Analysis:
- Write the net ionic equations involved in this experiment.
44 OCl-]]]]aqgggg++S O 232 - aq 2 OH--aq " 2 SO 42 - ()aq++Cl aq H O (^2) ]lg
- Show the steps necessary to produce the 0.5 M sodium hypochlorite solution.
The commercial bleach was 5.20% NaClO. The directions required making a 0.500 M
solution. MM of NaClO = 74.44 g ⋅mol–1. 300. mL of solution should contain 11.2 g
NaClO:
.'
''
'mL sol n..
mL sol nL sol n
L sol nmole NaClO
mole NaClOg NaClO
1300
10001
10 500
174 44
###= 11.2 g NaClOTo obtain 11.2 g of NaClO, using a 5.20% solution, would require
215 g of bleach solution, diluted to 300. mL with distilled water:
.
..'
'.g NaClO
g NaClOg sol n
1 g bleach sol n diluted to mL11 2
520100
# = 215 300- Show the steps necessary to produce the 300. mL of 0.500 M sodium thiosulfite in 0.200
M NaOH.
(a) 0.200 M NaOH
.'
'
'
'mL sol n..
mL sol nL sol n
L sol nmol NaOH
mol NaOHg NaOH
1300
10001
10 200
140 00
###= 2.40 g NaOH(b) 0.500 M Na 2 S 2 O 3
.'
''
'mL sol n..
mL sol nL sol n
L sol nmol Na S O
mol Na SOgNa S O
1300
10001
10 500
122 3158 12 2
23
###^23= 23.7 g Na 2 SO 3To make the solution, add 2.40 g of NaOH and 23.7 g Na 2 S 2 O 3 to a graduated cylin-
der. Dilute to 300. mL with distilled water.- Why must the volume of solution produced remain constant?
The amount of heat evolved should be proportional to the mole ratios of the reactants in-
volved. If varying volumes of reactants were allowed, calculations would need to be per-
formed for each mixture so that a valid comparison could be made. - Referring to the graph produced, what is the limiting reactant on the line with the positive
slope? With the negative slope?
Positive Slope: The mL and mole ratio for equivalency is 4 NaClO to 1 Na 2 S 2 O 3 (see #1
above). In examining the data table, it can be seen that the Na 2 S 2 O 3 is in excess, with the
NaClO being the limiting reactant; for example, in Trial 7, for 30 mL of 0.500 M NaClO,
Part III: AP Chemistry Laboratory Experiments