Name
Period
Date
Pre-AP Chemistry
Homework: Acids & Bases Problem Types
1. Provide a short summary of each problem type and work out the answer to the example.
Problem Type: Word
Equation(s)
Example
Answer
summary of what this
problem is
M1V1=M2V2
Dilution:
M1V1=M2V2
2.0 mL 0.50 M
(0.50)(2.0) = M2(40)
Water is added or
acid is diluted to
(0.50)(2.0)/(40) = 0.025 M
removed from one
a new volume of
chemical to change its
40.0 mL.
concentration. Often the
Calculate the
first step in a larger
new
problem.
concentration.
Concentration
pH=–log [H3O+1]
Calculate the pH pH = –log(0.0677) = 1.16941 = 1.17
Conversions:
pOH=–log[OH–1]
of a solution
Numbers for pH, pOH,
pOH+pH=14
with a [H3O+1]
+1
–1
-14
hydronium ion or
Kwater=[H3O ] [OH ]=10
of 0.0677 M.
hydroxide ion are used to
determine the other
values. Usually a small
part of larger problem
Disassociation: A
weak acid (or weak
base) dissolves or
“comes apart” in water
A single type of acid or
base reacts with water.
• HA+H2O⇄A–1+H3O+1
(acid)
•K=
0.10 M HCN
dissolves in
water. Calculate
the H3O+1
concentration.
Ka= 6.17×10-10
[A -1 ][H 3O+1 ]
[HA]
• BOH ⇄B+ + OH–1 (base)
[B+1 ][OH –1 ]
•K=
[BOH]
Neutralization: An
acid is mixed with a
base to form water.
MbVb=MaVa
We will focus on
strong acid + strong
base in this class.
2.
K=[CN–1][ H3O+1]/[HCN]
6.17×10-10= x2/0.10
(0.10)(6.17 ×10 -10 ) =x
2.803×10-3= x
[ H3O+1] = 2.8×10-3 M
HA+BOH = H2O + AB
Acid Coefficient
×
Base Coefficient
HCN(aq) + H2O(l) ⇄CN–1(aq) + H3O+1(aq)
0.10M
x
x
7.57 mL of 0.25
M KOH
neutralizes 20.0
mL of HCl.
Calculate the
concentration of
HCl
(1/1) • (0.25)(7.57) = Ma(20.0)
(0.25)(7.57)/(20) = Macid
0.095 M = Macid
How is a weak acid different from a strong acid when placed in water? How is the pH different?
A weak acid only partially breaks up in water to make some hydronium ions and some anions. It has a equilibrium constant value (K) which
shows the ratio of products (separate ions) to reactants (whole acid). Since it only partially separates, this K value is usually small. A strong acid
completely breaks up into hydronium ions and anions. Its K value is not given because it is just “really big.” Whatever the starting amount is of
acid, this instantly becomes the hydronium ion concentration, without much calculating. The weak acid process use a double-arrow, ⇄, while a
strong acid process uses a single arrow, à. 3.
a.
50.0 mL of 0.80 M Sulfurous acid (H2S) reacts with 10.0 mL of strong base, NaOH to form S–2 ions.
Write the balanced and formatted equation for this process.
H2S(aq) + 2NaOH(aq) ⇄ Na2S(aq) + 2H2O(l)
b.
______________________ What acid-base “problem type” is this?
Neutralization
c.
______________________ Calculate the concentration of the sodium hydroxide.
(acid coeff/base coeff)• MbVb=MaVa à (1/2)(Mb)(10.0mL)=(0.80M)(50.0mL)à
Mb=(0.80)(50)(2)/10 = 8.0 M NaOH
Homework: Acid-Base Problem Types
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4. 0.020 moles of HBrO is dissolved in 0.350 L of water. Kacid=2.0 × 10–9
a.
Write the balanced and formatted equation for this process.
HBrO(aq) +
Acid
H2O(l) ⇄ BrO–1(aq) +
Base
Conj. Base
K = [ BrO-1][ H3O+1]/[HBrO]
2.0×10–9=x2/(0.057)
(0.057)( 2.0×10–9) = x2
1.1×10–5 = x
H3O+1(aq)
conj. acid
b.
c.
Label the acid, base, conjugate acid, and conjugate base for this process.
Complete the ICE chart for this.
HBrO
BrO-1
0.020 moles ÷ 0.35 L = 0.057 M
Initial Concentration
0
≈ 0 (too small to change 0.020)
x = 1.1×10–5
Change
0.057 M
x =1.1×10–5 M
Equilibrium concentration
d.
H3O+1
≈0
x =1.1×10–5
x =1.1×10–5 M
______________________ What general “problem type” is the above work?
Disassociation (weak acid dissolving in water)
e.
pH
Calculate each value concentration or p-value.
pOH
4.97 = 5.0
f.
[H3O+1]
[OH–1]
–5
9.0
9.1×10–10 M
1.1×10 M
______________________ What general “problem type” is the above work?
Concentration conversions
g.
Predict how the moles of each chemical would be different after the reaction re-establishes equilibrium for a given change
(↑, ↓, =)
Change:
HBrO
BrO-1
H3O+1
↑
↓
{↑}because it was added)
• A strong acid is added
Adds H3O+1; Forces reaction to left;
• Water is added
=
=
=
↓
↑
{↓}(because the H3O+1 was
removed)
=
=
=
Just dilutes what’s there; the moles of each don’t
change.
• NaOH is added
The OH– reacts with the H3O+1 essentially removing
it from the reaction. Forces reaction to the right to
make up for lost H3O+1
• The solution is poured into a smaller beaker
Just moves chemicals to new location.
5.
a.
b.
Complete, balance and format these chemical equations. Label the type of process shown.
neutralization ___________
NaOH(aq) + HF(aq) à
H2O(l) +
NaF(aq)
disassociation ___________
HPO4–2(aq) + H2O(l) ⇄
PO4–3(aq) +
H3O+1(aq)
…of a weak acid
c.
neutralization ___________
d.
disassociation ___________
e.
disassociation ___________
f.
disassociation ___________
H2SO4(aq) + 2LiOH(aq)à 2H2O(l)
KOH(aq) à
K+1(aq)
+ Li2SO4(aq)
+
OH–1(aq)
+
H3O+1(aq)
+
OH–1(aq)
… of a strong base
HIO(aq) + H2O(l) ⇄ IO–1(aq)
…of a weak acid
NH3 + H2O(l)
⇄ NH4+1(aq)
… of a weak base
6.
HCOOH has a Ka value of 1.77×10-4. With only this information, what can you infer a solution made with this compound?
It must be a weak acid. Only weak acids/bases have K values. The a part stands for acid. It means that only part of the compound breaks up
(disassociates) when dissolved in water. It would have a low pH (less than 7), but you’d have to actually calculate it to determine the value. (If it
were strong acid, the concentration of the acid would convert instantly to the concentration of H3O+1.)
7.
a.
Water can split into hydrogen ions (H+1) and hydroxide ions. It’s Kwater value is 1.0×10–14
Write the balanced and formatted equation for this process.
H2O(l) ⇄ H+1(aq) + OH–1(aq)
b.
______________________ Write the equilibrium expression for this process.
Kw = [H+1][OH–1] = 1.0×10–14; remember that liquid water is always left out of equilibrium expressions.
c.
______________________ Calculate the pH, given this information.
x=[H+1]=[OH–1];
x2 = 1.0×10–14,
Homework: Acid-Base Problem Types
x= 1.0×10–7,
pH =–log(1.0×10–7) = 7
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