COMMON ACID –BASE INDICATORS AND THEIR pH RANGE ACID-BASE TITATION

Topic: Acid- base reactions

CONTENT:

1. COMMON ACID –BASE INDICATORS AND THEIR pH RANGE
2. ACID-BASE TITATION
3. HEAT OF NEUTRALIZATION (INTRODUCTORY) CONSTRUCTION OF WOODEN RETORT STAND

PERIOD 1:  1: COMMON ACID –BASE INDICATORS AND THEIR pH RANGE

Acid-base indicator is a weak organic acid or weak organic base (organic dyes) which shows colour change depending on the pH of the solution or organic dye which show one colour in acidic medium and another colour in basic medium. It can defined as an organic dye which as colour in ionized state and another colour in unionized state.Examples of indicators are methyl orange, phenolphthalein, methyl red etc.

CHOICE OF INDICATOR FOR ACID-BASE TITRATION

The choice of an indicator for acid-base titration depends on the pH of the mixture at equivalent point and the pK of the indicator. When a strong acid is titrated against a strong base, at the equivalent point, the salt formed does not under hydrolysis and so any suitable indicator can be used. The salt formed when strong acid is titrated against a weak base undergoes hydrolysis to produce excess hydrogen ions. This makes the solution acidic and any indicator that shows colour change in acidic medium is used in the case. Methyl orange changes colour within a pH of2.9—4.6and so suitable for this titration. The salt formed when weak acid is titrated against strong base undergoes hydrolysis to produce excess hydroxide ions making the solution alkaline. Phenolphthalein changes colour in basic medium and so suitable to the kind of titration.

Indicators are used in acid-base titration to show the end point when the acid would have materialised the base. Colour changes of common indicator are shown below

Indication	pH range	Colour in acid solution	Colour in the alkaline solution	Neutral or end point	Types of acid base titration
Litmus	50-8.0	Red	Blue	purple	Strong acid and strong base
Methyl Orange	2.9-4.6	Pink	Yellow	orange	S.A and S.B or Na2Co3 in H2O
Phenolphthalein	8.2 – 10	Colourless	Pink	colourless	S.A and W.A
Methyl red	4.4 – 6.3	Red	Yellow	orange	S.A and S.B

S.A = Strong acid, S.B = Strong base, W.A = weak base

 

 

This is volumetric analysis which aim at estimation of the qualities of certain substances in solution

PERIOD 2 AND 3: ACID- BASE TITRATION

This involves titrating a solution(usually the acid) from the burette into a fixed volume(20 or 25cm³) of base until the two solutions have completely reacted. The concentration of one of the solution is normally known. An indicator is used to determine the point of complete reaction.

For titration, more dilute solutions containing 0.1moldm^-3 or 0.05mold^-3 are used.

STANDARD SOLUTION

In any titration, standard solution must be used to react with a solution of unknown concentration.

A standard solution is a solution of known concentration.

PRIMARY STANDARD

A primary standard is a reagent that is extremely pure, stable and has no water of hydrate. A primary solution is used for the preparation of standard solution. There a primary standard solution is a solution which is prepared from a non-hygroscopic, non-deliquescent and pure substance. Examples are sodium trioxocarbonate(iv), sodium oxalate and benzoic acid

Features of a primary standard

1. High purity
2. Stability
3. High solubility
4. High relative molecular mass

SECONDARY STANDARD

Secondary standard solution refers to a solution that has that has its concentration measured titration with a primary standard solution.

The concentration of a solution is the amount of solute in 1dm³ or 1000cm³ the solution. It can be expressed as moldm^-3 or gdm^-3

The above diagram is an example of the apparatus used in acid-base titration. Some materials used in acid- base titration are:

1. Weighing bottle
2. chemical balance
3. pipette
4. Burette
5. retort stand
6. filter paper
7. funnel
8. white tile
9. standard volumetric flask
10. conical flask

PRECAUTIONS IN USING THE PIPETTE, BURETTE AND CONICAL FLASK

(a)PIPETTE: (i) Rinse the pipette with the solution it should be used to measure i.e. base

(ii) Avoid air bubbles in the pipette

(iii) Make sure that the mark to be read is at level with youreye.

Do not blow the last drop on the pipette.

(b) BURETTE:

(i) Rinse the burette with acid or allow it to drain after rinsing with distil water

(ii) Make sure the burette jet is filled

(iii) Make sure that the burette is not leaking

(iv)Take your burette readings with your eyes at the same level as the measures to avoid error due to parallax

(v) Remove the funnel before taking your reading

(vi)Avoid incorrect reading.

CONICAL FLASK

1. Do not rinse it with any of the solutions used in the titration but with distilled water.
2. Wash down with distilled water any drop of the solution that stick by the sides of the conical flask during titration.

Example of titration problem solved.

A is a solution of hydrochloric acid containing 4.00g in 1dm³ of solution B is a solution containing sodium hydrochlorid

• Put A into a burette and titrate with 25cm³ portion of B using methyl orange indicator. Tabulate you burette reaching. Determine the average volume of solution A used
• From your result and information provided
• Find concentration of A in moldm^-3
• Find concentration of B in moldm^-3
• Find the amount in mass of NaOH present in 1dm³ of solution B

Equation of reaction:

HCl +  NaOH          NaCl + H₂O

[H = 1, Cl= 355, K=39, Na= 23, O = 16]

STUDY ANSWER

Indicator used: methyl orange

Size of pipette: 25cm3

	Rough	Ist Titration	2nd Titration	3rd Titration
Final volume of acid A used in (cm3)	25.80	25.50	25.40	25.50
Initial volume acid used in(cm3)	0.00	0.00	0.00	0.00
Volume of solution A used (cm3)	25.80	25.50	25.40	25.50

 

Average litre volume (cm³) =

= = = 25.45cm³

B(i) Concentration of HCl of A in moldm^-3

Conc. in gdm^-3 = 4 gdm^-3

Molar mass of HCl=  1+35.5 =36.5

Con. In moldm^-3 of HCl = = 0.110 moldm^-3

b(ii) V_a= 25.45 cm³

Ca= 0.110 moldm^-3

V_b= 25cm³

C_b=? Of resultant concentration of effect of the two alkalisin solution B.

2HCl +            HCl + NaCl + 2H₂O

=

C_b V_b: na = Ca Va: nb

Make C_b the subject formula.

Cb=  =

0.0559 moldm^-3 (38.8 fig)

b(v) Amount in mass of NaOH on 1dm³  of solution B

Molar mass of NaOH = 23 +1 + 16=40g/mol

Amount in mass of NaOH in 1dm³ of solution B=

40 × 0.0559 moldm^-3

= 2.23gdm^-3

EVALUATION

State five precautions you need to take while using. (a) Pipette   (b) burette and 2 precautions while using (c) conical flask

PERIOD 4:   HEAT OF NEUTRALIZATION

When an aqueous solution of a acid is added to alkalis, there is an increase in temperature. This increase in temperature is called heat of neutralization.

Heat of neutralization of strong acid and strong alkalis are of the same value. The value of weak acids by strong base and strong acids by weak bases are different and varies due to the fact that energy for the complete ionisation of such weak bases and weak acids before neutralization is drawn from the solution.

PROCEDURE

Measure the temperature of the acid solution and that of the alkali solution before mixing. Find the average temperature of both acid and base.

Determine volumes of acid and alkali to mix in order to get a neutral, solution [by titration]. Mix these volumes (say 50cm³ of acid + 50cm³ of alkali; find the temperature of mixture; note the increase in temperature.

If t₁ = temperature of acid, and t₂= temperature of alkalis

Average of the two =  = t₃

If t₄ is the temperature of mixture, then the increase in temperature      = t₄ –t₃ =t₅

Assume that the solution, have the same density and specific heat as water.

Therefore, mass of mixture=100g

Specific heat of water = 4.2joulesg^-k

Heat of neutralization= specific heat x mass x temperature rise=4.2 x 100 × t₅joules

Heat evolved when 1 mole of each reacts=100 x t₅ x 4.2 x 10joulesmol^-1

EVALUATION

1. What is heat of neutralization?
2. State three indicators with their pH ranges.
3. What indicators are as applied in volumetric analysis.

 

GENERAL EVALUATION

OBJECTIVE TEST :

1. A student prepares 0.5M solution each of hydrochloric and ethanoic acids and then measured their pH the result show that

• pH values are equal
• The solution has a higher pH
• Sum of the pH value is 14
• Ethanoic acid solution has a higher pH

2. What volume of 0.5M H₂SO₄ will exactly neutralize 20cm³ of 0.lM NaOH solution?

• 0cm³ (b) 5.0 cm³  (c) 6.8 cm³  (d) 8.3 cm³  (e) 10.4 cm³

3. Filling the burette for titration involves these except.

• Wash with water (b) rinse with acid (c) eject air bubbles (d) read at eye level

4. In the titration of acid against base solutions, averaging must involve.

• Rough reading (b) concordant reading (c) higher reading (d) all the titres obtained

5. What mass of anhydrous sodium trioxocarbonate(IV), Na₂CO₃; present in 500cm³ of 0.1moldm^-3? (Na=23, C=12, O=16)

• 6g (b) 212g (c) 5.3g (d) 106g.

ESSAY QUESTIONS

1. (a) In an acid –base titration (i) which solution is normally put in the burette and why. (ii) To which of the solutions is the indicator added? (iii) Why is the conical flask rinsed with distilled water and not with the solution to be place in it?

• What is the colour of methyl orange in (i) distilled water (ii) lime water (iii) vinegar?

2. A is dilute tetraoxosulphate (vi) acid. B contains 1.50g of sodium hydroxide per 250cm³ of solution.25 cm³ of B required 15.5cm³ of A for complete neutralization

The equation for reaction:

H₂S0_4(aq) + 2NaOH _(aq)   Na₂SO_4(aq) + 2H_2(l)

• Calculate:
• The concentration of solution B in moldm^-3 [H=1, O= 16, Na=23]
• The concentration of A in moldm^-3
• The number of hydrogen ions in 1.0dm^-3 of solution A [Avogadro’s constant 6.0
• The volume of distilled water that should be added to 25cm³ of solution A in order to dilute it ten times

1. What would be the colour of methyl orange indicator in:

(i) Solution A  (ii) solution B (iii) titration mixtures of A and B at the end point.

3. Calculate the mass concentration in gdm^-3 of the ions in the following solutions:

(a) Cl^–ions in 0.2 moldm^-3 sodium chloride solution

(b) NO₃^–ions in 0.75 mol dm^-3trioxonitrate (v) acid.

(c) SO₄^2- ions in 2moldm^-3 potassium tetraoxosulphate (vi) solution. Cl= 35.5, N=14, O=16, S=32]

 

TOPIC: WATER

CONTENT

1. Structure of water, Hardness of water and removal of hardness of water
2. Solubility- Basic concept
3. Types of Solutions
4. Factors affecting solubility and uses of solubility curve.

PERIOD 1: STRUCTURE OF WATER, HARDNESS OF WATER AND REMOVAL OF HARDNESS OF WATER

A molecule of water consists of two atoms of hydrogen and one atom of oxygen chemically combined to form a covalent molecule. A molecule of water has a V-shape. The repulsion between the two lone pairs of electrons on the oxygen atom makes the structure of water molecular to be V- shaped

Structure of water

O

(a)Bonding    (b) Shape of water molecule

The shared pair of electrons between each hydrogen atom and oxygen is drawn more electronegative, making the oxygen atom to be partially negatively charged and hydrogen atoms to be partially positively charged. This leads to hydrogen bonding between the molecules of water the attendant relatively high melting point, high boiling point and low vitality.

Hydrogen bonding in water molecules

HARDNESS OF WATER

A hard water is water which will not readily form lather with soap. For example, river water, lake water, stream water and sea water. Hardness of water is due to the presence of calcium, magnesium and iron in water. There are two types of hard water: temporary hardness and permanent hardness.

TEMPORARY HARDNESS: This is caused by the presence of dissolved calcium or magnesium hydrogen trioxocarbonate(IV), HCO₃^– of Ca, Mg or Fe. Temporary hardness is so called because it is easily removed by boiling.

Removal of temporary hardness: Temporary hardness can be removed by:

1. Boiling: The calcium hydrogen trioxocarbonate(IV) which causes the temporary hardness is decomposed by heating.
2. Addition of calculated amount of slaked lime, CaCOH)₂

Ca(HCO₃)₂ + Ca(OH)₂ 2CaCO₃ + 2H₂O

Soluble    slightly soluble    insoluble

Effects of temporary hardness:

1. Furring of kettle: When a kettle or boiler has been used to boil hard water for some time, the inner surface becomes coated with a white fur-like layer. The layer is due to the gradual deposit of CaCO₃ from the decomposition of Ca(HCO₃)₂

 

 

 

 

2. Stalagmites and stalactites: These are pillars of limestone, CaCO_3.

Found in hot caves. They are formed when hard water flow temporary over the top of the cave and drop off water then becomes decomposed by the heat inside the cave, leaving deposits of CaCO₃ on the roof and floor of the cave. The deposits that grow downward are termed stalactites and the ones that grow upward are termed stalagmites. Pillars of CaCO₃ are formed when both stalactites and stalagmites meet.

PERMANENT HARDNESS: This is caused by the presence of dissolved calcium and magnesium ions in form of soluble tetraoxosulphate(VI) and chlorides and certainly cannot be removed by boiling. When water containing any of these substances is evaporated, a white solid deposit of calcium or magnesium tetraoxosulphate(VI) and/ or calcium trioxocarbonate(IV) is left behind.  Calcium trioxocarbonate(IV) causes the ‘furring’ in kettles that occur in hard water.  This furring may be removed by the addition of a dilute acid:

2H⁺   +  CO₃^2-_(aq)            CO_2(g)     +  H₂O_(l)

Blockages in hot water pipes are caused by similar process to the furring of kettles. A thick deposit of limescale builds up.

Removal of permanent hardness: permanent hardness as well as temporary hardness can be removed by:

1. Addition of washing soda, (Na₂CO₃.10H₂O) crystals

In each type, the calcium or magnesium ion, which actually causes the hardness, is removed as a precipitate and can, therefore, no longer cause hardness.

Na₂CO_3. 10H₂O + CaSO₄               CaCO₃ + Na₂SO₄

Soluble                 insoluble

Na₂CO₃ + MgSO₄                    MgSO₄ + Na₂SO₄

2.Addition of caustic soda, NaOH

2NaOH + CaSO₄                          Ca(OH)₂ + Na₂SO₄

Soluble             insoluble

2NaOH + MgSO₄ Mg (OH)₂ + Na₂SO₄

3. Ion- exchange method: Zeolite or permutit is an ion exchange resin used industrially and in the home for softening water. When the hard water is passed through the resin, the sodium ions in it goes into the solution and the calcium and magnesium ions take their place in the complex salt.

CaSO₄ + Na₂Y          CaY + NaS₂O₄

Soluble     resin             insoluble

Sodium aluminium trioxosilicate (iv)

 

EVALUATION

1. Why is water molecule V-shaped?
2. Name the constituent elements in water with their respective number of atoms.
3. What type of chemical combination exists between water molecules?

PERIOD 2: SOLUBILITY

Definition: The Solubility of a solute (solid) in a solvent (liquid) is the concentration of the saturated solution. Solubility can be defined thus:

1. It is the maximum amount of the solid that dissolves in 1 dm³ of the solution at a given temperature. It is expressed in mol.dm^-3.
2. It is the maximum mass, in grams, of the solid that dissolve in 100 g of the solvent at a given temperature.

When some common salts is added to a beaker of water and the mixture stirred, the salt gradually disappear, and the clear colourless mixture is obtained. The salt is said to have dissolved in the water. Thus, the salt that dissolved in the water is called solute and the water that does the dissolving is known as the solventand the product obtained by dissolving the salt in the water is called a solution.

Therefore, a solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated or supersaturated.

Determination of solubility of KNO₃ at 30^oC

Stage 1: Preparing of Saturated Solution

1. Put 100 cm³ of distilled water in a beaker, add the salt little at a time and warm on a Bunsen burner with stirring. Continue the addition of the salt with stirring and keeping the beaker warm at about 50^oC, until the salt can no longer dissolve.
2. Allow the saturated solution to cool to 30^oC; this is the saturated solution of the salt at 30^oC. It will be noticed that as the hot saturated solution cools, the excess solid in the solution separates out, to give a saturated solution at a lower temperature.
3. Stage 2: Evaporation of Saturated Solution to Dryness.
4. Weigh accurately a clean dry evaporating dish: x g.
5. Rinse a 25 cm3 pipette with the saturated solution, in order to warm it.
6. Pipette 25 cm³ of the saturated solution at 30Oc, as quickly as possible (without pipetting any undissolved solid) into the evaporating dish and re-weigh: y g
7. Place the dish on a steam bath and evaporate to dryness. Allow to cool in a charged desiccators, and then re-weigh.
8. Repeat the process of heating and cooling, until a constant mass is obtained: w g.

Calculation of the solubility at 30^oC

The following hypothetical values will be used to illustrate how to calculate the solubility of a salt.

Data required:

• Mass of the evaporating dish:                                               x = 15.20 g
• Mass of evaporating dish + 25 cm³ of saturated solution: y = 40.70 g
• Mass of dish + anhydrous salt:                                              w = 20.70 g

Solubility in mol.dm^-3

Mass of anhydrous salt  = (w – x)   = (20.70  – 15.20) g = 5.50 g

i.e. 25 cm³of saturated solution contain 5.50 g of salt,

Hence, 100 cm³of saturated solution contain 5.50  x 1000/25  = 220 g of salt.

To convert 220 g to moles

Amount (mol)  = (Mass of salt)/ (Molar mass).

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g.mol^-1

i.e. Amount, in moles = 220/ 101  = 2.18 moles

Therefore, Solubility of KNO₃ at 30^oC = 2.18 mol.dm^-3

Solubility in grams per 100 g of water

Mass of water used = (y – w)  = (40.70  – 20.70 )g = 20.0 g

Mass of salt used =      (w – x)  = (20.70  – 15.20) g = 5.50 g

i.e. 20.0 g of water at 30^oC saturated 5.50 g of salt

Therefore, 100 g of water will saturate 5.50 x 100/20 g of salt

i.e.                                          5.50 x 5 g = 27.5 g of salt.

Hence, the solubility of the salt at 30^oC is 27.5 g per 100 g of water.

WORKED EXAMPLES

1. If 11.87g of potassium trioxonitrate(V) were dissolved in 43.4g of distilled water at 60 , calculate the solubility of the solute in moldm^-3

Solution:

Molar mass of KNO₃=101g

1g of water = 1cm3 of water

Mass of dissolved KNO₃ = 11.87g

Concentration in moldm^-3. Of dissolved KNO₃

= 0.118moldm_-3

43.4g or cm³ of water at 60  dissolve 0.118 moldm_-3 of KNO₃

1000cm³ of water at 60  will dissolve:

= 2.7 moldm^-3 of KNO₃

Therefore, the solubility of KNO₃ at 60  in water is 2.7moldm^-3

EVALUATION

1. Explain the following terms: (a) solute (b) solvent (c) solution
2. What is solubility? List the different types of solution.
3. Calculate the solubility of Pb(NO₃)₂ in moldm^-3 if 12.2g of the solute were dissolved in 21g of distilled water 20 .

PERIOD 3: TYPES OF SOLUTION

A solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated, unsaturated or supersaturated.

1. A saturated solution is one which contains, at a given temperature, as much solute as it can hold in the presence of the dissolving substance. That is, a given solute dissolves in a given solvent, such that no more of the solute will dissolve in the solvent at the temperature of dissolution.
2. An unsaturated solution is one which contains less solute than it can hold at the given temperature.
3. A supersaturated solution is one which contains more of the dissolves substance than it can hold at a given temperature, when the saturated solution is in contact with the solid solution. Super saturated solution are unstable. The excess solute can be separated out by slightly shaking the solution or creating centres of crystallization for the excess solute.

Relationship Between solubility and crystallization

The different solubilities of substances are utilized in their purification by the process of crystallization. In manufacturing KNO₃, solutions of KCl and NaNO₃are mixed. The resulting solution is then concentrated at boiling point and the NaCl is then deposited. The equilibrium in the equation is displaced to the right.

K⁺ + Cl^– +  Na⁺ +  NO₃^–          K⁺ +  NO₃^– + NaCl

The solution is filtered hot and later cooled. KNO₃, with the lowest solubility, crystallizes out first and is purified by recrystallization :

KCl  +  NaNO₃⇌KNO₃ +   Na⁺ + Cl^–

A saturated solution is in dynamic equilibrium with the excess solid present in the solution. The dissolution equilibrium can be expressed in terms of an equilibrium constant, e.g. in the case of silver chloride

AgCl_(s)⇌Ag⁺_(aq)  + Cl^–_(aq)

EVALUATION

1. Define the term saturated solution.
2. Distinguish between saturated and unsaturated solutions.

PERIOD 4.  FACTORS AFFECTING SOLUBILITY OF SOLIDS IN LIQUIDS

The four factors that influence or affect the solubility of substances are:

1. Nature of solvent: Generally, ionic compounds are more soluble in polar solvent (e.g. water), than in non- polar increases the solubility of the solvent e.g. NaCl is more soluble in water than in ethanol.
2. Nature of the solute: Ionic compounds are more soluble than covalent compounds and vice-versa. For instance, NaCl is more soluble in water than sugar at room temperature.
3. Temperature: In most cases solubility increases with increase in temperature. In endothermic reactions solubility increases with increasing temperature and vice versa e.g. calcium oxide.
4. Common ion effect: The solubility of an ionic compound in water is affected by the presence of another compound if both compounds have the same cation or anions. For example, the solubility of NaCl in water is much higher than its solubility in dilute HCl because there is a common chloride in NaCl and HCl.

Uses of Solubility curves

1. Solubility curves enable pharmacists to determine the amount of solid drugs in a solution of drug mixture of solids
2. It helps separate and purify mixtures of solids.
3. Used to determine the most suitable solvents for the extraction of solutes from natural sources.

EVALUATION:

1. Mention three factors that affect the solubility of a solid in a liquid.
2. Explain the term common ion effect.

GENERAL EVALUTION

OBJECTIVE TEST:

1. A substance that dissolves in a solvent to form a solution is called (a) acid (b) suspension (c) glue (d) solute
2. The shape of a water molecule is: (a) K-shaped (b) V-shaped (c) N-shaped (d) S-shaped (e) Y-shaped
3. Which of the following birds water molecules together? (a) ionic bond (b) covalent bond (c) hydrogen bond (d) van der Waals forces
4. When a crystal was added to its solution, it did not dissolve and the solution remained unchanged, showing that the solution was. (a) concentrated (b) colloidal (c) saturated (d) unsaturated (e) supersaturated
5. Calculate the solubility of KCl in moldm^-3 at 30 if 20cm³ sample of a saturated solution of KCl contains 4.50g of the salt. (a) 2.5 (b) 3.0 (c) 3.02 (d) 2.03

ESSAY QUESTIONS:

1. If the solubility of KHCO₃ is 0.4 moldm^-3 at room temperature, at 20 (KHCO₃ =100g.mol^-1)
2. (a) What is solubility curve? (b) Sketch a curve to show how the solubility of a gas varies with an increase in temperature. (c) List two uses of solubility curve.
3. Name and explain any three factors that affect the solubility of a solute in a solvent.
4. Explain why a solid dissolves faster in hot water than cold water
5. (a) How is the solubility of a solute determined? (b) If the solubility of KHCO₃ at 20 is 3.10moldm^-3 determine whether a solution of KHCO₃ containing 303gdm^-3 at 20  is saturated or unsaturated.