Acid and Base Review Game

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Acid and Base
Review Game
Chemistry
Name the Acid
 HNO2

Nitrous Acid
 H2SO4

Sulfuric Acid
 HCl

Hydrochloric Acid
 HBr

Hydrobromic Acid
 HClO

Hypochlorous Acid
Name the Base
 NaOH
 Sodium Hydroxide
 Ca(OH)2
 Calcium Hydroxide
 Mg(OH)2
 Magnesium Hydroxide
 KOH
 Potassium Hydroxide
Write the Formula
 Nitric Acid
 HNO3
 Hydroiodidic Acid
 HI
 Phosphoric Acid
 H3PO4
 Acetic Acid
 HC2H3O2
Write the Formula
Sodium carbonate
 Na2CO3
Barium hydroxide
 Ba(OH)2
Lithium hydroxide
 LiOH
Arrhenius
 What is the definition of an
acid?
 A substance which produces H+ when put in water
 What is the definition of a
base?
 A substance which produces OH- when put in water
 If a soap has a hydrogen
ion concentration of
-6
2.0 x 10 M, what is the pH
of the solution? The pOH?
Is it an acid or a base?
 [H+] = 2 x 10-6M pH = -log(2 x 10-6) = 5.7
 pOH = 14 – 5.7 = 8.3
 Acid because pH < 7
 What is the hydrogen ion
concentration and hydroxide
ion concentration of a solution
with a pH of 2.3? Is the
substance an acid or base?




[H+] = 10-2.3 = 0.005 M
pOH = 14 – 2.3 = 11.7
[OH-] = 10-11.7 = 2 x 10-12 M
Acid because pH < 7
 What is the hydrogen ion
concentration of a sample of
phosphoric acid that has a pH of
4.9? What is the concentration
of phosphoric acid?
 [H+] = 10-4.9 = 1.3 x 10-5M
 Since there are 3 H’s in each H3PO4 … the concentration of phosphoric
acid would be a third of the concentration of H+
 1.3 x 10-5/3 = 4.2 x 10-6 M
Neutralization Reaction
What is the
neutralization reaction
for nitrous acid reacting
with potassium
hydroxide?
 HNO2 + KOH  KNO2 + H2O
Neutralization Reaction
 Write the complete
balanced equation for
the neutralization of
phosphoric acid with
calcium hydroxide
 2H3PO4 + 3Ca(OH)2  Ca3(PO4)2 + 6H2O
 How many grams of copper
(II) sulfate pentahydrate will
be needed to make 75 mL
of a 0.250 M solution?
 CM = n/V
 .25 = n/0.075
 n = 0.01875 mol x 249.5 = 4.7 g
How many kilograms of
sucrose C12H22O11 will
be needed to make 3.50
L of a 1.15 M solution?
 CM = n/v
 1.15 = n/3.5
 n = 4.025 mol x 342 = 1376.55 g / 1000 = 1.38 kg
What is the pH of a
solution made with
0.15 grams of barium
hydroxide in 4500 mL
of water?
 What is the molar
concentration of a solution
that contains 0.0750 moles
of NaHCO3 in a volume of
115 mL
 CM = n/v
 CM = 0.075/.115 = 0.65 M
 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?




CM = n/V
CM = 3/2 = 1.5M
1.5 = n/0.35
n = 0.53 mol
 Describe in your own words
how you would prepare 1.00L
of a 0.85 M solution of formic
acid HCO2H?




CM = n/v
0.85 = n/1
n = 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 normality?
 5.4g/98 = 0.055 mol
 CM = 0.055/0.25 = 0.22 M
 N = (eq)M = 2 x 0.22 = 0.44 N
 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
normality?
 CM = 0.94/0.45 = 2.1 M
 N = (eq)M = 1 x 2.1 = 2.1 N
 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?
 NaVa = NbVb
 (0.4)(50) = Nb(35)
 Nb = 0.57 N / 1 = 0.57 M
 Calculate how many
milliliters of 0.25 M
Ba(OH)2 must be added to
titrate 46 mL of 0.40 M
HClO4
 NaVa = NbVb
 0.4(46) = 0.5Vb
 Vb = 36.8 mL
 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) = Na(21.2)
 Na = 0.157 N / 1 = 0.157 M
What is the molarity
of a solution of
phosphoric acid with
a pH of 2.65?
 10-2.65 = 0.0022M / 3 = 7.62 x 10-4 M
 A 20 mL sample of an HCl
solution was titrated with
27.4 mL of a standard
solution of Ba(OH)2. The
concentration of the
standard is 0.0154 M. What
is the molarity of the HCl?
 (0.0308)(27.4) = Na(20)
 Na = 0.042N / 1 = 0.042M
Buffer
What is a buffer?
 A solution that resists changes in pH
Buffer
What are buffers
made of? Where can
you find them?
 Weak acid and its conjugate base
 Weak base and its conjugate acid
 You can find them in your BLOOD
 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?
 4.5(20) = M(60)
 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?
 1.5 (450) = M (300)
 2.25 M
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?
 50 g K2CrO4 / 194 = 0.26 mol
 1.4 = .26/v
 V = 0.184 L
What is the pH of a
solution made by putting
6.0 grams of phosphoric
acid in 250 L of water?




6 g H3PO4 / 98 = 0.0612 mol
[H3PO4] = 0.0612 / 250 = 2.45 x 10-4
[H+] = 3(2.45 x 10-4) = 7.35 x 10-4
pH = -log(7.35 x 10-4) = 3.13
 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

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