Tuesday 18 September 2012

MODULE ANSWER


MODUL ANSWER

Preparation of a solution by dilution (pg 128)

1(a) Standard solution is a solution in which its concentration is accurately known.

(b)        -molar mass of NaOH = 23+16+1 = 40 g mol-1
              - mol NaOH              = 500 × 1.0 /1000 = 0.5 mol
              - Mass of NaOH        = 0.5 mol × 40 g mol-1= 20.0 g

Preparation of 500 cm3  1.0 mol dm-3 sodium hydroxide
-          20.0 g                          weighing bottle
-          20.0 g                          distilled water
-          Volumetric flask
-          Rinse           distilled water   volumetric flask
-          Distilled water         volumetric flask                                calibration mark
-          Volumetric flask     stopper                                inverted


Calculate the volume of 1 mol dm-3 sodium hydroxide used:
-          M2  × V2      = 0.1 × 250 = 25 cm3
M1                   1


Preparation of 250 cm3 1.0 mol dm-3 sodium hydroxide
-          25 cm3
-          25 cm3         volumetric flask
-          Distilled water         volumetric flask                                calibration mark
-          Volumetric flask     stopper                                inverted


1(a) (i) weak acid       : An acid that partially ionizes in water to produce low concentration of hydrogen  ion, H+.

Strong acid                  : An acid that completely ionizes in water to produce high concentration of hydrogen ion, H+.

(ii)
-          higher
-          stong acid                  completely                         higher
-          weak acid                  partially
-          CH3COO-  +   H+

(iii)
-          higher                         lower
-          lower                          higher


b)
  -      molecules                  ionise                    molecules           molecules           neutral
   Hydrogen
-          ionises partially       ethanoate           hydrogen

2a(i) solution P
     (ii) solution U

b(i)         Q
(ii)         R
(iii)        T
(iv)       P
(v)        U
(vi)       S

c(i) P/Q/R and T/U
  (ii) P/Q

3.       80 g dm-3
4.       0.5 mol dm-3
5.       2 mol dm-3
6.       10 g
7.       0.4 mol dm-3
8.       50 cm3
9.       a)
 Number of mol of sulphuric acid
 = 50 × 1
      1000
 = 0.05 mol

H2SO4 → 2H+ + SO42-

From the equation,
1 mol of H2SO4: 2 mol of H+
0.05 mol of H2SO4 : 0.1 mol of H+

Number of mol of hydrocloric acid
 = 50 × 1
      1000
 = 0.05 mol

HCl → H+ + Cl-

From the equation,
1 mol of HCl: 1 mol of H+
0.05 mol of HCl : 0.05 mol of H+
The number of H+  in 50 cm3 of 1 mol dm-3 of sulphuric acid is twice of the number of in 50 cm3 of 1 mol dm-3 of hydrochloric acid.

Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid.
1 mol of sulphuric acid ionises to 2 mol of H+  whereas 2 mol of hydrochloric acid ionizes to 1 mol of H+ . The number of H+  in the same volume and same concentration is doubled in sulphuric acid compared to hydrochloric acid.


b) 100 cm3

Friday 7 September 2012

preparation standard solution


Standard solution


  •  A standard solution is a solution containing a precisely known concentration.




Methods of Preparing Salts


Methods of Preparing Salts
1.   The method used to prepare a salt will depend on its solubility in water, that  is, whether it is soluble or insoluble in water.




Q1.     Suggest suitable chemicals required to prepare the following salts and write down the chemical equations involved.
a.       Potassium nitrate
b.       Copper(II) sulphate
c.       Aluminium nitrate
d.       Iron(II) chloride
e.       Ammonium nitrate                       

PEKA 2 (TITRATION)


Topic :  Acids and Bases
                                                                             

Aim                           :To find the end point in the titration of hydrochloric acid and sodium hydroxide solution using an acid-base indicator

                                  
           
Materials                   :0.1 mol dm-3 hydrochloric acid, 0.1 mol dm-3 sodium hydroxide solution, phenolphthalein

Apparatus                :Burette, 25 cm3 pipette, pipette filler, retort stand and clamp, white tile, 250 cm3 conical flask

                                  

Procedure                 :                (Must draw the diagram and list out the procedure)

                                
Result                        :   
Titration set
Estimation
1
2
3
Final burette reading/ cm3





Initial burette reading/ cm3





Volume of hydrochloric acid needed/ cm3












 Discussion                :

Able to write the chemical equations for the reactions correctly    

Able to show the calculation

                                      Able to state a observation and inference  correctly
                                     
                                      Able to state precaution


Conclusion                :