In this experiment, you are to determine relative atomic mass of M.
Your are provided with two solutions:
BA1, 0.1M sulphuric acid
BA2, contains 5.55gl^- of M(OH)₂
Pipette 25cm3 of BA2 into a flask. Add 2 drops of phenolphalein indicator. Titrate it against BA1 from the burrete.
Volume of pipette used =25.0cm³

Table of results:

Volumes of BA1 used to calculate the average volume are 18.70cm3, 18.75cm3 and 18.75cm3, Average value of titre

(18.70+18.75+ 1875)cm³      = 18.73cm³

               3          

Questions:

1. (a) (i) Write the equation for the reaction between BA1 and BA2.
H₂S0₄(aq)+M(OH)₂ (aq)®MSO₄ (aq)+H₂0(l)
(ii) Calculate the moles of the acid. 0.1 mole of the acid are in 1000cm³
of solution = moles of acid in 18.73cm³ of solution is  = 0.00183 mole

 (iii) Determine moles of M(OH)2 that reacted. From the equation:
1 mole of acid reacts with 1 mole of M(OH)₂. i.e the mole ratio is 1:1
Therefore, moles of M(OH)2 in 25cm3 = 0.001813 mole

(b) (i) Calculate the morality of M(OH)2 0.001873 moles are in 25cm3 of solution
Number of moles in 1000cm³ of solution will be (0.001873x 1000) = 0.07492
25
Hence the molarity of M(OH)2 is 0.75M.
(ii) Find the molecular mass of M(OH)₂.
Molarity = Concentration in gl^-1

      Relative mass

So relative molecular mass = Concentration in gl^-1

                                    Molarity

                                        = 74

Hence the relative molecular mass of M(OH)₂ is 74.

(c) Determine the relative atomic mass of M

 

M(OH)₂ = 74

M+(16+1)x2 = 74

M+34 74

M = 74-34

M = 40.

 

Hence the relative atomic mass of M is 40.

 

BA1, 0.1M of the acid BA2, 0.0556M sodium hydroxide solution. Pipette 25cm³ of BA2 into a flask. Add 2 drops of methyl orange indicator. Titrate it against BA1 from the burette. Volume of pipette used  = 25.0cm³.

 

Table of results:

 

Burette readings	1	2	3	4
Final burette reading (cm3)	15.50	30.60	45.60	15.00
Initial burette reading (cm3)	00.00	15.50	30.60	00.00
Volume of BA1 used (cm3)	15.50	15.10	15.00	15.00

 

Volume of BA2 used to calculate the average volume are 15.10cm³, 15.00cm³, 15.00cm³. Average value of titre.

            =          15.10 + 15.00  +  15.00  cm³

                                           3

 

            =          15.03 cm³

 

Questions:

1.   (a)   Calculate the moles of the

            (i)    Sodium hydroxide 0.0556 moles of NaOH are in

       1000cm³ of solution will  be  0.0556 x 25

                                                                       1000   

                           = 0.00139 mole.

                (ii)    Acid

                        0.1 moles of the acid are in 100cm³  of solution,

moles of the acid in 15.03cm³  of solution is be 

0.1 x 15.03

                              1000                  = 0.00139 mole.

 

 

@

=  0.00139     1

    0.001505   1

            \mole ratio of reaction, alkali: Acid  =  1:1

 

      (c)  Determine the basicity of the acid. The basicity of the is 1.

 

3.         You are provided with the following (BA2, 0.1M sulphuric acid, BA1, solution of sodium hydroxide of unknown concentration). You are required to determine the concentration of sodium hydroxide. Pipette 25cm³ of BA1 into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate it against BA2. Volume of pipette used  =  25.0cm^3.

Burette readings	1	2	3
Final burette reading (cm3)	25.02	25.10	25.01
Initial burette reading (cm3)	00.00	00.00	00.00
Volume of BA1 used (cm3)	25.02	25.01	25.01

 

 

 

 

 

 

 

 

 

Titre values used to find the average volume of BA₂. 25.01cm³, 25.01cm³. Average volume of BA₂ 

=    25.01 + 25.01 cm³    = 25.01cm³

                2

1.   (a)   Write the equation for the reaction between BA₁ and BA₂.

            2NaOH(aq)  + H₂SO₄(aq)                Na₂SO₄ (aq)  + 2H₂O(1)

 

      (b)    Calculate the number of moles of

              (i)   BA₂  0.1  mole of H2₂SO₄ are in 1000cm³ of solution.

                   \ Moles of H₂SO₄ in 25.01cm³ of solution 

=  25.01 x 0.1     =  0.002501

      1000

Mole

 

              (ii)    BA1 mole ratio is 2BA₁:1BA₂ (alkali: acid)

Hence the number of moles of BA₁ = 2 x 0.002501mole      

  = 0.005 mole

 

(c)         Calculate the concentration of sodium hydroxide in

(i)   Moles per litre.

       0.005 mole of sodium hydroxide are in 25cm³ of solution

        Þ Moles of sodium hydroxide in 1000cm³ of solution 

              = 1000 x 0.005   = 0.2 mole

                         25

    Hence the concentration of sodium hydroxide is 0.2 molel^-1

 (ii)    grammes per litre (Na = 23,  O = 16,  H  = 1)

          Molar mass of NaOH

          =  23 + 16 + 1   =  40g.

          Concentration in grammes per litre

          =   0.02 x 40  = 8gl^-1.

 

         In this experiment you are to determine

(a)      Mole ratio of the reaction between an acid and alkali

(b)     Heat of neutralization of the reaction.

You are provided with

ü  BA₁ which is 1.6M sodium hydroxide

ü  BA₂ which is 1.6M hydrochloric acid

ü  Pipette 25cm3 of BA₁ into a conical flask

ü  Add 2-3 drops of methyl orange indicator.

ü  Titrate it against BA₂ from burette.

ü  Volume of pipette used  = 25.0cm³.

 

Table of results:

 

Burette readings	1	2	3	4
Final burette reading (cm3)	15.50	30.60	45.60	15.00
Initial burette reading (cm3)	00.00	15.50	30.60	00.00
Volume of BA1 used (cm3)	15.50	15.10	15.00	15.00

 

Titre values used to find the average volume are 25.00cm³, 25.00cm³.

Average value of titre  =   25.00 + 25.00   cm³   =  25cm³.

                                                      2

 

1.   (a)   Write the ionic equation of reaction between BA₁ and BA₂.

                OH^-(aq)  + H⁺(aq)                 H₂O(1)

 

    (b)   Calculate the number of moles of

(i)   Sodium hydroxide.

       1.6 mole of sodium hydroxide are in 1000cm³ of solution

        Þ Moles of sodium hydroxide in 25cm³ is  

              =  25 x 1.6    = 0.04 mole

                     1000

                (ii)    Hydroxide acid

                           1.6 mole of hydrochloric acid is in 1000cm³

 

  Þ Moles of hydrochloric acid in 25cm³ is  

                                         24 x 1.6

                                           1000        =   mole 0.04 mole.

              (c)     Find the mole ratio of reaction

                           Mole ratio of reaction

                             =    moles of acid   =  0.004  = 1

                                     Moles of alkali    0.004      1

                                Hence the mole ratio of reaction, acid: alkali  = 1:1

 

2.         Pipette volume of an acid that is equal to the average titre value you got into a plastic cup. Take the temperature of the acid. Pipette 25cm³ of sodium hydroxide into a beaker. Take the temperature of the alkali. Transfer the alkali solution to the acid solution in the cup. Record the temperature of the mixture.

 

Initial temperature of acid (0C)	20.0
Initial temperature of alkali (0C)	20.4
Temperature of the mixture (0C)	33.2

 

                (a)   Determine the

                        (i)   initial temperature of the reactants.

                                 (20.0 + 20.4)⁰C  = 20.2⁰C

                                                2

 

                       (ii)   temperature rise for the solution

                                  (33.2 - 20.2) ⁰C  = 13.0⁰C

 

                      (iii)   final volume of solution.

                                (..............) cm³  = 50cm³

 

                (b)   Calculate the heat of neutralization of the reaction (specific heat

        capacity of water  = 4.2lg^{-1 0}C-1, density of water  = lgcm^-3).

                     Heat evolved

                      = mc0

                        = 50 x 4.2 x 13

                        =  2730;

                      Mole of acid  = 0.04 mole

          From the ionic equation

          H⁺(aq) + OH^-(aq)             H₂O(1)

          Moles of water formed during neutralization  = 0.04 mole

          For 0.04 mole of water formed heat liberated is 2730J;

          Hence for 1 mole of water to be formed, heat liberated will be

              (1 x 2730);

                  0.04                 =   68250J  = 68.25J.

          Hence the heat of neutralization of the reaction is 68.25kJ.

          You are provide with BA₁ which is made by dissolving 28.6g

          of a metal carbonate.

 (M₂CO₃nH₂O) in water to make 1 litre of solution.

BA₂ which is 0.2M hydrochloric acid.

You are required to find the number of moles of water of crystallization in the metal carbonate.

Pipette 25cm³ of phenolphalein indicator. Titrate against BA₂.

Volume of pipette used 25.0cm³.

 

Table of results:

 

Burette readings	1	2	3
Final burette reading (cm3)	25.00	24.90	24.90
Initial burette reading (cm3)	00.00	00.00	00.00
Volume of BA1 used (cm3)	25.50	24.90	24.90

 

Titre values used to find the average volume 24.90cm³, 24.90cm³.

Average value of titre  =   24.9 + 24.9   cm³   =  24.90cm³.

                                                      2

 

1.  Calculate the number of moles of:

 

      (i)    acid 0.2 mole of HCI are in 1000cm³ of solution.

          \moles of HCI in 24.90cm³ of solution  =  0.2 x 24.9  = 0.00498 mole

           1000 

 

    (ii)   the metallic carbonate when the mole ratio of reaction is 1

carbonate:2 acid. If the mole ratio = 1:2, then the moles of the metallic carbonate  = 2 x 0.00498  = 0.00996 mole.

 

(b)   Calculate the concentration of the carbonate in moles per litre.

        0.00996 mole of the carbonate is in 25cm³ of solution.

     Þ moles of the carbonate in 100cm³ of solution  =   1000 x 0.00996

                                                                                                            25

     = 0.3984 mole. Hence the contra ration of the carbonate is 0.3983 mole l^-1.

 

(c)   Calculate the relative molecular mass of the metal carbonate.

        Molarity  =   Concentration in_gl^-1.

                                    R.M.M

     Þ R.M.M   =    28.6  @ = 72

                             0.3984 

     You are provided with

     BA1:0.1M HCI

     BA2:6gl^-1 of impure sodium carbonate.

     You are required to determine to determine the percentage of pure    

 

                Sodium carbonate.  Pipette 25cm³ of BA2 into a conical flask. Add

  2-3 drops of methyl orange. Titrate it against BA1 from burette.

 Volume of pipette used 25.0cm³. Table of results.

 

Burette readings	1	2	3
Final burette reading (cm3)	50.00	50.00	50.00
Initial burette reading (cm3)	21.50	21.70	21.70
Volume of BA1 used (cm3)	28.50	28.30	28.30

 

Titre values to find average volume 28.30cm³, 28.30cm³.

                Average value of titre  28.3 + 28.3  = 28.30cm³

                                                            2

(a)    (i)   Write equation for the reaction

               Na₂CO₃(aq) + 2HCI(aq)         2NaCI(aq)+ CO₂(g) + H₂O(1)

 

       (ii)  Calculate the number of moles of the acid.

              0.1 mole of the acid are in 1000cm³ of solution.

              Moles of the acid in 28.3cm³ of solution is

                                                         0.1 x 28.3  = 0.00283 mole

                                                                100

   (iii)     Determine the number of moles of the carbonate

              Mole ratio:2 acid: 1 alkali..

              Þ moles of the carbonate  =  0.00283  =  0.001415 mole.

                                                                                       2

          (b)         Calculate the concentration of the carbonate in

   (i)   Moles per litre

          0.001415 mole is in 25cm³ of solution

          Þ moles of the carbonate in 1000cm³ of solution

  =  0.001415 x 1000   = 0.0566 mole

                                25 

                \Concentration = 0.0566M

     (ii)   gl^-1

                  Molarity = gl^-1

                                 R.M.M.

                   gl^-1 = R.M.M x molarity

                  106 x 0.0566  = 5.9996_gI^-1