Solution Concentrations
58. Formula stoichiometry: 1 mol of BaCl2 contains 1 mol Ba
2+
–
ions and 2 mol Cl ions.
2+
–
–
So, the 0.12 M BaCl2 solutions contains 0.12 M Ba ion and 2 × (0.12 M) Cl ion = 0.24 M Cl ion.
60. Define the problem: Given the mass of the solute and the volume of the solution, find the molarity of
the solute, and the concentrations of the ions.
Develop a plan: Use the molar mass to find moles of solute, convert the volume to liters from
milliliters, and divide the moles of solute by the volume in liters to get molarity. Use formula
stoichiometry to find the concentrations of the ions.
Execute the plan:
(a) Find moles of solute: 6.73 g Na 2CO 3 ×
1 mol Na 2CO 3
−2
= 6.35 ×10 mol Na 2 CO3
105.9885 g Na 2CO 3
250 . mL solution×
Molarity =
1L
= 0.250 L solution
1000 mL
6.35 ×10−2 mol Na 2CO 3
= 0.254 M Na 2 CO3
0.250 L solution
Notice that the three sequential calculations shown above can be combined into one calculation, as
follows:
6.73 g Na 2 CO 3
1 mol Na 2CO 3
1000 mL
×
×
= 0.254 M Na 2CO 3
250. mL solution 105.9885 g Na 2 CO3
1L
This combined version prevents having to write unnecessary intermediate answers and helps cut
down on round-off errors in significant figures. Both ways give the right answer, but it is helpful
to consolidate your work, when you can.
(b) Na 2 CO3 has a formula stoichiometry that looks like this:
+
1 mol Na 2 CO3 : 2 mol Na ions: 1 mol CO3
+
2–
ions.
+
2–
So, the 0.254 M Na 2 CO3 contains 2 × (0.254 M) Na ion = 0.508 M Na ion and 0.254 M CO3 .
Check your answers: The number 0.0635 is about one quarter of .250; this value looks right. The
+
2–
concentration of Na is twice than the concentration of CO3 . These answers look right.
62. Define the problem: Given the volume of the solution and its molarity, find the mass of the solute in
the solution.
Develop a plan: Convert the volume of solution to liters from milliliters. Then use the molarity as a
conversion factor to determine moles of solute. Then use the molar mass to find grams of solute.
Execute the plan:
250. mL solution×
1L
0.0125 mol KMnO4 158.0339 g KMnO 4
×
×
= 0.494 g KMnO4
1000 mL
1 L solution
1 mol KMnO 4
Check your answer: The units cancel appropriately. The relative size of the answer seems appropriate.
64. Define the problem: Given the mass of the solute and the solution’s molarity, find the volume of the
solution.
Develop a plan: Use the molar mass to find determine moles of solute. Then use the molarity as a
conversion factor to determine volume of the solute in liters. Then convert the volume of solution to
milliliters from liters.
Execute the plan:
25 .0 g NaOH ×
1 mol NaOH
1 L solution
1000 mL
×
×
= 5.08 ×10 3 mL solution
39 .9971 g NaOH 0.123 mol NaOH
1L
Check your answer: The units cancel appropriately. The relatively large number of milliliters seems
appropriate, since this relatively dilute solution contains a relatively large mass of solute.
66. Define the problem: Given the volume of a concentrated solution, the concentrated solution’s molarity,
and the final volume of the dilute solution, find the concentration of the dilute solution.
Develop a plan: Moles of solute do not change when water is added. So we can equate the moles of
solute in the concentrated solution with the moles of solute in the dilute solution. Get the moles of
solute in the concentrated solution by using the molarity of the concentrated solution as a conversion
factor to convert the volume into moles. Divide the moles of solute by the new dilute solution’s
volume to get the dilute solution’s concentration. The logical plan outlined here is built into the
equation given in Section 5.6 for doing dilution calculations:
Molarity(conc) × V(conc) = Molarity(dil) × V(dil)
Execute the plan:
Molarity (conc) ×V (conc) (0.0250 M CuSO 4 conc)× (6.00 mL conc)
=
= 0.0150 M CuSO4
V(dil)
(10 .0 mL dil)
Check your answer: The dilute solution has a smaller concentration than the concentrated solution.
That is a sensible result, since water was added to make the new solution.
68. Define the problem: Given the desired volume and molarity of a dilute solution, determine which of
several dilution processes produces this solution.
Develop a plan: Looking at each choice, it’s easy to see that each solution results in a total volume of
1.00 L, which is the desired volume. So, we should focus our attention on which of the choices
provides a solution that contains the proper number of moles. Convert milliliters to liters. Then
calculate the moles in each choice by multiplying the volume in liters by the concentration of the
concentrated solution.
Execute the plan: We want 1.00 L of 0.125 M H2 SO4
0.125 mol H 2SO 4
= 0.125 moles of H2 SO4
1 L dil
So, determine moles H2 SO4 in each choice and compare it to the 0.125 moles H2 SO4 desired.
1.00 L dil ×
(a) 36 .0 mL conc×
1L
1.25 mol H 2SO 4
×
= 0.00450 mol H 2 SO 4
1000 mL
1 L conc
No
(b) 20 .8 mL conc×
1L
6.00 mol H 2SO 4
×
= 0.125 mol H 2SO 4
1000 mL
1 L conc
Yes
(c) 50 .0 mL conc×
1L
3.00 mol H 2SO 4
×
= 0.150 mol H 2 SO 4
1000 mL
1 L conc
No
(d) 500 . mL conc ×
1L
0.500 mol H 2SO 4
×
= 0.250 mol H 2SO 4
1000 mL
1 L conc
No
Only choice (b) will make the desired solution.
Note: You also could have used the Molarity(conc) × V(conc) = Molarity(dil) × V(dil) dilution
equation to solve this Question.
Check your answer: Compare moles of solute in the concentrated and dilute solutions:
Molarity(conc) × V(conc) = (6.00 M H2 SO4 conc) × (20.8 mL conc)×
1L
= 0.125 mol H2 SO4
1000 mL
Molarity(dil) × V(dil) = (0.125 M H2 SO4 dil) × (1.00 L dil) = 0.125 mol H2 SO4
Within three significant figures, the moles in the concentrated solution (0.125 mol) are the same as the
moles in the dilute solution (0.125 mol), as expected. This answer makes sense.
102. Too much water was added, making the solution too dilute. So, (d) the concentration of the solution
is less than 1 M because you added more solvent than necessary.
103.
Solution A has 12 particles in 500 mL.
Solution B has 6 particles in 500 mL.
Solution C has 3 particles in 500 mL.
Solution D has 8 particles in 500 mL.
Solution E has 3 particles in 250 mL.
(That’s the same concentration as a solution with 6 particles in 500 mL.)
Solution F has 5 particles in 250 mL.
(That’s the same concentration as a solution with 10 particles in 500 mL.)
(a) The most concentrated solution has the largest number of particles in a fixed, comparable
volume. So, the answer is Solution A, with 12 particles in 500 mL.
(b) The least concentrated solution has the smallest number of particles in fixed, comparable
volume. So the answer is Solution C with 3 particles in 500 mL.
(c) The two solutions with the same concentration are Solution B and Solution E, since 6 particles
in 500 mL has the same concentration as half that number of particles in half that volume.
(d) When solutions E and F are combined, the total number of particles is 8, and the total volume is
500 mL. So, the combined solution has the same concentration as solution D, with 8 particles in
500 mL.
(e) When solutions C and E are combined, the total number of particles is 6, and the total volume is
750 mL. So, the combined solution has the same concentration as a solution with 4 particles in
500 mL. None of the pictured Solutions is the right answer.
(f) Reducing the volume of Solution B from 500 mL to 250 mL, gives it 6 particles in 250 mL,
which has the same concentration as 12 particles in 500 mL, Solution B.
(g) Half of Solution A has 6 particles in 250 mL. Increasing the volume to 500 mL gives it the
same concentration as Solution B.
119. Define the problem: Given the molecular formula of caffeine and the typical mass of caffeine
dissolve in a fixed volume of coffee, determine the molarity of the solute in the solution and the
mass of the solid that does not dissolve.
Develop a plan: Use metric conversion to convert from milligrams to grams, then use the molar
mass to determine moles of caffeine dissolved in the coffee, then divide by the volume of the
solution in liters to get the molarity.
Execute the plan:
125 mg C8 H10 N4O2
1g
1mol C 8H 10 N 4O 2
1000 mL
×
×
×
= 0.0026 M C 8H 10 N 4O 2
250 mL
1000 mg 194.1902 g C8 H10 N 4 O2
1L
Notice, if we interprete the molar mass in terms of milligrams per millimole and the molarity as
millimoles per milliliter, the two milli- metric conversions can be avoided to get the same answer.
125 mg C8 H10 N4O2
1mmol C8 H10 N4 O2
×
= 0.0026 M C8 H10 N 4 O2
250 mL
194.1902 m g C8 H10 N4O2
Check your answers: The concentration is relatively small.