7 acid and bases

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CHAPTER 7 : ACID AND BASE T. 7.0

Arrhenius Definition!!

Chemistry Module 2009@ Hak Milik JPN Pahang

1.An acid is a chemical substance which ionises in water to produce hydrogen ion,H+.2. A base is a chemical substance which ionises in water to produce hydroxide ion, OH-

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ACIDS AND BASES

Analysing characteristics and properties of acids and bases

Synthesising the concept of strong acids, weak acids, strong alkalis and weak alkalis

Analysing Concentration of acids and alkalis

Analysing Neutralization

1. Define the Arrhenius acid and alkali.2. Define base. 3. Describe chemical properties of acids

1. Define strong acid and weak acid. 2. Define strong base and weak base. 3. Write 3 examples of strong acid, weak acid, strong alkali and weak alkali respectively.

1. Define molarity2. Describe methods for preparing standard solutions.3. State relationship between the numbers of moles with molarity.

1. Define neutralization reaction.2. Write ionic equation for neutralization reaction.

7.1 7.2

7.37.4

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T. 7.1


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CHARACTERISTIC AND PROPERTIES

OF ACIDS AND BASES

Define acid and base

AcidBase

The functions of water in determining properties of acid and alkali

Substance produced hydrogen ion H+ when dissolved/ionise in water.

Substance with opposite properties to acid and may react with acid to form salt and water.

Base that dissolved in water is called alkali

Water dissociated acid to form H+ ion H2OH2SO4 2H+ + SO4

2-

Water dissociated alkali to form OH-

H2ONaOH Na+ + OH-

NH3 + H2O NH4+ + OH-

The difference between base and alkaliProperties of

acid and alkali

Uses of acids and alkalis

Acid Alkali1. Sour Bitter2. Electricity conductor

Electricity conductor

3. React with base to form salt and water.

React with acid to form salt and water.

4. React with metal to form salt and hydrogen gas.

React with ammonium salt to form ammonia gas.

5. React with carbonate to form salt, water and CO2

React with halogen to form 2 types of salt.

Base AlkaliWater insoluble metalOxide or hydroxide.

Water soluble base forming OH-

ion

Definition of alkali

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T.7.2

*T.7.2.1


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STRONG ACIDS, WEAK ACIDS,

STRONG ALKALIS AND WEAK ALKALIS

Strong acid and weak acid

Strong alkalis and weak alkalis

Definition of strong acid

Definition of weak acid

Definition of strong alkali

Definition of weak alkali

H+ Cl-

Cl-

H+

H+

Cl-H+

Cl-

Cl-

H+

H+Cl-H+

CH3COO-H+

CH3COO-

CH3COOHCH3COOH

CH3COOH

CH3COOH

Na+ OH-

OH-

Na+

Na+

OH-Na+

OH-

OH-

Na+

Na+OH-

NH4+

NH4+

OH-

OH-

NH4OH

NH4OH

NH4OHNH4OHNH4OH

Acid that completely ionise in water to produce H+

Acid that partially ionise in water to produce H+

Alkali that completely ionise in water to produce OH-

Alkali that partially ionise in water to produce OH-

Relationship between pH and

molarity of acids and alkalis

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T. 7.3


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Relationship between pH and

molarity of acids and alkalis

Scale pH

To determine pH of acids and

alkali

1 2 3 4 5 6 7 8 9 10 11 12 13 14

NeutralAcidity increase Basicity increase

The smaller pH value, the higher concentration of H+

The higher pH value, the lower concentration of H+

Molarities(M)

PH scaleH2SO4

(Strong acid)

HCl(Strong acid)

CH3COOH(weak acid)

NaOH(strong alkali)

NH4OH(weak alkali)

0.1 M 1.0 1.0 3.0 14.0 11.00.01 M 1.2 2.0 3.5 13.0 10.50.001M 1.5 3.0 4.0 11.0 10.0

Acid or alkali

Universal indicator

Explanation

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*T.7.3.1


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CONCENTRATION OF ACIDS AND

ALKALIS

Concentration

Concentration in g dm-3 Concentration in mole dm-3 (M)

Solute mass (g)Concentration =

Volume of solution (dm3)

No. of mole of solute Molarity (M) =

Volume of solution ( dm3)

Or M = n/ (V/1000) or n = MV/1000Where V=volume in cm3

X Molar mass

÷ Molar mass

Preparation of a solution

Preparation of standard solution

Dilution Method

Relationship between pH and molarity of acid

and alkali

Calculation exercise for concentration

The molarity of an acid decreased the pH value

increase. The molarity of an alkali decreased, the

pH value decreased

pipette

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Volume and concentration of

substance.

Molarity (mol dm-3)

Concentration (g dm-3 )

No. of. MoleVolume(cm 3 )

Mass(g)

100 cm3 sulphuric acid 49.0 g dm-3 49/98 = 0.5 49.0

0.5 x 100/1000= 0.05

0.05 x 10000.5 0.05 x 98

50 cm3 hydrochloric acid 0.2 mol dm-3

2 mole copper(II) sulphate 1.5 mole

dm-3

50 cm3 aluminium chloride 0.5 mol dm-3

150 cm3 sodium hydroxide 4.0 g dm-3

1.5 mole ammonium hydroxide 1.0 mole

dm-3

250 cm3 potassium hydrogen sulphate

1.0 mol dm-3

25 cm3 copper nitrate 0.025 mole dm-3

*T.7.3.2


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(n = MV/1000)

Exercise 1

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Exercise 2. Complete the table below.[RAM.: H,1; S,32; O,16; Cl 35.5; Cu,64; Al,27; Na,23; N,14; K,39; Pb,207]

No.Amount and molarity of chemicals.

FormulaNo. of moles (n=MV/1000)

MassFormula

of ionNo. mol of ion

Molarity ion

1100 cm3 sulphuric acid 2.0 mol dm-3

H2SO4

2 x 100 1000= 0.2 mol

0.2 x 98H+ 2 x 0.2

= 0.44.0

SO42- 1 x 0.2

= 0.22.0

250 cm3

hydrochloric acid 0.2 mol dm-3

H+

325 cm3 copper(II) sulphate solution

1.5 mol dm-3

1.5 x 25 1000= SO4

2-

450 cm3 aluminium chloride 0.5 mol

dm-3 1.5

5150 cm3 sodium

hydroxide 0.1 mol dm-3

Na+

6100 cm3 ammonium hydroxide 1.0 mol

dm-3

7250 cm3 potassium hydrogen sulphate

1.0 mol dm-3

825 cm3 copper(II) nitrate 0.025 mol

dm-3


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T. 7.4

7.1 Definition of acid and alkali


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NEUTRALISATION

DefinitionExample

Experiment:Titration method

1. The reaction between acid and base to form salt and water.

2. The reaction between hydrogen ion (H+) from acid with hydroxide ion (OH-) from alkali to form water.

1. The reaction between hydrochloric acid and sodium hydroxide.

Equation: HCl + NaOH NaCl + H2OIonic equation:

H+ + OH- H2O

2. Reaction between sulphuric acid and calcium hydroxide.

Neutralisation between sulphuric acid and sodium hydroxide

Calculation using formula:

Va Ma = n a

Vb Mb n b

Application of neutralization in every day life

To build ion equation

H2O

H2O

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a. Acid : Chemical compound that DISSOLVES IN WATER

to produce HYDROGEN IONS.

HCl H + (aq) + Cl-(aq)

H2SO4 2H+(aq) + SO42-(aq)

CH3 COOH CH3 COO- (aq) + H+ (aq)

b. Alkali : Chemical compound that DISSOLVES IN WATER to produce HYDROXIDE IONS.

KOH (s) K+ (aq) + OH- (aq)

NaOH (s) Na+ (aq) + OH- (aq)

7.2 Properties of acid and alkali.

a. Acid - Tastes sour and turns moist blue litmus to red.

Chemical properties

(i) Reaction with metals that are more electropositive than hydrogen in the electrochemical series produces salt and hydrogen gas :

Acid + Metal Salt + Hydrogen

Example: Zn + 2HCl Zn Cl2 +H2

Mg + 2HNO3 Mg(NO3)2+ H2

Observation :………………………………………………………… Gas test : …………………………………………………………

(ii) Reaction with metallic carbonates produces salt, water and carbon dioxide


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Acid + Metallic Carbonate Salt + Water + Carbon dioxide

Example: CuCO3 + 2HNO3 Cu(NO3)2 + H2O + CO2

Observation :…………………………………………………………

Gas test : ……………………………………………………………..

(iii) Reaction with bases produces salt and water

Acid + Base/alkali Salt + Water

Example: ZnO + H2SO4ZnSO4 +H2O

b. Alkali - Tastes bitter,slippery and turns moist red litmus to blue.

Chemical properties

(i) Reaction with acids and alkalis / bases produces salt and water.

Example : NaOH + HCl NaCl + H2O

(ii) Reaction with ammonium salts produces ammonia gas.

Example: NH4+ + OH- NH3(g) + H2O

Activity 1

A) WRITE CHEMICAL FORMULAE FOR THE FOLLOWING COMPOUNDS:


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No Name of compound Chemical formula

No Name of compound Chemical formula

1 hydrochloric acid 16 magnesium oxide

2 nitric acid 17 calcium oxide

3 sulphuric acid 18 copper (II) oxide

4 sodium oxide 19 lead (II) oxide

5 Potassium hydroxide 20 sodium nitrate

6 Calcium hydroxide 21 Potassium sulphate

7 Barium hydroxide 22 sodium carbonate

8 Magnesium hydroxide 23 sodium chloride

9 Ammonium hydroxide 24 magnesium

10 hydroxonium ion 25 Zinc

11 hydroxide ion 26 sodium

12 hydroxyl ion 27 Carbon dioxide

13 sodium carbonate 28 Hydrogen gas

14 calcium carbonate 29 water

15 copper (II) carbonate 30 Magnesium nitrate

B) WRITE THE CHEMICAL EQUATIONS FOR THE FOLLOWING REACTIONS :


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1: Hydrochloric acid and zinc oxide :

2: Hydrochloric acid and sodium hydroxide:

3: Hydrochloric acid and magnesium ribbon:

4: Hydrochloric acid and sodium carbonate:

5: Sulphuric acid and zinc powder:

6: Sulphuric acid and zinc oxide:

7: Sulphuric acid and zinc carbonate:

8: Nitric acid and copper(II)oxide:

9: Nitric acid and sodium hydroxide:

10: Nitric acid and aluminium powder:

11. Nitric acid and copper(II)carbonate:

12 : Ethanoic acid and magnesium ribbon:

13: Ethanoic acid and calcium carbonate:


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14: Ethanoic acid and sodium hydroxide:

15: Ethanoic acid and sodium oxide:

7.3 Role of water in determining the properties of acid and alkali:

a. Role of water in determining the properties of acids:i. Acids in the absence of water or in other solvents: - does not show acidic properties

- acids remain in the form of molecules - no free mobile hydrogen ions - no acidic properties

ii. In the presence of water :- Molecules in acids will ionise to produce hydrogen ions

HCl(aq) → H+ + Cl-

- The presence of hydrogen ions are the causes for the acidic properties shown.

b.c. Role of water in determining properties of alkali :


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Draw molecules of hydrochloric acid in the absence of water

Draw H+ and Cl- ions in water

H + H +

Cl- Cl-

H + H +

Cl- Cl-

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i. In the presence of water , an alkali dissolves and ionises to produce hydroxide ions.

KOH K+ (aq) + OH-

(aq)

ii. In the absence of water or in organic solvents, no free and mobile hydroxide ions are produced, so the alkaline properties are not shown.

7.4 pH Concept

a. pH is a scale of numbers to measure the degree of acidity and alkalinity of an aqueous solution based on the concentration of hydrogen ions, H+ .

b. pH scale: 0-14:

pH value : 1 2 3 4 5 6 7 8 9 10 11 12 13 14

More acidic Neutral More alkaline

7.4 Acid and alkali strength

a) The strength of acids and alkalis depends on the degree of ionisation or dissociation.

b) Strong acid : Acid which dissociates completely in water to produce hydrogen ions.

HCl → H + (aq) + Cl-(aq)

Weak acid : Acid which ionises / dissociates partially in water to produce hydrogen ions.

CH3 COOH CH3 COO- (aq) + H+ (aq)

Example:

4 HCl molecules in water 4 CH3 COOH molecules in water

Due to H+ concentration in hydrochloric acid is higher than in ethanoic acid, at the same concentration , the pH of the strong acid is lesser than the pH of the weak acid.


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H2O

CH3 COOHCH3 COOH H + H +

CH3 COO- CH3 COO-

All 4 molecules ionises to produce 4 Cl-

ions and 4 H+ ions

Only some CH3COOH molecules ionise while others ions

remain as molecules.

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c) Strong alkali : Alkali which ionises completely in water to produce hydroxide ions.

KOH (aq) K+ (aq) + OH- (aq)

Weak alkali : Alkali which ionises partially in water to produce hydroxide ions.

NH4OH(aq) NH4+ (aq) + OH-(aq)

At the same concentration, the pH value of a weak alkali is

_______________than a strong alkali because the concentration of its

hydroxide ions is _______________. This is because a weak alkali ionises

____________but a strong alkali ionises____________.

Classification into strong and weak acid (a) Fill in the empty spaces with the correct answer.

Strong acid Weak acid

symbol Name Symbol Name

1. HCl HCOOH Methanoic acid

2.

3.

7.5 Acid and alkali concentration

a. Concentration – The measurement of substance dissolved in an amount of solvent.

c. pH of substance solution depends on :

i. Degree of dissosiation (The strength of acid or alkali) ii. Molarity (concentration in mol dm-3 )

d. Preparation of standard solution :Standard solution is a solution of which its concentration is accurately

known.

e. Dilution of acid/alkali :

M1 V1= M2 V2 -M1 – Molarity before dilution


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-V1 – Volume before dilution-M2 – Molarity after dilution-V2 – Volume after dilution

e. g : Find the volume of 2.0 mol dm3 sulphuric acid, H 2SO4 needed

to prepare 100 cm3 of 1.0 mol dm3 sulphuric acidBefore dilution After dilutionM 1

(mol dm-

3)

V1

(cm3 )M2

( mol dm-3)V2

(cm3)

2.0 ? 1.0 100V1 = M 2V 2

M1

= 1 x 1002

=50 cm3

Try these yourself:Calculate the volume needed to prepare each of the following dilute solutions

a. 50 cm3 of 0.1 mol dm3 sodium hydroxide, NaOH solution from 2.0 moldm3 solution hydroxide ,NaOH solution

b. 100cm3 of 0.5 mol dm3 potassium manganate (VII) ,KMnO4 solution from 1.0 mol dm3 potassium manganate (VII) solution.


Before dilution After dilution

M1 V1 M2 V2

2 ? 0.1 50

Before dilution After dilutionM 1 V1 M2 V2

1 - 0.5 100

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c. 5.0 dm 3 of 1.0 mol dm3 nitric acid, HNO3 from18.0 mol dm -3 nitric acid

7. 7 Neutralisation :

a. Neutralisation is the reaction between an acid and a base to form salt and water.Example :

HCl + NaOH → NaCl + H2O

2HNO3 + MgO → Mg (NO3)2 + H2O

b. At end point, all H+ ions from the acid are neutralised by OH- ions from the alkali or vice-versa .

pH of the titration product is 7 (neutral).

c. Titration process- a method to determine the degree of neutralisation between acid and alkali .

d. Two ways to determine the end –point:i. Using an indicatorii. Using electric conductivity

e. Steps in determining the degree of neutralisation:

S1 Write the balanced equation.S2 Write the information from the question above the equation.S3 Write the information from the chemical equation below the equation

(information in relations to the number of moles of substances involved).S4 Change the information to mole. S5 Use the relationship between the number of moles of the substances in S3.

OR


Before dilution After dilution

M1

(mol dm-3 )V 1

(cm3)M2

(mol dm-3)V2

(cm3)18 1 5 dm3

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MaVa = a Ma = acid concentration , Mb = alkali concentration MbVb a = coefficient of acid in the balanced equation

b = coefficient of alkali in the balanced equation Preparation of standard solutions

M1 : the molarity od the solution before water is addedV1 : the volume of the solution before water is addedM2 : the molarity of the solution after water is addedV2 : the volume of the solution after water is added

Calculate the volume of a concentrated solution needed to prepare each of the following dilute solutions:

1. 50 cm 3 of 0.1 mol dm -3 sodium hydroxide, NaOH solution from 2.0 mol dm -3 sodium hydroxide, NaOH solution.

2. 100 cm 3 of 0.5 mol dm -3 potassium manganate(VII), KMnO4 solution from 1.0 mol dm -3 potassium manganate(VII), KMnO4 solution.

3 5 .0 dm -3 of 1.0 mol dm -3 nitric acid, HNO3 from 18.0 mol dm -3 nitric acid, HNO3 .

Concentration of acid and alkali ;

Definition !!


1.Concentration – Quantity of solute in a given volume of solution usually in dm3

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M 1 x V 1 = M 2 x V 2

Concentration = mass(g)(g / dm3 ) volume (dm3)

÷ molar mass X molar mass

Concentration = no. of moles(mol)(mol / dm3 ) volume (dm3 )

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2. Unit Molarity (mol dm3 ) @ M

Answer all questions:

1. Write a balance equation for the neutralization of each of the following.

i. Nitric acid, HNO3 and calcium hydroxide, Ca(OH)2 solution

ii. Sulphuric acid, H2SO4 and barium hydroxide, Ba(OH)2 solution

iii. Ethanoic acid, CH3COOH and potassium hydroxide, KOH solution.

iv. Phosphoric acid, H3PO4 and ammonia, NH3 solution

2. Calculate the molarity for 10 g dm -3 sodium hydroxide solution, NaOH.[ Relative atomic mass: H,1;O,16;Na,23]

3. Calculate the concentration in g dm -3 for 0.5 mol dm -3 sodium hydroxide solution, NaOH.[ Relative atomic mass: H,1;O,16;Na,23]

4. Calculate the molarity of solution that is produced when 2.8 g of pottasium hydroxide ,KOH is dissolved in 100 cm3 of water.[ Relative atomic mass: H,1;O,16;Na,23]


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5. Calculate the molarity of a pottasium hydroxide solution,KOH, if 200 cm 3 of 2 mol dm -3 potassium hydroxide solution is added to 200 cm 3 of water.

6. 60 cm 3 of sodium hydroxide solution, 0.5 mol dm -3 is diluted with 30 cm 3 of water. Calculate the molarity of the solution produced.

7. 50 cm 3 of hydrochloric acid, HCl, 2 mol dm -3, reacted with excess zinc powder. Calculate the volume of hydrogen gas released under room conditions.[ Molar volume: 24 dm 3 at room condition ]

8. Calculate the molarity of 25 cm 3 of sulphuric acid, H2SO4 , which reacted completely with 1.2g magnesium.[Relative atomic mass:Mg,24]

9. 3.25g of zinc, Zn, reacted completely with hydrochloric acid, HCl, with molarity of 0.5 mol dm -3. Calculate the volume of acid used.[ Relative atomic mass: Zn,65]

10. 50 cm 3 of sulphuric acid, 0.5 mol dm -3 , reacts completely with 25 cm 3 sodium hydroxide solution, NaOH. Calculate the molarity of sodium hydroxide solution.


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11. If 25 cm 3 of hydrochloric acid, HCl, 0.05 mol dm-3, is neutralized completely by calcium hydroxide solution, Ca(OH)2 , 0.1 mol dm-3 , what is the volume of calcium hydroxide solution, Ca(OH)2, used ?

TRY THIS YOURSELF !!

1. The equation shows the reaction between sulphuric acid and sodium hydroxide.Persamaan menunjukkan tindak balas antara asid sulfurik dan natrium hidroksida.

H 2SO4 + 2NaOH Na 2SO4 + 2H 2O

What is the volume of 1.0 mol dm-3 sodium hydroxide solution which can be neutralise 25.0 cm 3 of 1.0 mol dm -3 sulphuric acid ? Berapakah isipadu larutan natrium hidroksida 1.0 mol dm -3 yang boleh meneutralkan 25.0 cm 3 asid sulfurik 1.0 mol dm -3 .

A. 12.5 cm3

B. 25.0 cm3

C. 50.0 cm3

D. 75.0 cm3

2. Which of the following is true about an alkali?Antara pernyatan berikut yang manakah benar tentang alkali?

A. An alkali is not corrosiveB. An alkali is a base that is soluble in waterC. A strong alkali has a low pH valueD. A weak alkali has a high degree of ionisation.

3. Which of the following substances is acidic?Antara bahan berikut yang manakah bersifat asid?

A. AmmoniaB. Potassium oxideC. Carbon dioxideD. Sodium hydroxide


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4. A dibase acid ,H 2J has the concentration of 0.5 mol dm -3 . Letter J is not the actual symbol of the element . What is the volume of potassium hydroxide,KOH 1.0 mol dm-3 that can neutralise 25.0 cm-3 of the H 2J acid solution ? Satu dwibes ,H 2J mempunyai kepekatan 0.5 mol dm-3

Huruf J bukan symbol sebenar unsure itu. Berapakah 1.0 mol dm-3 yang dapat meneutralkan 25.0 cm3 larutan asid H2 J itu?

A. 6.25 cm3

B. 12.50 cm3

C. 25.00 cm3

D. 50.00 cm3

5. Which of the following statements is true about all bases?Antara pernyataan berikut , yang manakah benar tentang semua bes?

A. React with acidsB. Dissolve in waterC. Contain hydroxide ionsD. Have alkaline properties

6. A student is stung by an insect with an alkaline sting.Which of the following substances is the most suitable to be applied to the part stung to treat the student?

Seorang murid disengat oleh serangga yang mempunyai sengatan yang beralkali. Antara bahan berikut,apakah yang paling sesuai disapukan pada tempat yang disengat untuk merawat murid itu?

A. VinegarB. EtanolC. Ubat gigiD. Cooking oil

STRUCTURAL QUESTION

SPM 20061. (a) 8g of solid sodium hydroxide, NaOH, is dissolved in distilled

water to produce a solution of 1000 cm 3.The NaOH solution produced has the concentration of 8.0 g dm -3 and molarity of 0.2 mol dm -3.

(i) State the meaning of the concentration for the solution produced.


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[ 1 mark ]

(ii) State the meaning of the molarity for the solution produced.

[ 1 mark ]

(iii) Write the formula that represents the relationship between the number of mole (n), molarity (M) and volume (V) for the solution.

[ 1 mark ]

(iv) Substitute the actual values of the number of mole, molarity and volume of the NaOH solution into the formula in 4(a)(iii).[Relative molecular mass of NaOH = 40 ]

[ 1 mark ]

(b) Diagram 4.1 shows the preparation of the standard solution of NaOH, 0.2 mol dm -3 .


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DIAGRAM 4.1

(i) What are the two parameters that shoulb be measured accurately to prepare the standard solution.

Parameter I _________________________

Parameter II _________________________ [ 2 marks]


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(ii) After all the NaOH solution is poured into the volumetric flask, the beaker and the filter must be rinsed several times with distilled water.After each rinse, all of this water is transferred into the volumetric flask.Give one reason for doing this.

[ 1 mark ]

(iii) What step should be taken to ensure that the meniscus level of the standard solution is exactly in line with the graduation mark on the volumetric flask ?

[ 1 mark ]

(iv) A volumetric flask is more suitable to be used in the preparation of the standard solution rather than a beaker.Why ?

[ 1 mark ](v) Why is the volumetric flask stoppered after the standard solution is

prepared ?

[ 1 mark ]2. Figure 1 shows the apparatus arrangement of an experiment in the laboratory to

determine the end point of the reaction between nitric acid, 2 mol dm-3 with 25 cm3 of potassium hydroxide solution.

Figure 1


25

Nitric acid

Potassium hydroxide solution+ phenolphthalein indicator

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The results of the experiment are shown in Table 2

Volume of nitric acid (cm 3) ExperimentFinal reading (cm3 ) 12.60 25.05 37.50Initial reading (cm3) 0.00 12.60 25.05Volume of nitric acid used (cm3)

Table 2

a. Nitric acid is a monoprotic acid. What is meant by monoprotic acid ?

[ 1 mark ]b. Name the type of reaction occurred in this experiment.

[1 mark ]

c. Write the balance chemical equation for the reaction between nitric acid and potassium hydroxide.

[ 3 marks ]

d. What is the colour of the potassium hydroxide solution when added with a few drops of phenolphthalein?

[ 1 mark ]

e. State the change on the colour of the solution in potassium hydroxide solution when the end point is reached.

[ 1 mark ]

f. (i) Complete the Table 2. [ 1 mark ](ii) Calculate the average volume of nitric acid used.

[ 1 mark ](iii) Calculate the concentration of the potassium hydroxide solution.


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[ 1 mark ]


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7 acid and bases

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