10 ml of sulphuric acid is dissolved in 90 ml of water then calculate the concentration of sulphuric acid
Answers
Answer:
Explanation:
The idea here is that you should expect the concentration of the acid to decrease as a result of the dilution because the amount of acid remains unchanged, but the volume of the solvent increases.
When you dilute a solution, the increase in volume that results from adding solvent must match the decrease in concentration. In other words, the concentration of the solution decreases by the same factor as its volume increases.
the dilution factor
DF
=
c
concentrated
c
diluted
=
V
diluted
V
concentrated
−−−−−−−−−−−−−−−−−−−−−−−−
Now, you know that after you add the acid to the water, and keep in mind that you must never add water to acid, the total volume of the solution will be
V
diluted
=
25.0 mL
+
500. mL
V
diluted
=
525 mL
This means that the volume of the solution decreased by a factor of
DF
=
525
mL
25.0
mL
=
21
You can thus say that the concentration of the solution must decrease by a factor of
21
.
Since
DF
=
% concentrated
% diluted
⇒
% diluted
=
% concentrated
DF
you will have
% diluted
=
36.0 %
21
=
1.71
%
−−−−−−
The answer is rounded to three sig figs.
You can double-check the result by using the solution's mass by volume percent concentration to figure out the mass of sulfuric acid present in the initial sample
25.0
mL solution
⋅
= 36.0% w/v
36.0 H H
2
SO
4
100
mL solution
=
9.00 g H
2
SO
4
After the dilution, the mass of sulfuric acid present in
'
100
m
L
of the diluted solution will be
100
mL solution
⋅
9.00 g H
2
SO
4
525
mL solution
=
1.7143 g H
2
SO
4
Therefore, you will once again have
% H
2
SO
4
=
1.71% w/v
−−−−−−−−−−−−−−−−−−−−
Answer:
Explanation:
The idea here is that you should expect the concentration of the acid to decrease as a result of the dilution because the amount of acid remains unchanged, but the volume of the solvent increases.
When you dilute a solution, the increase in volume that results from adding solvent must match the decrease in concentration. In other words, the concentration of the solution decreases by the same factor as its volume increases.
the dilution factor
DF
=
c
concentrated
c
diluted
=
V
diluted
V
concentrated
−−−−−−−−−−−−−−−−−−−−−−−−
Now, you know that after you add the acid to the water, and keep in mind that you must never add water to acid, the total volume of the solution will be
V
diluted
=
25.0 mL
+
500. mL
V
diluted
=
525 mL
This means that the volume of the solution decreased by a factor of
DF
=
525
mL
25.0
mL
=
21
You can thus say that the concentration of the solution must decrease by a factor of
21
.
Since
DF
=
% concentrated
% diluted
⇒
% diluted
=
% concentrated
DF
you will have
% diluted
=
36.0 %
21
=
1.71
%
−−−−−−
The answer is rounded to three sig figs.
You can double-check the result by using the solution's mass by volume percent concentration to figure out the mass of sulfuric acid present in the initial sample
25.0
mL solution
⋅
= 36.0% w/v
36.0 H H
2
SO
4
100
mL solution
=
9.00 g H
2
SO
4
After the dilution, the mass of sulfuric acid present in
'
100
m
L
of the diluted solution will be
100
mL solution
⋅
9.00 g H
2
SO
4
525
mL solution
=
1.7143 g H
2
SO
4
Therefore, you will once again have
% H
2
SO
4
=
1.71% w/v
−−−−−−−−−−−−−−−−−−−−