Question1) burette of 50cm
capacity pipette, either 20cm or
25 (using d usual apparatus for
titratione) Reagents for
qualitative work
(i) red and blue litmus paper;
(ii) aqueous ammonia
(iii) dilute hydrochloric acid
(iv)dilute sodium hydroxide
solution
(v)barium chloride solution
(vi)dilute trioxonitrate
(V) acid
(vii)silver trioxonitrate
(V) solution [AgNO3]
(viii)lime water methyl orange
indicator
Question 2) A. 150cm of
tetraoxosulphate
(VI) solution in a corked flask or
bottle,labelled 'A' containing
2.8cm of concentrated H2SO4
(about 98% w/w)per dm of
solution.
B. 150cm of NaOH solution, in a
corked flask or bottle labelled 'B'
containing 3.9g of NaOH per dm
of solution. C. 10cm of (NH4)
2SO4 solution in a bottle labelled
'C' containing 66g of (NH4)2SO4
per dm of solution.
D. 10cm of FeCl3 solution in a
bottle labelled D containing 40g
of FeCl3 per dm of solution.
ANSWER
> QUANTITATIVE ANALYSIS <
TEST
Solution C + aqueous NaOH +
Heat
-----------------------
Portion of Solution C + BaCl2 +
dil. HCl.
-------------
Portion of Solution D + NaOH in
drops. Then in excess.
----------
Portion of Solutio D + NH3(aq) in
drop then iq excess.
--------
Portion of Solution D + HNO3 +
AgNO3 solution.
---------
Mixture frm above + HNO3 in
excess.
------
OBSERVATIONS
Evolution of colourless gas with
characteristics choking smell
which turns moist red litmus
paper blue and form white
foam with conc. HCl acid vapour.
-----------------
white ppt insoluble in dilute HCl
--------
Reddish brown ppt insoluble
----------
Reddish brown ppt insoluble
--------
White ppt insoluble
--------
White ppt dissolved to give a
colourless solution.
-----------------
INFERENCE
NH3 gas from NH4+ is present.
-------
SO42-, SO32-, CO32- are present
--------
Fe3+ is present
-------
Fe3+ is present
--------
Cl- ion is present
------
Cl- ion is confirmed.
(N.B u write a test scroll to its
observation and then its
inference. 'cos it is to be written
in tabular form i.e make a 3
columns: then write
TEST| OBSERVATION|INFERENCE.)
,
====================
> VOLUMENTRIC ANALYSIS <
A solution acid H2X 2.45g in
500cm3 of solution. Solution B
contain 3.90g of NaOH in 1dm3
of solution. Put A into the
burrette and titrate with 20cm3
or 25cm3 portion of B using
methyl orange as indicator.
Record the value of your
pipette. Tabulate your readings
and calculate the average titre
value of acid used. From ur
results and information
provided. Cal. the:
i) Concentration of solution B in
mol/dm3
ii) Concentration of A in mol/
dm3
iii) Molar Mass of acid H2X and
hence determine the relative
mass of X.
H2X + 2NaOH===> Na2X + 2H2O.
Burrete reading (cm3)
Final
Initial
Volume
Rough | 1st | 2nd | 3rd
28.50 | 26.40 | 25.60 | 26.50
1.00 | 2.00 | 1.00 | 2.00
27.50 |24.40 |24.60 | 24.50
(N.B folow d order i.e final
reading, initial reading and
volume of A used).
Average Volume of Acid used =
(24.40+24.60+24.50) /3
= (73.50/3) =24.50cm3.
A=H2X B=NaOH
2.45g ==> 500cm3
xg ==>1000cm3
:. x= (2.45 *100)/500
=4.90g/dm3.
i.) Cb = mas concentration/molar
concentration
=(3.90/40)mol/dm3
Cb=0.0975mol/dm3.
ii.) CaVa/CbVb =na/nb ; Va
=24.50cm3, Cb=0.0975mol/dm3,
Vb=25cm3, na=1, nb=2.
(Ca*24.50)/(0.0975*25) Ca=
(0.0975*25)/(24.50*2). Ca
=2.4375/49.00
Ca=0.0497mol/dm3.
iii.) Ca = mass concentration /
molar concentration
0.0497=4.90/molar mass
molar mass =4.90/0.0497
molar mass=98.6g/mol
H2X =98.6
(1*2) + X =98.6
X= 98.6 -2
= 96.6
NOTE:-
1. Even if you don't Undestand
the Question and Answer, Don't
worry yourself, Just chill until
you see the
Question, Its gonna be easy
with the above Post.
2. Number 3 Question is not
Included because we can't get
it.. Try and Get it done yourself
in the Exam Hall.