies of some typical ions are listed here. The table illustrates the point that as the atomic numbers increases, so do the ionic size, leading to a decrease in absolute values of enthalpy of hydration." (Hill & Holmon, 1995)

The following image shows the hydration of ions in Na+, CH3OH + H2O, CHCL3 and also demonstrates the ion dipole, H bond dipole, ion induced dipole, and dipole induced dipole and dispersion:

(Hill & Holmon, 1995)

(Hill & Holmon, 1995)

2.4 EVALUATE, INTERPRET AND PERFORM A RANGE OF CALCULATIONS FROM SOLUBILITY DATA AND SOLUBILITY CURVES FOR SOLIDS AND GASES.

SOLUBILITY CURVES FOR A NUMBER OF SOLIDS

7. Solubility curves for a number of solids are given in the above graph. Using the graph calculate the following.

a. What mass of K2Cr2O7 will dissolve at 800C

The mass of K2Cr2O7 that will dissolve in 100g of H2O = 58g

b. What happens when this solution K2Cr2O7 is cooled to 740 C

Solubility reduces to approximately 52g per 100g H2O

c. What mass of KClO3 dissolves at 580 C

The mass of KClO3 that will dissolve in 100g of H2O = 23g

d. What is the solubility of the following at 500 C

i. KNO3 Solubility = 87g

ii. KCl Solubility = 43g

iii. K2Cr2O7 Solubility = 30g

e. A solution is saturated at 600 C with KNO3 and NaCl . A 100g sample of solution is cooled from 600C to 200C. What mass of crystals will form?

Looking at the graph, KNO3 at 200, solubility reduces to 30g therefore 30g of KNO3 will form. At 200 NaCl solubility reduces to 35g.

Therefore total mass of crystals that will form at 200 will be:

If 100 g of each substance was dissolved in 100g of H2O, the solubility per 100g of H2O at 20o would reduce to 30g for KNO3 and 35g for NaCl. Therefore if 100g (200g total) of both substances were used, only 65g would be dissolved in the solution thus 135g of crystals will form in total and this will be made up of the following:

KNO3 - 70g

NaCl - 65g

Total mass of crystals = 135g

Total solubility of crystals per 100g H2O = 65g (KNO3 30g and NaCl 35g)

3.1 DEFINE THE MEANING OF ACID, BASE AND ALKALI (USING BRONSTED LOWRY THEORY).

8. State and explain the definition of Bronstead – Lowry of acids and bases

Show by writing appropriate equations how:

HSO4‾ion and NH3 can act as B / L bases

HCl and CH3COOH can act as B /L acids

According to The Bronstead-Lowry Theory devised in 1923, "an acid is a proton donor – a base is a proton acceptor" (Hill & Holmon, 1995)

When acids and bases react with each other, the acid on one side of the equation "loses" a proton and becomes the (conjugate) base on the opposite side. As such, the base on one side gains the proton and becomes the (conjugate) acid on the opposite side of the equation.

The speed of which the transfer takes place denotes whether it is a weak/strong Acid/Base. Rapid transfer defines strong Acid/Base slow transfer defines weak Acid/Base.

HSO4‾ion and NH3 can act as B / L bases

Both HSO4‾ and NH3 are amphoteric which means they can act as an acid or a base. Water (H2O) is also amphoteric.

HSO4- + H+ H2SO4

When acting as a base, the Hydrogen Sulphate is the conjugate base of Sulphuric Acid and gains the proton to change from the base on the left side of the equation to become the acid on the opposite side. This proton swap can shift each way.

NH3 + H+ NH4+

Due to the Ammonia having a lone pair of electrons on the outer shell of the Nitrogen atom it can donate this pair and as such is a base. As this is slow reaction ammonia is classed as a "weak" base.

b) HCl and CH3COOH can act as B /L acids

When HCl is added to CH3COOH, the increase of H+ ions is counteracted by association with the excess of acetate ions to form unionised CH3COOH. Thus the added H+ ions are neutralized and the pH of the buffer solution remains virtually unchanged.

However owing to the increased concentration of CH3COOH, the equilibrium shifts slightly to the right to increase H+ ions. This is known as a "buffer" solution and has a marginal increase of pH of the buffer solution on addition of HCl.

3.2 INTERPRET COMMON ACID/BASE REACTIONS (IN AQUEOUS SOLUTION AND GAS PHASE) AS REPRESENTED IN CHEMICAL EQUATIONS.

9. Interpret the reactions of a strong acid (e.g. HCl, HNO3) with the following giving balanced equations

a. i. Aluminium

When HCl is added to aluminium it starts to dissolve and creates AlCl3 (aluminium chloride), it also creates an exothermic reaction where the H2O becomes hot and evaporates.

The balanced equation for this reaction is:

2Al + 6HCl 2AlCl3 + 3H2

ii. Lithium - Li + 2 HCl = LiCl2 + H2

When Lithium (Li) and HCl are mixed, there is a violent exothermic reaction that takes place with the H2O creating lithium chloride and hydrogen gas.

The balanced equation for this reaction is:

Li + 2 HCl = LiCl2 + H2

iii. Magnesium -

Mg+ 2HNO3-->Mg(NO3)2 + H2

b. i. Phosphorous

- 20 P + 40 HNO3 + H20 = 20 H3PO4 + 40 NO

ii. Sulphur

- 10 S + 40 HNO3 = 10 H2SO4 + 40 NO2 + H20

c. i KOH

HNO3 + KOH H2O + KNO3

ii. Na2CO3 –

Na2CO3 + 2 HNO3 H2CO3 + 2 NaNO3

10. Describe the reactions of a strong alkali ( e.g. NaOH, KOH ) with the following giving balanced equations

a. i. Aluminium

In normal circumstances Al doesn’t react with water due to having an impermeable protective layer of aluminium hydroxide which is formed by oxidisation with the air. When sodium Hydroxide is mixed with aluminium this dissolves the protective layer and a reaction takes place. This reaction takes place at the beginning relatively slowly. The water acts as an acid and dissolves the aluminium completely.

6H2O+2Al+2NaOH 2NaAl4+3H2

ii. Lithium

li(NH3)4 li+4NH3

iii. Magnesium

mg + NaOH mgNa +OH

b i. Chlorine

Cl2 + NaOH NaCl + NaOCl + H2O

ii. Sulphur

S + Na2CO3 SO3 + Na2C

c i. HCl

HCl + NaOH NaCl + H2O

ii. H2SO4

K20 + H2SO4 H2O + K2SO4

3.3 DEFINE THE MEANING OF NEUTRAL SOLUTION, ACIDIC SOLUTION AND ALKALINE SOLUTION.

11. Define the meaning of the following using your knowledge of chemistry and giving at least one example for each

a. a neutral solution

A neutral solution is a solution that has a pH level of around 7. This solution would be neither acidic nor alkaline. The H+ and OH- ions are balanced. Saliva and blood are both pH neutral.

an acidic solution

An Acidic solution is an aqueous solution where the H+ ions are more than the OH- ions. This makes the solution acidic and the pH level is less than the neutral value of 7. Examples of acids would be Vinegar (Acetic acid) and car batteries (Sulphuric Acid).

an alkaline solution

An Alkaline solution is an aqueous solution that has had bas solids dissolved in it. The OH- Ions outnumber the H+ ions and as such has a pH level above 7. Sodium Hydroxide and Potassium Hydroxide are two examples of alkaline’s.

3.4 EXPLAIN THE CHEMICAL PROPERTIES OF ACIDIC SOLUTIONS AND ALKALINE SOLUTIONS AND REPRESENT THESE WITH CHEMICAL EQUATIONS.

12. Explain the chemical properties of acidic solutions and alkaline solutions taking into consideration their chemistry giving at least three examples for each and represent these with chemical equations.

In acidic solutions, the H+ ions outnumber the OH- ions thus the balance is in favour of the positive. This makes the substance corrosive and a conductor of electricity due to the mobile ions. pH <7

Acids react with reactive metals to form salts and Hydrogen gas. Acids react with metal carbonates to create salt, CO2 and H2O. All acids are soluble in water. pH >7

Examples:

HCl + NaOH NaCl + H2O

2HCl + CaCO3 H2O + CO2 + CaCl2

2 HCl + Ba(OH)2 BaCl2 + 2 H2O

Alkaline solutions are corrosive have a soapy feel, contain hydroxyl ions OH- and react with acids to form salt and H2­O.

Examples:

Ca(OH)2 + 2 HNO3 Ca(NO3)2 + 2 H2O

Mg + 2H2O Mg(OH)2+ H2Ca(OH)2

2 NaOH + H2SO4 Na2(SO4) + 2 H2O

3.5 EXPLAIN THE MEANING OF PH SOLUTION, AND INTERPRET PH VALUES.

13. Explain what you understand by the meaning of the term pH of a solution in terms of H ion concentration and any other features you are familiar with and interpret the following pH values of solutions.

The pH is the measurement of the activity of Hydrogen Ions (H+) in a solution that will define if it is an acid, alkaline or neutral substance. The pH is measured against an internationally agreed scale. The measurements are taken using a variety of methods, in aqueous solutions, indicators such as Methyl Orange or Phenol Red. These indicators react with the acidity/alkalinity of the solution and change colour when they reach a known pH level.

The actual pH scale is devised from the logarithm calculation of the Hydrogen ion activity, pH=log10 (aH+). The pH of H2O at 25c is 7 which is recognised as neither Acidic nor Alkaline so a state of neutral. However when exposed to air and temperature, water becomes slightly acidic as it absorbs carbon dioxide and creates bicarbonate and hydrogen atoms.

The higher the H+ ions in a solution, the more reactive this becomes and as such the more acidic but lower pH number, the less H+ ions, the more alkaline and thus a higher pH number. With a change in number is a multiplication in strength by 10. pH 4 will be ten times more acidic than 5, 5 ten times more acidic than 6 etc.

Examples of pH levels:

4.5 – Acidic – A pH level of 4.5 would indicate a very strong acid. Tomatoes have a pH level of 4.5 and are responsible for giving consumers heartburn.

7.0 – Neutral – A pH level of between 6.6 and 7.3 is considered "neutral" so neither acidic nor alkaline. Saliva has a pH level of between 6.6 and 7.3 as is blood; however skin is slightly acidic to perform the defence of the body.

11.6 – Alkalinity – Strong alkaline contain a lot more OH- ions which react with acids. Sodium Hydroxide (Caustic soda) is an example of a substance with a pH of 11.6. When mixed with H2O, the NaOH becomes ionised and an exothermic reaction takes place making this a dangerous substance.