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?

Zn + H2CO3 -> Zn(CO3) + H2

V2 = 4.0 g = 0.004 L

H2: MM = 2.02 g/mol

n = m / MM

= 4.0 / 2.02

= 1.98 mol

C2 = n / V

= 1.98 / 0.004

= 495 M

C1 = 2.5 M

V1 = ?

C1V1 = C2V2

V1 = C2V2 / C1

= (495)(0.004) / 2.5

= 0.792 L