Solutions & Acid
and Base Review
Game
Chemistry
Name the Acid
 HBr
Name the Acid
 HBr

Hydrobromic Acid
Write the Formula
Lithium hydroxide
Write the Formula
Lithium hydroxide
 LiOH
Name the Base
 Mg(OH)2
Name the Base
 Mg(OH)2
 Magnesium Hydroxide
Properties
 What is the “technical”
definition of an acid in
regards to ?
Arrhenius
 What is the “technical”
definition of an acid in
regards to ?
 A substance which contains more H+ than OH-
Write the Formula
 Hydroiodidic Acid
Write the Formula
 Hydroiodidic Acid
 HI
Properties
 What is the “technical”
definition of a base
Arrhenius
 What is the “technical”
definition of a base?
 A substance which contains more OH- than H+
Name the Base
 KOH
Name the Base
 KOH
 Potassium Hydroxide
 If a soap has a hydrogen
ion concentration of
-6
2.0 x 10 M, what is the pH
of the solution? Is it an acid
or a base?
 If a soap has a hydrogen
ion concentration of
-6
2.0 x 10 M, what is the pH
of the solution? Is it an acid
or a base?
 [H+] = 2 x 10-6M pH = -log(2 x 10-6) = 5.7
 Acid because pH < 7
 What is the hydroxide ion
concentration of a solution
with a pH of 2.3? Is the
substance an acid or base?
 What is the hydroxide ion
concentration of a solution
with a pH of 2.3? Is the
substance an acid or base?
 pOH = 14 – 2.3 = 11.7
 [OH-] = 10-11.7 = 2 x 10-12 M
 Acid because pH < 7
Name the Acid
 H2SO4
Name the Acid
 H2SO4

Sulfuric Acid
 What is the hydrogen ion
concentration of a sample of
phosphoric acid that has a pH of
4.9?
 What is the hydrogen ion
concentration of a sample of
phosphoric acid that has a pH of
4.9?
 [H+] = 10-4.9 = 1.3 x 10-5M
Neutralization Reaction
Write the balanced
neutralization reaction
for nitrous acid reacting
with potassium
hydroxide?
Neutralization Reaction
Write the balanced
neutralization reaction
for nitrous acid reacting
with potassium
hydroxide?
 HNO2 + KOH  KNO2 + H2O
Name the Base
 Ca(OH)2
Name the Base
 Ca(OH)2
 Calcium Hydroxide
Neutralization Reaction
 Write the complete
balanced equation for
the neutralization of
nitric acid with calcium
hydroxide
Neutralization Reaction
 Write the complete
balanced equation for
the neutralization of
phosphoric acid with
calcium hydroxide
 2HNO3 + Ca(OH)2  Ca(NO3)2 + 2H2O
 How many grams of copper
(II) sulfate pentahydrate will
be needed to make 75 mL
of a 0.250 M solution? (The
molar mass of copper (II)
sulfate pentahydrate is
249.5 g/mole)
 What mass of copper (II)
sulfate pentahydrate will be
needed to make 75 mL of a
0.250 M solution? (The
molar mass of copper (II)
sulfate pentahydrate is
249.5 g/mole)
 M = moles/liters
 .25 = moles/0.075
 n = 0.01875 mol x 249.5 = 4.7 g
Name the Acid
 HNO2
Name the Acid
 HNO2

Nitrous Acid
How many grams of
sucrose, C12H22O11 ,will
be needed to make 3.50
L of a 1.15 M solution?
How many kilograms of
sucrose, C12H22O11 ,will
be needed to make 3.50
L of a 1.15 M solution?
 M = moles/liters
 1.15 = moles/3.5
 n = 4.025 mol x 342 = 1376.55 g
What is the pH of a
solution made with
0.15 grams of sodium
hydroxide in 4500 mL
of water?
 What is the pH of a solution made with
0.15 grams of sodium hydroxide in
4500 mL of water?
 NaOH
 0.15 g / 40 = 0.00375 moles
 M = 0.00375/4.5 = 8.33X10-4 [OH-]
 pOH = -log (8.33x10-4)
 pOH = 3.08
 pH = 14-3.08 = 10.92
Write the Formula
 Sulfuric Acid
Write the Formula
 Phosphoric Acid
 H2SO4
 What is the molarity of a
solution that contains
0.0750 moles of NaHCO3
in a volume of 115 mL
 What is the molarity of a
solution that contains
0.0750 moles of NaHCO3
in a volume of 115 mL
 M = moles/liters
 M = 0.075/.115 = 0.65 M
Write the Formula
 Nitric Acid
Write the Formula
 Nitric Acid
 HNO3
 A solution has 3.00 moles of
solute in 2.00 L of solution,
what is its molar
concentration? How many
moles would there be in 350
mL of solution?
 A solution has 3.00 moles of
solute in 2.00 L of solution,
what is its molar
concentration? How many
moles would there be in 350
mL of solution?




M = moles/liters
M = 3/2 = 1.5M
1.5 = moles/0.35
moles = 0.53 mol
Write the Formula
Sodium carbonate
Write the Formula
Sodium carbonate
 Na2CO3
 Describe how you would
prepare 1.00L of a 0.85 M
solution of formic acid
HCO2H?
 Describe how you would
prepare 1.00L of a 0.85 M
solution of formic acid
HCO2H?




M = moles/liters
0.85 = moles/1
moles = 0.85 mol x 46 = 39.1 g
Take 39.1 grams of formic acid and dissolve in a little bit of
water. Put into a 1.00 L volumetric flask and fill to the line with
water
 What is the molarity of a
sulfuric acid solution which
contains 5.4 grams of sulfuric
acid in 250 mL of water?
 What is the molarity of a
sulfuric acid solution which
contains 5.4 grams of sulfuric
acid in 250 mL of water?

5.4g/98 = 0.055 mol
 M = 0.055/0.25 = 0.22 M
 What is the molarity of a
potassium hydroxide solution
which contains 0.94 moles of
potassium hydroxide in 450
mL of water?
 What is the molarity of a
potassium hydroxide solution
which contains 0.94 moles of
potassium hydroxide in 450
mL of water?
 M = 0.94/0.45 = 2.1 M
Write the Formula
Barium hydroxide
Write the Formula
Barium hydroxide
 Ba(OH)2
 In the titration of 35 mL of liquid
drain cleaner containing NaOH,
50 mL of 0.4M HCl must be
added to reach the equivalence
point. What is the molarity of the
base in the cleaner?
 In the titration of 35 mL of liquid
drain cleaner containing NaOH,
50 mL of 0.4M HCl must be
added to reach the equivalence
point. What is the molarity of the
base in the cleaner?
 MaVa = MbVb
 (0.4)(50) = Mb(35)
0.57 M
 Calculate how many
milliliters of 0.50M NaOH
must be added to titrate 46
mL of 0.40 M HClO4
 Calculate how many
milliliters of 0.50M NaOH
must be added to titrate 46
mL of 0.40 M HClO4
 MaVa = MbVb
 0.4(46) = 0.5Vb
 Vb = 36.8 mL
Name the Base
 NaOH
Name the Base
 NaOH
 Sodium Hydroxide
 A 15.5 mL sample of 0.215 M
KOH was titrated with an acetic
acid solution, It took 21.2 mL of
the acid to reach the equivalence
point. What is the molarity of the
acetic acid?
 A 15.5 mL sample of 0.215 M
KOH was titrated with an acetic
acid solution, It took 21.2 mL of
the acid to reach the equivalence
point. What is the molarity of the
acetic acid?
 (0.215)(15.5) = Ma(21.2)
 0.157 M
What is the molarity
of a solution of acetic
acid with a pH of
2.65?
What is the molarity
of a solution of acetic
acid with a pH of
2.65?
 Molarity of Acid = [H+]
 10-2.65 = 0.0022M
 A 20 mL sample of an HCl
solution was titrated with
27.4 mL of a standard
solution of NaOH. The
concentration of the
standard is 0.0154 M. What
is the molarity of the HCl?
 A 20 mL sample of an HCl
solution was titrated with
27.4 mL of a standard
solution of NaOH. The
concentration of the
standard is 0.0154 M. What
is the molarity of the HCl?
 (0.0154)(27.4) = Ma(20)
 = 0.021 M
 I want to dilute 20 mL of a 6M
solution of acetic acid to a
3.8M solution of acetic acid.
How much water should I add
to the 6M acetic acid to
achieve this?
 I want to dilute 20 mL of a 6M
solution of acetic acid to a
3.8M solution of acetic acid.
How much water should I add
to the 6M acetic acid to
achieve this?
 6(20) = 3.8V
 V = 31.6 mL
 ADD 11.6 mL of water
 I mix 20 mL of 4.5M
NaCl with 40 mL of
water. What is the new
concentration of the
NaCl?
 I mix 20 mL of 4.5M
NaCl with 40 mL of
water. What is the new
concentration of the
NaCl?
 4.5(20) = M(60) Note: final volume is 20+40 mL
water added
 M = 1.5 M
 A 450 mL solution of 1.5 M
HCl is sat out over night. 150
mL of the water evaporated.
What is the new
concentration of the HCl?
 A 450 mL solution of 1.5 M
HCl is sat out over night. 150
mL of the water evaporated.
What is the new
concentration of the HCl?
 1.5 (450) = M (300) NOTE: final vol is 450 – 150 evaporated = 300
 2.25 M
What does the term
“strong acid” mean?
What does the term
“strong acid” mean?
 Strong acid means that the acid
dissociates completely in water
 I used 50 grams of
potassium chromate to
make a 1.4M solution.
What is the volume of the
solution?
 I used 50 grams of
potassium chromate to
make a 1.4M solution.
What is the volume of the
solution?
 50 g K2CrO4 / 194 = 0.26 mol
 1.4 = .26/liters
 V = 0.184 L
Write the Formula
 Acetic Acid
Write the Formula
 Acetic Acid
 HC2H3O2
Name the Acid
 HCl
Name the Acid
 HCl

Hydrochloric Acid
 What is the pH of a solution
made by putting 3.25 grams
of strontium hydroxide in
5000 L of water?
 What is the pH of a solution
made by putting 3.25 grams
of strontium hydroxide in
5000 L of water?





3.25 g Sr(OH)2 / 121.6 = 0.0267 mol
[Sr(OH)2] = 0.0267/ 5000 = 5.34 x 10-6M
[OH-] = 2(5.34 x 10-6) = 1.07 x 10-5 M
pOH = -log(1.07 x 10-5) = 4.97
pH = 14 – 4.97 = 9.03