Home Technical Mathematics Languages Science Social Science Art Literature Arabic Islamic Studies C.R.K HistoryMusicVisual Art Clothing/Textile Home Management Shorthand
 Chemistry Paper 1 (Practical) ,May/June 2008
 Questions: 1 2 3 4 5 6 Main
Weakness/Remedies
Strength

Question 4

D is  0.095 mol dm-3 HCI

E is a solution prepared by diluting 10 cm3 of a saturated solution of  NaHCO3 at 25° C to 1000 cm3 with distilled water.

(a) Put D into the burette and titrate it against 20.00 cm3 or 25.00 cm3 portions of E using methyl orange as indicator.  Repeat the titration to obtain consistent titres.  Tabulate your burette readings and calculate the average volume of D used.

The equation for the reaction involved in the titration is

NaHCO3(aq) + HC1(aq) → NaC1(aq) + CO2(g) +H2 O(l)    [10 marks]

(b) From your results and the information provided above, calculate the

(i)  concentration of E in  mol dm-3;
(ii) solubility of  NaHCO3. [5 marks]

(c)  State (i)    the effect of the gas released during the reaction on damp litmus   papers.
(ii)   one chemical test to confirm the gas released.  [4marks]

(d) (i)  Name one chemical industry that uses this gas as one of its raw materials.
(ii) List two other processes that can release this gas into the atmosphere.  [3 marks]

_____________________________________________________________________________________________________
OBSERVATION

The question was attempted by many candidates and the performance was good.

In (a), candidates were able to correctly carry out the titration experiment, tabulate the burette readings and calculate the average volume of acid used.  However, some of them lost marks because they lacked the knowledge of concordancy in precision measurement and cancelled out their tables to retabulate in order to conform with the supervisor’s titre values.

In (b)(i), most candidates calculated correctly the concentration of E in moldm-3 as follows:

(i)   CDVD     =     1
CE  VE           1

CE     =     CDVD
VE

=     0.095 xVD
VE

=     P moldm-3  say

OR
1000 cm3  of D  contain 0.095 moles of HCI
\ VD  of D will contain 0.095 x VD  moles of HCI
1000

From the equation,

1mole of HCI  reacts with 1 mole of NaHCO3

∴   0.095 x VD   mole  HCI  will react with    0.095 x VD   moles of   NaHCO3
1000                                                             1000

Hence,

20/25 cm3 of E contain  0.095 x VD  moles of NaHCO3
1000

∴     1000 cm3 of E will contain  0. 095 x VD  X   1000
1000               20/25
=      0. 095 x VD  =   P moldm-3  say
20/25
In (b)(ii), only few candidates correctly calculated the solubility of  NaHCO3 thus:

The dilution factor of solution E is 10:1000 = 1:100
Solubility of   NaHCO3   =   100 x  P moldm-3   =    Q moldm-3
OR
CfVf  =   C1V1

C1     =    CfVf
V1
=    P x 1000
10
=   100 x P  = Q Say

In (c)(i), candidates correctly stated that damp blue litmus paper would turn red/claret/pink or there is no effect on damp red litmus paper.

In (c)(ii) candidates stated that the gas when passed into lime water turns it milky.

In (d), candidates were able to correctly name one chemical industry that uses the gas and listed two other processes that release the gas into the atmosphere to include:

(i)   manufacture of Na2CO3/solvay process/food/drinks industry (aerated drinks, soda water/beer/dry ice)

(ii)  -    respiration
-    burning of fossil fuel
-    decay of organic matter
-    volcanic eruption
-    fermentation of sugars etc.