ACID-BASE REACTION: Calculation

Week:                                                                   Date:                                                     Time:

Period:                                 Duration: 1 HR 20 MIN.                                                  Average age of learners: 16YEARS

Subject:                               CHEMISTRY                                                                         Class: SS TWO

Topic:                                    ACID – BASE REACTIONS

Sub topic:  CALCULATION IN VOLUMETRIC ANALYSIS

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: beaker, beam balance

Entry behavior: The students have been density.

Behavioural objective: At the end of the lesson the students should be able to:

1.       Prepared a standard solution

2.       Record a reading from titration.

3.       Solve problems related to titration

CONTENT

CALCULATION IN VOLUMETRIC ANALYSIS

Calculation in volumetric analysis involve:

i.                     Balanced  equation of the reaction.

ii.                   Volume of solution used and

iii.                  Concentration of the standard solution

Amount(mol)  = molar concentration (moldm-3) x volume (dm3)

                    n =  c x v

An expression in term of concentration of A and B is  =  = mole ratio

PREPARATION OF STANDARD SOLUTION

A standard solution of a substance can be prepared, if the substance can be obtained in pure state, by weighing accurately a definite mass of the substance, dissolving it in a suitable solvent, usually water, and making up the solution to a known volume in a volumetric flask.

When a substance is not available in the pure form a solution of approximate concentration is first prepared, and then standardized (i.e, its actual concentration determined) by titrating against a standard solution of a pure substance. Standard solution of common acids and alkalis cannot be prepared by direct weighing.

Reasons: Some of them are volatile and corrosive liquids, for example HCl, HNO3 and NH3; while H2SO4 is hygroscopic. Other are solids that are deliquescent and hence absorb moisture and carbon(iv) oxide, from the atmosphere, e.g KOH and NaOH pellets.

A substance which can be obtained in a high state of purity, anhydrous, non-deliquescent, non-hygroscopic, readily soluble in water or in the required solvent, and hence suitable for the preparation of a standard solution is called a primary standard. Example are anhydrous sodium trioxocarbonate (iv), Na2CO3, and benzoic acid, C6H5COOH. A substance which is hydrated, but does not efflorescent can be used in the preparation of a standard solution. Such a substance is also a primary standard. An example is the dibasic organic acid called hydrated ethanedioc acid (H2C2O4), oxalic acid (H2C2O4. 2H2O).

HOW TO CALCULATE VOLUME OF SOLUTION TO BE DILUTED

Concentrated acid or alkalis are always diluted with the appropriate solvent before being used for titration or otherwise. The volume , Vo, of such a commercial product required to prepare a dilute solution of volume Vd and of a molar concentration Cd is given by  Vo =  

Where  and d is the % by mass and density respectively.

Example

Calculate the molar concentration of 20.2cm3 of conc. H2SO4 of specific gravity 1.80gcm-3 and 98% of acid (H2SO4 = 98).

Solution

Vd = 1dm3, Vo = 20.2cm3; d = 1.80gcm-3 ; P = 98%; Cd =?

Vo =      ; =   =   =   = 3.636 moldm-3

 

 

 

RECORDING IN TITRATION

Titration work could be recorded thus:

1.       State the size of pipette used in cm3.

2.       Name the indicator used

3.       Record your titration in tabular form as shown below:

Burette reading

Rough or Trial (cm3)

1st titre (cm3)

2nd titre (cm3)

3rd titre (cm3)

Final reading

 

 

 

 

Initial reading

 

 

 

 

Volume of acid used

 

 

 

 

Find the average volume of acid used from any two or more titre values that do not differ by more than 0.20 cm3 (Rough titre value may be used in averaging as long as the difference of any two titre values is within 0.20cm3).

Example

1.A solution of trioxonitrate(v) acid contained 0.67g in 100cm3. 31.0cm3 of this solution neutralized 25cm3 of a sodium trioxocarbonate (iv) solution. Calculate the concentration of the  trioxocarbonate (iv) solution (HNO3 =63, Na2CO3 = 106).

Solution

100cm3 dissolves 0.67g of HNO3

1000cm3  will dissolves   of HNO3 i.e 6.7g

Molar conc.  of  A =   =  = 0.106moldm3

Equation of reaction

2HNO3 +  Na2CO3               2NaNO3 + CO2  + H2O

  2         :      1                              2         :   1      :  1

CA = 0.106M , VA = 31.0cm3  , na = 2 ; CB = ? , VB = 25.0cm3  , nb = 1

Using   =      ;  =

=  =   = 0.0657 moldm-3

2.A is a solution of an impure H2SO4 containing 16g in 1dm3 of solution. B is a solution of pure NaOH containing 1.5g in 250cm3. If 25.0cm3 portion of B required an average of 15.50cm3 for complete neutralization. Calculate the:

(a) Conc. of solution B in moldm-3    (b) Conc. of A in moldm-3      (c) Conc. of the pure acid in solution A in gdm-3

(d) number of H+ in 1.0dm3 of solution A  (NA = 6.02x1023 (e) % of impurity in solution A.

Solution

(a)250cm3 dissolves 1.5g of NaOH

1000cm3  will dissolves   of NaOH i.e 6.0g

Molar mass of NaOH = Na + O + H = 23 + 16+1 = 40gmol-1

Molar conc.  of NaOH in  B =   =  = 0.150moldm3

(b)Equation of reaction

2NaOH +  H2SO4               Na2SO4 +  2H2O

  2         :      1                              1         :   2     

CA = ? , VA = 15.50cm3  , na = 1 ; CB = 0.150M, VB = 25.0cm3  , nb = 2

Using   =      ;  =

=  =   = 0.121 moldm-3

(c ) molar mass of H2SO4 = 2H + S + 4O = [(2x1) + 32 + (4x16)] = 2+32+64 = 98gmol-1

Mass conc. = molar conc. x molar mass

                       =  0.121 moldm-3 x  98gmol-1  = 11.9gdm-3

(d) H2SO4             2Haq+  +  SO4aq2-

        1                            2                1

1 mol of H2SO4 produces 2 mol of Haq+

0.121 mol of H2SO4 produces 0.242 mol of Haq+

1 mol of Haq+ contain  6.02x1023 ions

0.242 mol of Haq+ contain  6.02x1023 X 0.204 ions = 1.46x1023 ions

(e)mass conc. of impure acid in solution A =16gdm-3

mass conc. of pure acid in solution A =11.9gdm-3

mass conc. of the impurity  =16 – 11.9gdm-3 = 4.10

% impurity =  X 100

                     =  X 100 = 25.6%

% purity = 100 - % impurity

               = 100 – 25.6  = 74.4%

(3) A is a solution containing 6.00gdm-3 of HCl. B also contains 12.0gdm-3 of hydrated sodium trioxocarbonate (iv), Na2CO3.X H2O. A student titrated 25cm3 portion of B with A using methyl orange as the indicator, and found that the average volume of solution A required was 13.10cm3 from the above results.

(a) Write a balance equation of reaction  (b) Calculate (i) the conc. of anhydrous  in Na2CO3 solution B. (ii) the value of X, and hence the % of water of crystallization in the solution B.

Solution

(a). 2HCl + Na2CO3.X H2O              2NaCl + CO2 + X H2O + H2O

(b). mass conc. of HCl in A = 6.0gdm-3

Molar mass of HCl = H + Cl = 1 + 35.5 = 36.5gmol-1

Molar conc.  of HCl =   =  = 0.164moldm3

CA = 0.164M , VA = 13.10cm3  , na = 2 ; CB = ? , VB = 25.0cm3  , nb = 1

Using   =      ;  =

=  =   = 0.043 moldm-3

ii. mass conc of hydrated  Na2CO3 in sol. B = 12.0gdm-3

molar conc. of B = 0.043 moldm-3

molar mass of hydrated Na2CO3 =  =  = 279.1gmol-1

relative molar mass of Na2CO3.X H2O = 279.1

                                                106 + 18X  = 279.1

                                                                18X = 279.1 – 106

                                                                18X = 173.1

                                                                X =  = 9.61 = 10

% of water crystallization =

                                             = X 100  =   = 60.0%

PRESENTATION

i.The teacher explains how to prepared a standard solution.

ii. The teacher shows the students to tabulate there reading in titration

iv. The students draw the table in there notes

iii. The teacher guides the students in solving some problems in titration

EVALUATION

 The teacher evaluates the lessons by asking the following questions:-

1.       Explain how to prepared a standard solution

2.       What are the information require in recording a reading from titration.

3.       State two precautions taken each when using burette and conical flask

Iron reacts with H2SO4 according to the equation:Fe(s) + H2SO4( aq)                      →        FeSO4aq+H2(g). Calculate  the mass of FeSO4 that would be produced by 0.5mole of Fe (H=1, S=32 Fe=56).

ASSIGNMENT

1a.          State an indicator suitable for the titration and give a reason for your answer in each case:

(i)                  dilute HCl and NaOHaq.

(ii)                                      dilute CH3COOH and KOHaq.

      (iii)           dilute HCl and NH3aq.

 (b)         Calculate the volume of water that would be added to 50cm3 of 0.01 moldm-3 of HCl to dilute it to 0.01 moldm-3.

2a.          In a titration experiment 22.50cm3 of an acid solution A containing 10.6g of NaHSO4 per dm3 reacted with 25.0cm3 of solution B containing Xg of NaOH per dm3. The equation for the reaction is from the information given above, [H=100; 0=16.0, Na =23.0; S= 32.0]  calculate the: i. Concentration of A in moldm-3.

ii.Concentration of B in moldm-3.               iii. Value of x                iv. Mass of Na2 SO4 formed during the reaction.

bi.Name the suitable indicator for the titration experiment.

State the apparatus used to measure the volume of solution:     I.A     II.B 

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