Preparation of Solutions
189
moles of solute
molarity of solution = M = liter of solution
It is simple to prepare solutions of known molarity from solids and nonvolatile liquids that can be weighed on an analytical balance, and then dissolved
and diluted to a known volume in a volumetric flask. When a reagent-grade
sample is accurately weighed and diluted with care, as in the following problem, the resulting solution is said to be A standard solution.
PROBLEM:
Prepare 250.0 ml of a 0.1250 M AgNO 3 solution, using solid AgNO 3 .
SOLUTION:
From the given volume and concentration you can calculate how many grams of
AgNO 3 to weigh out:
wt. of AgN0 3 needed = (0.2500 liter)( 0 . 1250
m
°'^.
AgN °
3 ) (l69.9
\
liter
A
mole AgNO
= 5.309 g AgN0 3
Transfer the 5.309 g AgNO 3 to a 250.0 ml volumetric flask, dissolve it in some
distilled water, then dilute to the mark (see p 86). Shake vigorously to get a
uniform solution. Don't add 250.0 ml of water to the weighed sample, because the
resulting solution may actually be larger or smaller than 250.0 ml due to interaction
of solute and solvent.
Many crystals contain "water of crystallization," which must be included in
the weight of the material weighed out for preparing a solution. Allowance is
made for this by using the molecular weight of the hydrate in your calculations,
not the mole weight of the anhydrous form.
PROBLEM:
Prepare 100.0 ml of 0.2000 M CuSO 4 , starting with solid CuSO 4 • 5H 2 O.
SOLUTION:
From the given volume and concentration of CuSO 4 , you can calculate the moles
of CuSO 4 required. Furthermore, the formula shows that 1 mole of CuSO 4 -5H 2 O
is required per mole of CuSO 4 . Thus the weight (W) of CuSO 4 -5H 2 O needed is
W = (0.1000 liter) ( 0.2000
mote
,
g CuS °<) ( , ™'* CuSO -SH,O\ /
g CuSO 5H,O \
V
liter
/ V
mole CuSO 4 / V
mole CuSO 4 -5H 2 O/
= 4.992 g CuSO,-5H 2 0 needed
Transfer the 4.992 g to a 100 ml volumetric flask, dissolve it in some distilled
water, then dilute to the mark. The fact that some of the water in the solution
189
moles of solute
molarity of solution = M = liter of solution
It is simple to prepare solutions of known molarity from solids and nonvolatile liquids that can be weighed on an analytical balance, and then dissolved
and diluted to a known volume in a volumetric flask. When a reagent-grade
sample is accurately weighed and diluted with care, as in the following problem, the resulting solution is said to be A standard solution.
PROBLEM:
Prepare 250.0 ml of a 0.1250 M AgNO 3 solution, using solid AgNO 3 .
SOLUTION:
From the given volume and concentration you can calculate how many grams of
AgNO 3 to weigh out:
wt. of AgN0 3 needed = (0.2500 liter)( 0 . 1250
m
°'^.
AgN °
3 ) (l69.9
\
liter
A
mole AgNO
= 5.309 g AgN0 3
Transfer the 5.309 g AgNO 3 to a 250.0 ml volumetric flask, dissolve it in some
distilled water, then dilute to the mark (see p 86). Shake vigorously to get a
uniform solution. Don't add 250.0 ml of water to the weighed sample, because the
resulting solution may actually be larger or smaller than 250.0 ml due to interaction
of solute and solvent.
Many crystals contain "water of crystallization," which must be included in
the weight of the material weighed out for preparing a solution. Allowance is
made for this by using the molecular weight of the hydrate in your calculations,
not the mole weight of the anhydrous form.
PROBLEM:
Prepare 100.0 ml of 0.2000 M CuSO 4 , starting with solid CuSO 4 • 5H 2 O.
SOLUTION:
From the given volume and concentration of CuSO 4 , you can calculate the moles
of CuSO 4 required. Furthermore, the formula shows that 1 mole of CuSO 4 -5H 2 O
is required per mole of CuSO 4 . Thus the weight (W) of CuSO 4 -5H 2 O needed is
W = (0.1000 liter) ( 0.2000
mote
,
g CuS °<) ( , ™'* CuSO -SH,O\ /
g CuSO 5H,O \
V
liter
/ V
mole CuSO 4 / V
mole CuSO 4 -5H 2 O/
= 4.992 g CuSO,-5H 2 0 needed
Transfer the 4.992 g to a 100 ml volumetric flask, dissolve it in some distilled
water, then dilute to the mark. The fact that some of the water in the solution