Name _____Mr. Perfect_____________________________ Date ____Sp 17________
1. A 0.50 M solution of hydrazoic acid, HN3, is 99.3 % associated in water. Calculate the
acid ionization constant (Ka) for this acid. (10 pts)
I
C
E
Dissociation
0.7 % =
HN3
0.50
-x
0.50-x
𝑥
× 100
0.50
⇄ H+ + N30
0
+x
+x
x
x
𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 =
0.7%
× 0.50 = 3.5𝑥10−3 𝑀 = [𝐻 + ]
100
Plug in and solve for Ka
𝑥2
(3.5𝑥10−3 )2
𝐾𝑎 =
=
= 𝟐. 𝟓𝒙𝟏𝟎−𝟓
−3
0.50 − 𝑥 (0.50 − 3.5𝑥10 )
2. At 25 °C, the solubility of Lead(II) Chloride is 3.96 g/L. Calculate the solubility
product constant (Ksp) of Lead(II) Chloride at this temperature. (5 pts)
Mw = 277 g/mol
I
C
E
PbCl2(s) ⇄ Pb2+(aq) + 2Cl-(aq)
--0
0
--+s
+s
--s
2s
Solve for molar solubility
3.96
𝑔
𝑚𝑜𝑙
×
= 1.43𝑥10−2 𝑀
𝐿
277 𝑔
Ksp = s(2s)2 = 4s3 = 4(1.43𝑥10−2)3 = 1.17 x 10-5
3. Calculate the solubility in g per L of a solution of Barium Sulfate. (5 pts)
Ksp = 1.07 x 10-10
Mw = 233 g/mol
I
C
E
Ksp = s2
or
BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq)
--0
0
--+s
+s
--s
s
1.07 x 10-10 = s2
2
(𝑚𝑜𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑏𝑖𝑙𝑖𝑡𝑦) 𝑠 = √1.07𝑥10−10 = 1.03𝑥10−5 𝑀
Solubility = 1.03x10-5 mol/L x 233 g/mol = 2.41x10-3 g/L
Chemistry 102 Exam 2
Name _____Mr. Perfect_____________________________ Date ____Sp 17________
4. Calculate the pH of 10.0 mL of a 0.25 M weak-base, nicotine (C10H14N2), solution
upon titration with the following increments of a 0.15 M strong-acid, HCl: (20 pts)
Ka = 1.0 x 10-8
Weak Base: 0.010 L x 0.25 mol/L = 0.0025 mol base
a) 0.00 mL of acid (before addition = ICE Box)
A- + H2O ⇄ HA + OHI 0.25
--0
0
C -x
--+x
+x
E 0.25-x
x
x
−14
𝐾𝑤 1𝑥10
𝐾𝑏 =
=
= 1𝑥10−6
𝐾𝑎
1𝑥10−8
[𝐻𝐴][𝑂𝐻 − ]
𝐾𝑏 =
[𝐴− ]
𝑜𝑟
1𝑥10
−6
𝑥2
𝑥2
=
≈
0.25 − 𝑥 0.25
x = 5.0 x 10-4 = [OH-]
pH = 14 – 3.30 = 10.70
pOH = -log(5.0 x 10-4) = 3.30
b) 10.00 mL of acid (Before equivalence point)
0.010 L x 0.150 mol/L = 0.0015 mol acid = [HA]
0.0025 mol base – 0.0015 mol acid = 0.0010 mol base remains = [A-]
Use the Henderson-Hasselbach equation:
𝑝𝐻 = 8.00 + 𝑙𝑜𝑔
(0.0010)
= 𝟕. 𝟖𝟐
(0.0015)
b) 16.67 mL of acid (at equivalence point)
0.01667 L x 0.15 mol/L = 0.0025 mol acid ÷ 0.02667 L = 9.37 x 10-2 M = [HA]
I
C
E
HA
⇄
9.37x10-2
-x
9.37x10-2-x
H+ + A0
0
+x
+x
x
x
𝐾𝑎 = 1.0𝑥10−8 ≈
x = 3.06 x 10-5 M = [H+]
𝑥2
9.37𝑥10−2
pH = -log(3.06 x 10-5) = 4.51
Chemistry 102 Exam 2
Name _____Mr. Perfect_____________________________ Date ____Sp 17________
5. Calculate the pH of a solution containing 0.15 M HC2H3O2, acetic acid, and 0.25 M
NaC2H3O2, sodium acetate. Under these conditions, would this solution act as a buffer?
Clearly state the buffer zone for this weak acid. (10 pts)
Ka = 1.8 x 10-5
pKa = -log(1.8 x 10-5) = 4.74
Buffer zone = 3.74 – 5.74
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔
[𝐴− ]
(0.25)
= 4.74 + log
= 𝟒. 𝟗𝟔
[𝐻𝐴]
(0.15)
Yes, this mixture would result in a buffer solution.
6. Calculate the pH of a 0.25 M NaIO solution. (10 pts)
Ka = 2.3 x 10-11
I
C
E
IO- + H2O ⇄ HIO + OH0.25
--0
0
-x
--+x
+x
0.25 – x
x
x
𝐾𝑤
1.0𝑥10−14
𝑥2
𝑥2
−4
=
=
4.34𝑥10
𝐾
=
≈
𝑏
𝐾𝑎
2.3𝑥10−11
0.25 − 𝑥 0.25
check
x = 1.04x10-2 M = [OH-]
(1.04x10-2/0.25) x 100 = 4.17% assumption is valid
𝐾𝑏 =
pOH = -log(1.04x10-2) = 1.98 pH = 14 – 1.98 = 12.02
7. A solution containing the following indicator, with a Ka of 0.02, turns orange at the
equivalence point of a titration. Determine the range of pH values this indicator would be
useful. Could this indicator be useful for a weak base-strong acid titration? Clearly state
why or why not? (10 pts)
HIn ⇄ H+ + In(Red)
(Yellow)
pKa = -log(0.02) =1.7
Appropriate Indicator range = 0.7 -2.7 = buffer zone for the indicator
For a weak base:
HA ⇄ H+ + A-
at the equivalence point
The equivalence point would be below 7, thus this is an appropriate indicator for a weak
base-strong acid titration.
Chemistry 102 Exam 2
Name _____Mr. Perfect_____________________________ Date ____Sp 17________
8. For the following reactions, predict whether a decrease in temperature would be
favorable or unfavorable. (10 pts)
∆G = ∆H - T∆S
(∆G < 0 is favorable)
a) 2H2O2(aq) → 2H2O(l) + O2(g)
+∆S unfavorable
b) PCl3(l) + Cl2(g) → PCl5(s)
-∆S
favorable
c) CO2(s) → CO2(g)
+∆S
unfavorable
d) H2O(g) → H2O(l)
-∆S
favorable
9. Complete the following table. (10 pts)
[H+]
[OH-]
1.91 x 10-5
5.25 x 10-10
pH
4.72
Acid or Base?
Acid
6.7 x 10-3
1.49 x 10-12
2.17
acid
6.3 x 10-9
1.59 x 10-6
8.2
base
3.1 x 10-8
3.2 x 10-7
7.51
base
10. Predict if a precipitate will form from when 125 mL of 0.375 M Ca(NO3)2 is mixed
with 245 mL of 0.255 M NaF. Show all work to receive full credit. The Ksp for CaF2 is
3.2 x 10-11 (10 pts)
Ca(NO3)2(aq) + 2NaF(aq) → CaF2(s) + 2NaNO3(aq)
CaF2(s) ⇄ Ca+2(aq) + 2F-(aq)
[𝐶𝑎+2 ] =
[𝐹 − ] =
(0.125 𝐿)(0.375 𝑀)
= 1.27 𝑥10−1 𝑀
(0.125 𝐿 + 0.245 𝐿)
(0.245 𝐿)(0.255 𝑀)
= 1.62 𝑥10−1 𝑀
(0.125 𝐿 + 0.245 𝐿)
Qsp = [Ca+2][F-]2 = (1.27 x 10-1)(1.62 x 10-1)2 = 3.33 x 10-3
Qsp (3.33 x 10-3) > Ksp (3.2 x 10-11) ; therefore a precipitation will form.
Supersaturated
11. (Extra Credit) Predict the effect on the solubility (if any) for the following reaction
upon addition of a base. (5 pts)
Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq)
Addition of a base would add excess hydroxide to the products. The system will shift the
equilibrium towards the reactants (shift to the left) in order to accommodate this change.
“Le Châtlier’s Principle”
Chemistry 102 Exam 2