Unit 5 Solution Problems Type 1: Concentration from Moles and Volume 1. Copper (II) sulfate, an important copper in salt, is used in electroplating cells, and to kill algae in swimming pools and water reservoirs. What is the molar concentration of an electroplating solution in which 1.50 mol of copper (II) sulfate are dissolved in what to make 2.00 L of solution? n = 1.5 mol copper (II) sulfate V= 2.00L C = n / v = 1.5 mol / 2.00L = 0.75 mol/L or 0.75 M 2. What is the molar concentration of a solution in which 0.240 mol of washing soda, Na 2 CO 3 · H 2 O is dissolved in water to make 480 mL of a solution for softening wash water? n = 0.240 mol Na 2 CO 3 V = 480mL = 0.48L C = n / v = 0.240mol / 0.48L = 0.5 mol/L or 0.5 M 3. Iron (II) sulfate finds use in mixing colours in dyeing and in making ink. What is the molar concentration of an ink solution that contains 0.210 mol of iron (II) sulfate dissolved to form 840 mL of solution? n = 0.210 mol iron (II) sulfate V = 840 mL = 0.84L C = n / v = 0.210 mol / 0.84L = 0.25 mol/L or 0.25 M 4. Since a saturated solution of calcium chloride does not freeze until -55°C, calcium chloride and be used to melt ice on roads and walks. What is the molar concentration of a saturated solution in which 35.55 mol of CaCl 2 is dissolved in water to make 5.00 L of solution? n = 35.55 mol CaCl 2_ V = 5.00L C = n / v = 35.55 mol / 5.00L = 7.1 mol/L or 7.1 M
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Unit 5
Solution Problems
Type 1: Concentration from Moles and Volume
1. Copper (II) sulfate, an important copper in salt, is used in electroplating cells, and to kill
algae in swimming pools and water reservoirs. What is the molar concentration of an
electroplating solution in which 1.50 mol of copper (II) sulfate are dissolved in what to
make 2.00 L of solution?
n = 1.5 mol copper (II) sulfate
V= 2.00L
C = n / v
= 1.5 mol / 2.00L
= 0.75 mol/L or 0.75 M
2. What is the molar concentration of a solution in which 0.240 mol of washing soda, Na 2CO3
· H2O is dissolved in water to make 480 mL of a solution for softening wash water?
n = 0.240 mol Na2CO3
V = 480mL = 0.48L
C = n / v
= 0.240mol / 0.48L
= 0.5 mol/L or 0.5 M
3. Iron (II) sulfate finds use in mixing colours in dyeing and in making ink. What is the molar
concentration of an ink solution that contains 0.210 mol of iron (II) sulfate dissolved to
form 840 mL of solution?
n = 0.210 mol iron (II) sulfate
V = 840 mL = 0.84L
C = n / v
= 0.210 mol / 0.84L
= 0.25 mol/L or 0.25 M
4. Since a saturated solution of calcium chloride does not freeze until -55°C, calcium chloride
and be used to melt ice on roads and walks. What is the molar concentration of a saturated
solution in which 35.55 mol of CaCl2 is dissolved in water to make 5.00 L of solution?
n = 35.55 mol CaCl2_
V = 5.00L
C = n / v
= 35.55 mol / 5.00L
= 7.1 mol/L or 7.1 M
5. Sulfuric acid is an important laboratory reagent as well as a very important chemical. One
of its many industrial uses is an electrolyte I lead storage (car batteries. Calculate the molar
concentration of a battery acid solution which contains 9.25 mol of H 2SO4 dissolved to
form 1.80 L of solution.
n = 9.25 mol
V = 1.80 L
C = n / v
= 9.25 mol / 1.80 L
= 5.14 mol/L or 5.14 M
Type 2: Concentration from Mass and Volume
6. A given sample of household ammonia contains 156 g of NH3(g) dissolved in water to form
2.00 L of solution. What is the molar concentration of the household ammonia solution?
NH3: m = 156 g
MM = 17 g/mol
V = 2.00L
n = m / MM
= 156 g / 17 g/mol
= 9.18 mol NH3
C = n / v
= 9.18 mol / 2.00L
= 4.6 mol/L or 4.6 M
7. When 11.0 g of glacial (pure) acetic acid is dissolved in water to make 250 mL of vinegar
solution, what is the molar concentration of the vinegar?
CH3COOH: m = 11.0g
MM = 60.05 g/mol
V = 250mL = 0.25L
n = m / MM
= 11.0g / 60.05 g/mol
= 0.1832 mol CH3COOH
C = n / v
= 0.1832 mol / 0.25 L
= 0.733 mol/L or 0.733 M
8. What is the molar concentration of 500 mL of a solution that contains 12.7 g of swimming
pool chlorinator, Ca(OCl)2?
Ca(OCl)2: m = 12.7 g
MM = 142.9837 g/mol
V = 500 mL = 0.5L
n = m / MM
= 12.7 / 142.9837
= 0.08882 mol
C = n / v
= 0.8882 / 0.5
= 0.178 mol/L or 0.178 M
9. A solution for water proofing concrete may be prepared by dissolving 200 g of ammonium
stearate in water to make 5.00 L of solution. Determine the molar concentration of the
solution.
C18H39NO2: m = 200g
MM = 301.5 g/mol
V = 5.00L
n = m / MM
= 200 / 301.5
= 0.663 mol
C = n / v
= 0.663 / 5.00
= 0.133 mol/L or 0.133 M
10. A car battery terminal protective coating can be prepared by dissolving 240.0 g of sodium
silicate (water glass) in water to make 250 mL of solution. What is the molar
concentration?
Na2SiO3: m = 200g
MM = 122.06 g/mol
V = 250 mL = 0.25L
n = m / MM
= 200 / 122.06
= 1.638 mol
C = n / v
= 1.638 / 0.25
= 6.552 mol/L or 6.552 M
Dilution and Concentration Problems
1. For each of the following solutions, tell how many grams of solute would be necessary for its
preparation.
a) 0.10 L of 0.10M AgNO3
V = 0.10L
C = 0.10M
MM AgNO3 = 169.848 g/mol
n = C x V
= (0.10)(0.10)
= 0.01 mol AgNO3
m = n x MM
= (0.01)(169.848)
= 1.699 g
b) 5.0 mL of 0.05M NaCN
V = 5.0mL = 0.005 L
C = 0.05 M
MM NaCN = 49.01 g/mol
n = C x V
= (0.05)(0.005)
= 0.00025 mol
m = n x MM
= (0.00025)(49.01)
= 0.012 g
c) 0.10 L of 0.10M barium chloride
V = 0.10 L
C = 0.10M
MM BaCl = 172.78 g/mol
n = C x V
= (0.10)(0.10)
= 0.01 mol
m = n x MM
= (0.01)(172.78)
= 1.73 g
d) 250 mL of 0.0014M KMnO4
V = 250m L = 0.25L
C = 0.0014 M
MM KMnO4 = 157.9889 g/mol
n = C x V
= (0.0014)(0.25)
= 0.0035 mol
m = n x MM
= (0.0035)(157.9889)
= 0.55 g
2. You dissolve 0.395g of KMnO4 in enough water to give 250 mL of solution. What is
the molar concentration of KMnO4?
KMnO4: m = 0.395g
MM = 157.9943g/mol
V = 250 mL = 0.25L
n = m / MM
= 0.395 / 157.9943
= 0.0025 mol
C = n / v
= 0.0025 / 0.25
= 0.10 mol/L or 0.01 M
3. How many grams of Na2CO3 are required to make 2.0 L of 1.5M Na2CO3?
C = 1.5M
V = 2.0L
MM Na2CO3 = 105.958 g/mol
n = C x V
= (1.5)(2.0)
= 3 mol
m = n x MM
= (3)(105.958)
= 317.9 g
4. What would the molar concentration of the solute in each of the following solutions?
a) 0.50 L containing 5.6g of Na2ClO4
V = 0.50 L
m = 5.6 g
MM = 145.388 g/mol
n = m / MM
= 5.6 / 145.388
= 0.038 mol
C = n / V
= 0.038 / 0.50
= 0.076 mol/L or 0.08 M
b) 0.10 L containing 2.3g of KNO3
V = 0.10 L
m = 2.3 g
MM = 101.07 g/mol
n = m / MM
= 2.3 / 101.07
= 0.023 mol
C = n / V
= 0.023 / 0.10
= 0.23 mol/L or 0.23 M
c) 0.25 L containing 1.5g of C4H8O
V = 0.25 L
m = 1.5 g
MM = 72.11 g/mol
n = m / MM
= 1.5 / 72.11
= 0.021 mol
C = n / V
= 0.021 / 0.25
= 0.84 mol/L or 0.84 M
d) 50 mL containing 0.55g of NaOH
V = 50 mL = 0.05L
m = 0.55 g
MM = 39.99 g/mol
n = m / MM
= 0.55 / 39.99
= 0.0137 mol
C = n / V
= 0.0137 / 0.05
= 0.27 mol/L or 0.27 M
e) 1.55 L containing 153g of Na2CO3
V = 1.55 L
m = 153 g
MM = 105.958 g/mol
n = m / MM
= 153 / 105.958
= 1.44 mol
C = n / V
= 1.44 / 1.55
= 0.929 mol/L or 0.93 M
5. Sucrose, common table sugar, has the formula C12H22O11 . If you add one spoonful
(3.4g) to 250 mL of coffee, what is the molar concentration of the sugar?
MM = 342.23 g/mol
m = 3.4 g
V = 250mL = 0.25L
n = m / MM
= 3.4 / 342.23
= 0.0099 mol
C = n / V
= 0.0099 mol / 0.25
= 0.0396 mol/L or 0.04 M
6. An experiment calls for you to use 300 mL of 1.00M NaOH, but you are only given a
large bottle of 3.00M NaOH. Explain how you would make up the 1.00M NaOH in the
desired volume.
Since the experiment calls for 300 mL of 1.00 M of NaOH, Then a concentration of 3.00 M NaOH
would only call for 100 mL because it’s a 3 : 1 ratio.
7. You need 1.00L of a 0.0100M K2Cr2O7 solution. You have some 0.100M of K2Cr2O7
available. How much of the more concentrated solution do you need and how much water
must be added to give finally 1.00 L of 0.0100M K2Cr2O7?
C1 = 0.100 M
C2 = 0.0100 M
V2 = 1.00 L
V1 = ?
CIVI = C2V2
V1 = C2V2 / C1
= (0.01)(1) / 0.1
= 0.01 L
100 mL concentration
= 900 mL H2O
8. This is a tough one! What volume of concentrated aqueous sulfuric acid, which is 98%
H2SO4 by mass and has a density of 1.84 g/cm3, is required to make 10.0 L of 0.200 M
H2SO4?
C2 = 0.200 M
V2 = 10.0 L
V1 = ?
C1V1 = C2V2
= (0.200)(10.0)
= 2 mol
H2SO4
n = 2 mol
MM = 98 g/mol
m = n x MM
= 2 (98)
= 196 g
m = .106 / .98
= .108 L
mass % = solute / solution x 100%
98% = .106 L / m x 100%
1.84 g / 1 mL = 196 / x
x = 106.5 mL or .106 L
More Concentration and Dilution Questions
1. Calculate the mass of the solute needed to make each of the following solutions:
a) 250 mL of a 1.25 mol/L lithium bromide solution
V = 250mL = 0.25L
C = 1.25 mol/L
MM LiBr =86.845
n = C x V
= (1.25)(0.25)
= 0.3125 mol
m = n x MM
= (0.3125)(86.845)
= 27.1
b) 50.0 mL of a 2.30 mol/L aluminum chlorate solution
V = 50.0mL = 0.05L
C = 2.30 mol/L
MM AlCl3 = 132.33 g/mol
n = C x V
= (2.30)(0.05)
= 0.115 mol
m = n x MM
= (0.115)(132.33)
= 15.2 g
2. 12.0 L of hydrogen chloride gas measured at 17°C and 110 kPa is dissolved in enough
water to produce 500 mL of solution. What is the concentration of hydrochloric acid
solution?
V = 12.0 L
T = 17°C = 290 K
P = 110 kPa
R = 8.31430 L kPa mol K
n = RT / PV
= (8.31430)(290) / (110)(12.0)
= 1.8266 mol
C = n / V
= 1.8266 / 12.0
= 0.152 mol/L or 0.152 M
3. Calculate the mass of aluminum sulfate required to prepare 300 mL of 0.220 mol/L
solution.
V = 300 mL = 0.3 L
C = 0.220 mol/L
MMAl2S3 = 150.17 g/mol
n = C x V
= (0.220)(0.3)
= 0.066 mol
m = n x MM
= (0.066)(150.17)
= 9.9 g
4.What volume of 0.0300 mol/L sodium sulfate solution can be prepared from 145 g of
Na2SO4?
C = 0.0300 mol/L
m = 145 g
MM Na2SO4 = 142.01 g/mol
n = m / MM
= (145)(142.01)
= 1.021 mol
V = n / C
= 1.021 / 0.0300
= 34.03 L
5. Calculate the volume of a stock solution (original solution) that must be used to make
each of the following:
a) 500 mL of a 0.750 mol/L solution of sulfuric acid (stock solution is 18.0
mol/L)
C1 = 18mol/L
C2 = 0.750 mol/L
V2 = 500 mL = 0.5L
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.750)(0.5) / 18
= 0.021 L
b) 100 mL of a 3.40 mol/L solution of ammonium hydroxide (stock solution is
15.0 mol/L)
C1 = 15.0 mol/L
C2 = 3.40 mol/L
V2 = 100mL = 0.1 L
V1 = ?
V1 = C2V2 / C1
= (3.40)(0.1) / 15.0
= 0.023 L
c) 250 mL of a 0.120 mol/L solution of acetic acid (stock solution is 18.0
mol/L
C1 = 0.120 mol/L
C2 = 18.0 mol/L
V2 = 150 mL = 0.15L
V1 = ?
V1 = C2V2 / C1
= (18.0)(0.15) / 0.120
= 22.5 L
6. What volume of stock solution would be needed to prepare each of the following
solutions?
a) 250 mL of a 1.00 mol/L hydrochloric acid solution; stock HCl is 12.0 mol/L
C1 = 12.0 mol/L
C2 = 1.00 mol/L
V2 = 250 mL = 0.25L
V1 = ?
V1 = C2V2 / C1
= (1.00)(0.25) / 12.0
= 0.021 L
b) 500 mL of a 4.30 mol/L sulfuric acid solution; stock sulfuric acid is 18.0 mol/L
C1 = 18.0 mol/L
C2 = 4.30 mol/L
V2 = 500 mL = 0.5L
V1 = ?
V1 = C2V2 / C1
= (4.30)(0.5) / 18.0
= 0.119 L
7. Describe how to prepare 100 mL of a 0.0250 mol/L zinc chloride solution using a 2.50
mol/L stock solution of zinc chloride.
You would find the volume of stock solution that is needed to prepare 0.1 L of a 0.0250 mol/L
using a 2.50 mol/L stock solution. Doing this you would use the forumula C1V1 = C2V2 and
rearrange equal to V1 (V1 = C2V2 / C1). You would then plug in the given measurements to find the
volume of the stock solution. You will then be able to prepare 100mL of a 0.0250 mol/L.
Dissociation: Reactions and Equations
1.Write the dissociation reactions for the following in water:
a) manganese (II)sulfide
MnS -> Mn
+2(aq) + S
-2(aq)
b) lithium sulfate
Li2S -> Li
+1 (aq) + S
-2 (aq)
c) copper I bromide
CuBr -> Cu
+1(aq) + Br
-1(aq)
d) sodium acetate
Na(CH3COO) -> Na
+1(aq) + CH3COO
-1(aq)
e) potassium hydroxide
K(OH) -> K
+1(aq) + OH
-1(aq)
2. Write the dissociation equation for the following in water:
a) sodium chloride
NaCl -> Na+1
(aq) + Cl-1
(aq)
b) barium chloride
BaCl2 -> Ba2+
(aq)+ Cl -1
(aq)
c) calcium sulfate
CaSO4 -> Ca 2+
(aq) + SO42-
(aq)
d) strontium hydroxide
Sr(OH)2 -> Sr 2+
(aq) + OH -1
(aq)
e) copper (I) iodide
CuI -> Cu+1
(aq) + I-1
(aq)
f) ammonium sulfide
(NH4)2S -> NH4
+1 (aq) + S
-1(aq)
g) potassium nitrate
KNO3 -> K+1
(aq) + NO3-1
(aq)
h) calcium hydroxide
Ca(OH)2 -> Ca2+
(aq) + OH-1
(aq)
i) sodium acetate
Na(CH3COO) -> Na+1
(aq) + CH3COO -1
(aq)
Equations: Molecular, Ionic and Net Ionic
1. Will a precipitate form between the following solutions when they are added
together? If YES, give the name of the precipitate.
a) Pb(NO3)2 + MgSO4 PbSO4(s) + Mg(NO3)2(aq)
Yes / Lead(II) Sulfate
b) Pb(NO3)2 + NaCl PbCl2(s) + NaNO3(aq)
Yes / Lead(II) chloride
c) CaCl2 + Na2 SO4 CaSO4(s) + NaCl(aq)
Yes / Calcium Sulphate
d) Ca(CH3COO)2 + AgNO3 Ca(NO3)2(aq) + AgCH3OO(s)
Yes / Silver(I) Acetate
e) Pb(NO3)2 + KBr PbBr2(s) + KNO3(aq)
Yes / Lead(II) Bromide
f) AgNO3 + NaCl AgCl(s) + NaNO3(aq)
Yes / Silver(I) Chloride
2.Write the molecular, ionic and net ionic equation (if possible) for the following
reactions:
a) hydrochloric acid and sodium fluoride HCl(aq) + NaF(aq) HF(aq) + NaCl(aq)
No Net Equation
b) ammonium carbonate and barium chloride (NH4)2CO3(aq) + BaCl2(aq) 2 NH4Cl(aq) + BaCO(s)
2 NH4+1
(aq) + CO3-2
(aq) + Ba+2
(aq) + 2 Cl-1
(aq) 2 NH4+1
(aq) + 2 Cl-1
(aq) + BaCO3(s)
CO3-2
(aq) + Ba+2
(aq) BaCO3(s)
c) copper (II) chloride and sodium hydroxide CuCl2(aq) + 2 NaOH(aq) Cu(OH)2(s) + 2 NaCl(aq)
Cu+2
(aq) + Cl-1
(aq) + 2 Na+1
(aq) + 2 OH-1
(aq) Cu(OH)2(s) + 2 Na+1
(aq) + 2 Cl-1
(aq)
Cu+2
(aq) + 2 OH-1
(aq) Cu(OH)2(s)
d) iron (II) sulfate and sodium phosphate 3 FeSO4(aq) + 2 Na3PO4(aq) Fe(PO4)2(s) + 3 Na2SO4(aq)
3 Fe+2
(aq) + 3 SO4-2
(aq) + 2 Na+1
(aq) + 2 PO4-3
(aq) Fe3(PO4)2(s) + 3 Na
+1(aq) + 3 SO4
-2(aq)
3 Fe+2
(aq) + 2 PO4-3
(aq) Fe3(PO4)2(s)
e) ammonium perchlorate and copper (II) nitrate 2 NH4ClO4(aq) + Cu(NO3)2(aq) 2 NH4NO3(aq) + Cu(ClO4)2 (s)