moles nd molarity
DESCRIPTION
molarty calculationTRANSCRIPT
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)
2 NH4+1
(aq) + 2 ClO4
-1(aq)
+ Cu
+2(aq) + NO3
-1(aq) 2 NH4
+1(aq) + 2 NO3
-1(aq) + Cu(ClO
4)2(s)
2 ClO4-1
(aq) + Cu+2
(aq) Cu(ClO4)2(s)
f) ammonium sulfide and sodium chloride (NH4)2S(aq) + 2 NaCl(aq) 2 NH4Cl(aq) + Na2S(s)
NH4+1
(aq) + S-2
(aq) + 2 Na+1
(aq) + 2 Cl-1
(aq) 2 NH4+1
(aq) + 2 Cl-1
(aq) + Na2S(s)
S-2
(aq) + Na+1
(aq) Na2S(s)
g) potassium chloride and sodium nitrate KCl(aq) + NaNO3(aq) KNO3(aq) + NaCl(aq)
No Net Equation
h) calcium chloride and sodium carbonate CaCl2(aq) + Na2(CO3)(aq) CaCO3(s) + 2 NaCl(aq)
Ca+2
(aq) + Cl-2
(aq) + Na+1
(aq) + CO3-2
(aq) CaCO3(s) + 2 Na+1
(aq) + 2 Cl-1
(aq)
Ca +2
(aq) + CO3-2
(aq) CaCO3(s)
i) lithium chlorate and ammonium chloride LiClO3(aq) + NH4Cl(aq) LiCl(aq) + NH4ClO3(aq)
No Net Equation
j) iron (III) sulfate and lead (II) chlorate Fe2(SO4)3(aq) + 3 Pb(ClO3)2(aq) 2 Fe(ClO3)3(aq) + 3 PbSO4(s)
Fe+3
(aq) + SO4-2
(aq) + 3 Pb+2
(aq) + 3 ClO3-1
(aq) 2 Fe+3
(aq) + 2 ClO3-1
(aq) + 3 Pb(SO4)(s)
SO4-2
(aq) + 3 Pb+2
(aq) 3 Pb(SO4)(s)
k) aluminum bromide and cadmium nitrate AlBr3(aq) + CdNO3(aq) Al(NO3)3(aq) + NaCl(aq)
No Net Equation
l) zinc chloride and sodium phosphate 3 ZnCl(aq) + Na3PO4 Zn3PO4(s) + 3 NaCl(aq)
3 Zn+1
(aq) + 3 Cl-1
(aq) + Na+1
(aq) + PO4-3
(aq) Zn3PO4(s) + 3 Na+1
(aq) + 3 Cl-1
(aq)
3 Zn+1
(aq) + PO4-3
(aq) Zn3PO4(s)
m) potassium iodide and ammonium nitrate KI(aq) + NH4NO3(aq) KNO3(aq) + NH4I(s)
K+1
(aq) + I-1
(aq) + NH4+1
(aq) + NO3-1
(aq) K+1
(aq) + NO3-1
(aq) + NH4I(s)
I-1
(aq) + NH4+1
(aq) NH4I(s)
n) calcium chloride and manganese (II) iodide CaCl2(aq) + MnI2(aq) CaI2(aq) + MnCl2(aq)
No Net Equation
o) ammonium phosphate and magnesium nitrate (NH4)3PO4(aq) + 3 Mg(NO3)2(aq) 3 NH4NO3(aq) + Mg3(PO4)2(s)
NH4+1
(aq) + PO4-3
(aq) + 3 Mg+2
(aq) + 3 NO3-1
(aq) 3 NH4+1
(aq) + 3 NO3-1
(aq) + Mg3(PO4)2(s)
PO4-3
(aq) + 3 Mg+2
(aq) Mg3(PO4)2(s)
pH Questions
A. Calculate the pH of the following solutions: (assume 100% dissociation)
1) 1.8 x 10-4
mol/L of H2SO4 - log (1.8 x 10
-4)
= 3.75
2) 2.7 x 10-3
mol/L of H3PO4 (actually a weak acid) - log (2.7 x 10
-3)
= 2.6
3) 1.9 x 10-7
mol/L of Mg(OH)2 - log (1.9 x 10
-7)
= 6.721
pH = 14 – 6.721
= 7.28
4) 5.0 x 10-4
mol/L of Mn(OH)3 (actually a weak base) - log (5.0 x 10
-4)
= 3.301
pH = 14 – 3.301
= 10.7
Lemon Juice, pH 2.1
Cow’s Milk, pH 6.5
Shampoo, pH 6.7
Egg White, pH 7.8
B.
1. Calculate the pH of
a) a 2.0 x 10-3
mol/L nitric acid solution - log (2.0 x 10
-3)
= 2.7
b) a 2.0 x 10-5
mol/L sodium hydroxide solution - log (2.0 x 10
-5)
= 4.699
pH = 14 – 4.699
= 9.30
2. Arrange the following substances in order of increasing solubility:
Cow’s milk, pH 6.5
Egg white, pH 7.8
Shampoo, pH 6.7
Lemon Juice, pH 2.1
3. Predict whether the pH of an aqueous solution of each of the following will be
above 7, below 7, or equal to 7.
a) acetic acid = below 7 b) sugar = equal to 7 c) ammonia = above 7
4. Calculate the pH of an aqueous solution containing
a) 2.5 x 10-5
mol/L of HCl - log (2.5 x 10
-5)
= 4.60
b) 1.0 x 10-3
mol/L of NaOH - log (1.0 x 10
-3)
= 3
pH = 14 – 3
= 11
5. Calculate the pH of each of the following aqueous solutions
a) 4.2 x 10-6
mol/L HNO3
- log (4.2 x 10-6
) = 5.38
b) 5.5 x 10-5
mol/L Ba(OH)2
- log (5.5 x 10-5
) = 4.26
pH = 14 – 4.26
= 9.74
Titration Problems
1. As part of a project, a student has to accurately determine the concentration of a
solution of hydrochloric acid. She titrates 100 mL of the solution of hydrochloric
acid of unknown concentration with 0.250 mol/L sodium hydroxide solution. She
determines that 40.0 mL of the NaOH solution will neutralize the acid. What is the
concentration of the hydrochloric acid?
HCl + NaOH = H(OH) + NaCl
V1 = 100 mL = 0.1 L
C2 = 0.250 mol/L
V2 = 40.0 mL = 0.04 L
C1 = ?
C1V1 = C2V2
C1 = C2V2/V1
= (0.250)(40.0) / 0.1
= 100 mol/L or 100 M
2. The concentration of hydrochloric acid is standardized (that is, it can be determined)
using pure sodium carbonate. If 30.0 mL of the acid reacts completely with 0.50 g of
the sodium carbonate, what is the concentration of the acid?
HCl + Na2(CO3) = H(CO3) + Na2Cl
V2 = 30.0 mL = 0.03 L
V1 = 0.50 g = 0.50 mL = 0.0005 L
Na2(CO3): m = 0.50 g
MM = 105.958 g/mol
n = m /MM
= 0.50 / 105.958
= 0.0047 mol
C1 = n / V
= 0.0047 / 0.0005
= 9.4 mol/L or 9.4 M
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (9.4)(0.0005) / 0.03
= 0.16 mol/L or 0.16 M
3. How many mL of 0.10 mol/L NaOH are needed to neutralize 25.0 mL of 0.15 mol/L
of H2SO4?
NaOH + H2SO4 = H(OH) + Na(SO4)
C1 = 0.10 mol/L
C2 = 0.15 mol/L
V2 = 25.0 mL
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.15)(25.0) / 0.10
= 37.5 mL or 0.0375 L
4. What is the concentration of a nitric acid solution in which 25.0 mL of the nitric acid
solution is completely neutralized by 40.0 mL of a 0.10 mol/L NaOH solution.
NaOH + HNO3 -> H(OH) + Na(NO3)
V2 = 25.0 mL = 0.025 L
V1 = 40.0 mL = 0.04 L
C1 = 0.10 mol/L
C2 = ?
C1V1 = C2V2
C2 = C1V1/V2
= (0.10)(0.04) / 0.025
= 0.16 mol / L or 0.16 M
5. The concentration of an acid solution is determined by using pure Na2CO3 . If 47.2
mL of nitric acid solution is just neutralized by 0.500 g of anhydrous sodium
carbonate, what is the concentration of the acid in mol/L?
V2 = 47.2 mL = 0.0472 L
V1 = 0.500 g = 0.0005 L
Na2CO3: MM = 105.99 g/mol
n = m / MM
= 0.500 / 105.99
= 0.0047 mol
C1 = n / V
= 0.0047 / 0.0005
= 9.4 M
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (9.4)(0.0005) / 0.0472
= 0.099 mol/L or 0.1 M
6. How many mL of 1.50 mol/L HCl will neutralize a solution that contains 32.0 g of
NaOH?
C1 = 1.50 mol/L
V2 = 32.0 g = 32.0 mL = 0.032 L
NaOH: MM = 40 g/mol
n = m / MM
= 32.0 / 40.0
= 0.8 mol
C2 = n / V
= 0.8 / 0.032
= 25 M
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (25)(0.032) / 1.50
= 0.53 L or 530 mL
7. A chemical storage truck carrying sulfuric acid is in an accident. A laboratory
analyzes a sample of the spilled acid and finds that 20 mL of the acid is neutralized
by 60 mL of 4.0 mol/L NaOH solution. What is the concentration of the acid?
NaOH + HSO4 -> H(OH) + Na(SO4)
V2 = 20.0 mL = 0.02 L
V1 = 60.0 mL = 0.06 L
C1 = 4.0 mol/L
C2 = ?
C1V1 = C2V2
C2 = C1V1/V2
= (4.0)(0.06) / 0.02
= 12 mol / L or 12 M More Titration Problems
1. How many milliliters of 0.250M HCl would be required to completely neutralize
2.50g of NaOH?
NaOH + HCl NaCl + H2O
C1 = 0.250 M
V2 = 2.50 g = 2.50 mL = 0.0025 L
NaOH: m = 2.50 g
MM = 39.9 g/mol
n = m / MM
= 2.50 / 39.9
= 0.063 mol
C2 = n / V
= 0.063 / 0.0025
= 25.2 mol/ L or 25.2 M
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (25.2)(0.0025) / 0.250
= 0.252 L or 2.52 mL
2. Sodium carbonate, Na2 CO3, is a good compound to use to standardize acid
solutions.
Na2CO3 + 2 HCl NaCl + H2O + CO2
1 : 2 : 1 : 1 : 1
If 42.43 mL of HCl solution is used to titrate 0.251 g of Na2CO3 to the equivalence
point, what is the molar concentration of the acid?
V1 = 42.43 mL = 0.04243 L x 2 = 0.08468 L
V2 = 0.251 g = 0.000251 L
Na2CO3:MM = 105.9886 g/mol
n = m / MM
= 0.251 / 105.9886
= 0.00237 mol
C2 = n / V
= 0.00237 / 0.000251
= 9.4422 M
C1 = ?
C1V1 = C2V2
C1 = C2V2 / V1
= (9.4422)(0.000251) / 0.08468
= 0.0279 mol/L or 0.028 M
3. Malic acid is a naturally occurring acid found in fruits, especially in apples. It
reacts with NaOH according to the equation
C4H6O7 + 2NaOH Na 2C4H 4O5 + 2H2O
1 : 2 : 1 : 2
How many milliliters of 0.520M NaOH are required to react completely with
1.34 g of malic acid?
C1 = 0.520 M x 2 = 1.04
V2 = 1.34 g = 0.00134 L
C4H6O7: MM = 166 g/mol
n = m / MM
= 1.34 / 166
= 0.008 mol
C2 = n / V
= 0.008 / 0.00134
= 5.97 M
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (5.97)(0.00134) / 1.04
= 0.007 L = 7 mL
4 An acid such as HCl can be standardized by using it to titrate a base such as
Na2CO3. If you find that 0.250g of Na2CO3 requires 25.76 mL of HCl for
titration of the equivalence point, what is the exact molar concentration of the HCl.
Na2CO3 + 2 HCl 2 NaCl + CO2 + H2O
1 : 2 : 2 : 1 : 1
V2 = 0.250g = 0.00025 L
V1 = 25.76 mL = 0.02576 L
Na2CO3: MM = 105.99 g/mol
n = m / MM
= 0.250 / 105.99
= 0.0023 mol
C2 = n / V
= 0.0023 / 0.00025
= 9.2 M
C1 = ?
C1V1 = C2V2
C1 = C2V2 / V1
= (9.2)(0.00025) / 0.02576
= 0.09 mol/L or 0.09 M x 2 = 0.18 M
More Solutions Problems
1. For each of the following solutions, tell how many grams of solute would be
necessary for its preparation:
a)0.25 L of 0.050M Na2C 2O4
n = C x V
= (0.050)(0.25)
= 0.0125 mol
m = n x MM
= (0.0125)(133.958)
= 1.674 g
b)0.125 L of 0.015M K2Cr 2O7
n = C x V
= (0.015)(0.125)
= 0.0019 mol
m = n x MM
= (0.0019)(294.09)
= 0.000006 g
c) 0.50 L of 0.010M KHCO3
n = C x V
= (0.010)(0.50)
= 0.005 mol
m = n x MM
= (0.005)(100.08)
= 0.5004 g
2. If you require 500 mL of 0.15M NaOH, how many milliliters of 1.00M NaOH
must you dilute?
V2 = 500 mL = 0.5 L
C2 = 0.15 M
C1 = 1.00 M
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.15)(0.5) / 1.00
= 0.075 L
3. If you need 100 mL of 0.15M of CuSO4, how many milliliters of 0.50M CuSO4
must you dilute? How many milliliters of water must be added? How many grams of
CuSO4 does the dilute concentration contain?
C2 = 0.15 M
V2 = 100 mL = 0.1 L
C1 = 0.50
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.15)(0.1) / (0.5)
= 0.03 L = 30 mL must be diluted
= 70 mL H2O
70 mL of water must be added
M = C x V
= 0.15 M x 100 mL
= 15 mL = 15 g of CuSO4
4. Chlorine, Cl2, can be made by treating HCl or a chlorine salt in an acid solution
with a strong oxidizing agent, MnO2 for example. The balanced, net ionic equation
for the reaction is
4H+1
+ 2Cl-1
+ MnO2 Mn+2
+ 2H2O + Cl2
4 : 2 : 1 : 1 : 2 : 1
If you have 125 mL of 0.100M solution of HCl and excess of MnO2, how many
grams of Cl2 can be formed? How many grams of MnO2 would be necessary to
complete the reaction?
limiting reactant = HCl
HCl: V = 125 mL = 0.125L
n = c x v
= (0.100)(0.125)
= 0.0125 mol Cl
Cl – Cl2
2 : 1 Ratio
2 / 0.0125 : 1 / X
X = 0.00625 mol Cl2
m = n x MM
= (0.00625)(37.46)
= 0.000167g Cl2
Cl – MnO2
2 : 1 Ratio
2 / 0.0125 : 1 / X
X = 0.00625 mol MnO2
m = n x MM
= (0.00625)(86.91)
= 0.543 g MnO2
5. How many milliliters of 0.250M HCl would be required to completely
neutralize 36.5mL of 0.100 M NaOH?
C1 = 0.250M
C2 = 0.100M
V2 = 36.5 mL
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.100)(36.5) / 0.250
= 14.6 mL
6. Potassium acid phthalate KHC8H4O4 is used to standardize solution of bases.
The acidic anions react with bases according to the net ionic equation,
HC8H4O4- + OH
- H2O + C8H4O4
2-
If a 0.902g sample of potassium acid phthalate is dissolved in water and titrated
to the equivalence point with 39.45 mL of NaOH, what is the molar concentration of
the NaOH?
V1 = 0.902 g = 0.000902 L
V2 = 39.45 mL
KHC8H4O4: MM = 204.22 g/mol
n = m / MM
= 0.902 / 204.22
= 0.0044 mol
C1 = n / V
= 0.0044 / 0.000902
= 4.878 M
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (4.878)(0.000902) / 39.45
= 0.0001 mol/L or 0.0001 M
7. One reason for the widespread use of platinum is its relative chemical inertness.
It will dissolve, however, in “aqua regia” a mixture of nitric and hydrochloric acid.
3 Pt + 4 HNO3 + 18 HCl 3 H2PtCl6 + 4 NO + 8 H2O
3 : 4 : 18 : 3 : 4 : 8
a) If you have 10.0g of Pt, how many grams of chloroplatinic acid H2PtCl6 (MW= 410),
can be produced?
3 : 3 Ratio
10 g = 10 g
You will have 10 g of chloroplatinic acid
b) How many grams of nitrogen oxide, NO, would be produced from 10.0g of Pt?
n = M / MM
= 10g / 195.06
= 0.051266277 g/mol Pt
3 : 4 ratio
3 / 0.051266277 : 4 / X
X = 0.068355036 mol NO
MMNO = 30.0061 g/mol
MNO = n x MM
= 0.068355036 mol x 30.0061 g/mol
= 2.05 g NO
You will have 2.05 g NO
c) How many milliliters of 10.0M nitric acid would be required to complete the reaction
with 10.0g of Pt?
n = M / MM
= 10g / 195.06
= 0.051266277 mol Pt
3 : 4 ratio
3 / 0.051266277 : 4 / X
X = 0.068355036 mol NO
MMHNO3 = 63.01284 g/mol
MHNO3 = n x MM
= 0.068355036 mol x 63.01284 g/mol
= 4.3 g HNO3 = 4.3 mL HNO3
You will need 4.3 mL HNO3 to complete the reaction
Solutions: Review Question of all Topics
1. 7.2 g of copper (II) sulfate is dissolved in enough water to make 200 mL of solution.
What is the molarity of the solution?
Cu(SO4)2: m = 7.2 g
MM = 255.597 g/mol
V = 200 mL = 0.2 L
n = m / MM
= 7.2 / 255.597
= 0.028 mol
C = n / V
= 0.028 / 0.2
= 0.14 mol/ L or 0.14 M
2. How many grams of sodium nitrate are required to prepare 500 mL of a 0.20 M
solution?
Na2(SO4): C = 0.20 M
MM = 141.9 g/mol
V = 500 mL = 0.5 L
n = C x V
= (0.20)(0.5)
= 0.1 mol
m = n x MM
= (0.1)(141.9)
= 14.19 g
3. What volume of a 0.50 M solution of potassium bromide would be needed to obtain
2.38 g of potassium bromide?
KBr: C = 0.50 M
MM = 119.002 g/mol
m = 2.38 g
n = m / MM
= 2.38 / 119.002
= 0.019 mol
V = n / C
= 0.019 / 0.05
= 0.38 L
4. 80 mL of water are added to 20 mL of a 1.5 M solution of potassium bromide. What is
the final molarity?
V1 = 20 mL = 0.02 L
V2 = 80 mL = 0.08 L
C1 = 1.5 M
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (1.5)(0.02) / 0.08
= 0.375 M
5. 30mL of 0.6 M potassium bromide is diluted to 300 mL. What is the final
concentration of the solution?
V1 = 30 mL = 0.03 L
V2 = 300 mL = 0.3 L
C1 = 0.6 M
C2 = ?
C1V1 = C2V2
C2 = C1VI / V2
= (0.6)(0.03) / 0.3
= 0.06 M
6. How many moles of barium nitrate are there in 20 mL of a 0.40 M solution of barium
nitrate?
Ba(NO3)2: C = 0.40 M
MM = 261.29 g/mol
V = 20 mL = 0.02 L
n = C x V
= (0.40)(0.02)
= 0.008 mol
7. 20 mL of 0.40 M sodium hydroxide solution was added to 60 mL of water. Determine
the concentration of the diluted solution.
Na(OH) + H(OH) = Na(OH) + H(OH)
V1 = 20 mL = 0.02 L
C1 = 0.40 M
V2 = 60 mL = 0.06 L
C2 ?
C1V1 = C2V2
C2 = C1V1 / V2
= (0.40)(0.02) / 0.06
= 0.13 mol/L or 0.13 M
8. 100 mL of 0.800 M hydrochloric acid are reacted with 0.65 M potassium hydroxide
solution until the resulting solution is just neutral.
a) What is the volume of the potassium hydroxide that was added?
HCl + K(OH) -> KCl + H(OH)
V2 = 100 mL = 0.1 L
C2 = 0.800 M
C1 = 0.65 M
V1 = ?
C1V1 = C2V2
V1 = C2V2 / C1
= (0.800)(0.1) / 0.65
= 0.12 L
b) What is the pH of the resulting solution? 7
9. 12.5 mL of 0.500 M nitric acid was neutralized by 50.0 mL of a solution of sodium
hydroxide. What is the concentration of the base?
Na(OH) + HNO3 -> H(OH) + Na(NO3)
C2 = 0.500 M
V2 = 12.5 mL = 0.0125 L
V1 = 50.0 = 0.05 L
C1 = ?
C1V1 = C2V2
C1 = C2V2 / V1
= (0.500)(0.0125) / 0.05
= 0.13 mol/L or 0.13 M
10. 32 mL of 0.10 M of sodium hydroxide exactly neutralized 25 mL of a solution of
sulfuric acid. What is the concentration of the acid?
Na(OH) + H2SO4 -> H(OH) + NaSO4
V1 = 32 mL = 0.032 L
C1 = 0.10 M
V2 = 25 mL = 0.025
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (0.10)(0.032) / 0.025
= 0.13 g/mol or 0.13 M
11. 13.7 mL of 0.065 M nitric acid was required to neutralize 25.0 mL of potassium
hydroxide solution. What is the molarity of the base?
HNO3 + K(OH) -> H(OH) + K(NO3)
V1 = 13.7 mL = 0.0137 L
C1 = 0.065 M
V2 = 25.0 mL = 0.025 L
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (0.065)(0.0137) / 0.025
= 0.036 mol/L or 0.036 M
12. 17.8 mL of 0.124 M solution of sulfuric acid exactly neutralized 50.0 mL of sodium
hydroxide solution. Calculate the concentration of the base.
Na(OH) + H2SO4 -> H(OH) + Na(SO4)
V1 = 17.8 mL = 0.0178 L
C1 = 0.124 M
V2 = 50.0 mL = 0.05 L
C2 = ?
C1V1 = C2V2
C2 = C1V1 / V2
= (0.124)(0.0178) / 0.05
= 0.044 mol/L or 0.044 M
13. Making a KOH solution required 1.4 g of solute to be dissolved in 12 L of solvent.
n = m / MM
= 1.4 / 56.09
= 0.025 mol
C = n / V
= 0.025 / 12
= 0.002 mol/L or 0.002
14. Zinc reacts with carbonic acid to produce zinc carbonate and hydrogen gas. Suppose
you wanted to obtain 4.0 g of hydrogen gas. What volume of 2.5 M carbonic acid would
you need to use with excess zinc?