- Calculate the mass of NH 3 in the Ni(NH 3 )nCl 2 sample.
..
.
mol NH
mol NH
gNH
1 gNH
0 00329 17 04
1 3 0 0561^3
3
3
# =
- Calculate the mass percent of NH 3 in Ni(NH 3 )nCl 2.
n
%.%
.
.
gNi NH Cl
gNH
100 43 2
0 130
0 0561
32
==^3 #
^h
Mass Percent of Ni2+in Ni(NH 3 )nCl 2
Recap: 50. mL samples of known concentration of [Ni(NH 3 )n]2+were made up by
using NiSO 4 ⋅6 H 2 O. The concentrations were then plotted versus their absorbance.
The unknown solution’s absorbance was then extrapolated from this graph and used
in the calculation for the mass percent of Ni2+.
7. Calculate the number of moles of NiSO 4 ⋅6 H 2 O added to form the standard solution.
.
.
g NiSO H O
g NiSO H O
mol NiSO H O
1
0 3000
262 88
(^61)
6
(^426)
42
: # 42
:
: = 0.0041 mol of NiSO
4 ⋅6 H 2 O
- Calculate the number of moles of Ni(NH 3 )nCl 2 in the standard solution.
. mol of NiSO H O
mol of NiSO H O
mol Ni NH Cl
1
0 001141
1
6 1
6
(^42) n
42
32
: |
---|
^h |
= 0.001141 mol Ni(NH 3 )nCl 2 |
- Calculate the concentration of Ni(NH 3 )nCl 2 in the standard solution.
.
..
mL.
mol Ni NH Cl
L
mL
50 0 M
0 001141
1
1000
n 0 0228
32
# =
^h
- Calculate the absorbance of the standard (known) and unknown solutions of
[Ni(NH 3 )n]2+from their measured transmittance.
absorbance of standard solution = 2.000 – log (32.6) = 0.487
absorbance of unknown solution of Ni(NH 3 )nCl 2 = 2.000 – log (22.4) = 0.650 - Calculate the concentration of the unknown solution of Ni(NH 3 )nCl 2 , given that
c
A b
c
ε A
s
s
syn
==syn
where As= absorbance of [Ni(NH 3 )n]2+from standard solution, Asyn= absorbance of
[Ni(NH 3 )n]2+from synthesized Ni(NH 3 )nCl 2 , cs= concentrations of standard solution
and csyn= concentrations of synthesized solution.
..
L..
mol M
1
0 650 0 0228
0 487
##^1 =0 0304
- Calculate the concentration of the Ni2+(aq)ion in the unknown solution.
.
L.
mol Ni NH Cl
mol Ni NH Cl
0 0304 mol Ni M
1
n (^1) 0 0304
n
32
32
2
=
^ +
^
h
h
- Calculate the mass of Ni2+(aq)ion in 50.0 mL of the unknown solution.
.. .'
L
mol Ni
mol Ni
gNi mL sol n
mL
0 0304 L
1
58 69
1
50 0
1000
(^21)
2
2
###= 0.0892 g Ni2+
Part III: AP Chemistry Laboratory Experiments