Chemistry 213 s2i
Wednesday, March 13th
Katie Seehusen
[email protected]
Agenda:
1. New Material
2. Test Review!
What’s on this test?
• Identify acids and bases and their conjugate acid/base pairs
• What are the strong acids, what are the strong bases?
• Calculating the pH of a strong acid or base
• Calculating the pH of a weak acid or base
• Determining whether a salt solution will be acidic, basic, or neutral
• Calculating pH when two solutions have a common ion
• Calculating the pH of a buffer system
• Solubility Product
First, let’s go over the new concept, solubility product, or Ksp. This is for ionic substances (salts)
that do NOT completely dissociate. Hint: If you are given a Ksp value, this means it does not
completely dissociate.
1. If the solid CuCl (s) à Cu2+ + Cl- has a Ksp of 1.72 x 10-7, what is the solubility of the
substance?
Steps:
[!"#$%&'(]
a) Write out the Ksp expression ([!"#$%#&%']).
Ksp= [Cu2+][Cl-] **solids not included!
b) Write out the equilibrium table.
CuCl
-
⇔
Cu2+
0
+x
x
c) Solve for x. This is your solubility in M.
1.72 x 10-7 = x2
x= 4.15 x 10-4 M
Cl0
+x
x
Let’s do another:
2. If the solubility of FeS is 8.94 x 10-10 M, calculate the Ksp.
FeS
-
⇔
Fe2+
0
+x
x
S20
+x
x
X= 0.42
Ksp= x2 = (8.94 x 10-10)2= 8.0 x 10-19
Now, let’s review for the test!
We’re going to do a large problem that incorporates all of the acid/base concepts.
3. a) If the acid HCN is put into solution, write out the dissociation of the acid including H2O,
and identify the acid, base, conjugate acid, and conjugate base.
HCN + H2O ⇔ H3O+ + CNHCN= acid
H2O= base
H3O+= conjugate acid
CN-= conjugate base
b) Write the Ka expression
𝐾𝑎 =
𝐶𝑁 − [𝐻3𝑂+]
[𝐻𝐶𝑁]
c) If the initial concentration of HCN= 0.44 M in solution, what will the pH of the solution be?
Ka= 6.2 x 10-10. Think: is this a strong acid?
HCN
0.44
-x
0.44-x
H2 O
-
⇔
H3O+
0
+x
x
CN0
+x
x
𝐶𝑁 − [𝐻3𝑂+]
𝑥!
𝐾𝑎 =
= [𝐻𝐶𝑁]
0.44 − 𝑥
6.2 x 10-10= x2/0.44
x=1.65 x 10-5
x is equal to [H3O+]
pH= -log (1.65 x 10-5)= 4.78
d) A new solution, KCN, is prepared. Will this salt solution be acidic, basic, or neutral?
KCN à K+ + CNH+ + OHso KOH and HCN will form. Because KOH is a strong base, the solution of just KCN in water
will be basic.
e) If a 0.21 M solution of KCN was added to the solution of HCN above, what will the pH be?
Solve this using the equilibrium table.
HCN
0.44
-x
0.44-x
𝐾𝑎 =
H2 O
-
H3O+
0
+x
x
⇔
CN0.21
+x
0.21 + x
𝐶𝑁 − [𝐻3𝑂+]
(𝑥)(0.21 + 𝑥) 0.21𝑥
= =
[𝐻𝐶𝑁]
0.44 − 𝑥
0.44
x= 1.30 x 10-9 so pH= 8.89
f) Now, using the new equation 𝑝𝐻 = 𝑝𝐾𝑎 + log
!!
!"
, calculate the pH and make sure they are
the same.
pKa= -log (Ka)= -log (6.2 x 10-10)= 9.20
𝐴!
𝑝𝐻 = 𝑝𝐾𝑎 + log
𝐻𝐴
. 21
. 44
𝑝𝐻 = 8.89 **Good because this is the same answer as part e!!!
𝑝𝐻 = 9.20 + log
f) What is the pH of a solution of 0.11 M NaOH? Think: is this weak or strong?
pOH= -log (.11)= 0.96
pH= 14-.96= 13.04
g) What will the pH be of the solution with KCN and HCN (letter e) be if the solution of 0.11 M
NaOH is added (assume solution volume is 1 L)? Hint: you will have to use an equilibrium table
!!
AND use the new equation 𝑝𝐻 = 𝑝𝐾𝑎 + log !"
HCN
0.44
-0.11
0.33
OH0.11
-0.11
0
𝑝𝐻 = 9.20 + log
. 32
. 33
⇔
H2O
-
CN0.21
+0.11
0.32
pH= 9.19
The point of parts e through g were to see that even though the solution of just NaOH had a pH
of 13.04, the pH of the buffer system only went up by 0.3 (9.19-8.89). This shows that the
system resisted a change in pH
Now, let’s go over some of the problems from his previous year exam.
1. The pH of a 0.10 M solution of a weak base is 9.82. What is the Kb for this base?
A) 2.1 × 10-4
B) 8.8 × 10-8
C) 6.6 × 10-4
D) 4.4 × 10-8
E) 2.0 × 10-5
B
0.10
-x
0.10-x
B + H2O ⇔ HB+ + OH-
H2 O
-
⇔
pH= 9.82 so pOH= 14-9.82= 4.18
[OH]= 10-4.18= 6.6 x 10-5 M à This is what x is equal to!
Kb= x2/0.10= (6.6 x 10-5)2/0.10= 4.4 x 10-8
HB+
0
+x
x
OH0
+x
x
2. A substance that is capable of acting as both an acid and as a base is __________.
A) miscible
B) amphoteric
C) saturated
D) conjugated
E) autosomal
3. Nitric acid is a strong acid. This means that __________.
A) HNO3 cannot be neutralized by a weak base
B) aqueous solutions of HNO3 contain equal concentrations of H+(aq) and OH-(aq)
C) HNO3 does not dissociate at all when it is dissolved in water
D) HNO3 dissociates completely to H+(aq) and NO3(aq) when it dissolves in water
E) HNO3 produces a gaseous product when it is neutralized
4. At equilibrium, __________.
A) the rates of the forward and reverse reactions are equal
B) the rate constants of the forward and reverse reactions are equal
C) all chemical reactions have ceased
D) the value of the equilibrium constant is 1
E) the limiting reagent has been consumed
5. An aqueous solution contains 0.10 M NaOH. The solution is __________.
A) highly colored
B) neutral
C) very dilute D) basic
E) acidic
6. What is the concentration (in M) of hydronium ions in a solution at 25.0 °C with pH = 4.282?
A) 1.66 × 104
B) 9.71
C) 5.22 × 10-5
D) 4.28
E) 1.92 × 10-10
+
-4.282
[H3O ]= 10
7. Of the following solutions, which has the greatest buffering capacity?
A) 0.543 M NH3 and 0.555 M NH4Cl Because concentration of conjugate acid/base is high
B) They are all buffer solutions and would all have the same capacity.
C) 0.234 M NH3 and 0.100 M NH4Cl
D) 0.100 M NH3 and 0.455 M NH4Cl
E) 0.087 M NH3 and 0.088 M NH4Cl
8. A 0.1 M solution of __________ has a pH of 7.0.
A) NH4Cl
B) KF
C) NaNO3
D) NaF
E) Na2S
Because a solution of NaNO3 in water will form NaOH (a strong base) and HNO3 (a strong acid).
Since both are strong, the solution will be neutral, or of pH= 7.0