Name ______Mr. Perfect_______________________________________ Date ____F 16____________
1. A 0.35 M solution of a mono-protic weak acid (HA) is 20 % ionized. Calculate the Ka for this
acid. (10 pts)
HA ⇄ H+ + A20 % =
𝑥
× 100
0.35
x = 0.07 M
𝐾𝑎 =
𝑥2
(0.07)2
=
= 𝟏. 𝟕𝟓 𝒙 𝟏𝟎−𝟐
0.35 − 𝑥 0.35 − 0.07
2. Calculate the pH of a 0.012 M benzoic acid, HC7H5O2, solution. Hint: benzoic acid is a monoprotic acid. (10 pts)
Ka = 6.5 x 10-5
HC7H5O2 ⇄ H+ + C7H5O2I
0.012
0
0
C
-x
+x
+x
E 0.012 – x
x
x
2
2
𝑥
𝑥
𝐾𝑎 = 6.5 𝑥 10−5 =
≈
𝑥 = 8.83 𝑥 10−4 𝑀 = [𝐻 +]
0.012 − 𝑥 0.012
8.83 𝑥 10−4
𝐶ℎ𝑒𝑐𝑘
𝑥100 = 7.4 % 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑠 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑!
0.012
Must Solve the Quadratic: x2 + 6.5 x 10-5x -7.8 x 10-7 = 0
−6.5 𝑥 10−5 ± √(6.5𝑥10−5 )2 − 4(1)(−7.8𝑥10−7 )
= 8.51 𝑥 10−4
2(1)
pH = -log(8.51x10-4) = 3.07
3. Calculate the pH of a 0.250 M HC5H5NCl salt solution. Hint: HC5H5N+ is the conjugate acid
of the weak base C5H5N. (10 pts)
𝑥=
Kb = 1.7 x 10-9
Strong Electrolyte (100 % dissociation)
HC5H5N+
I 0.25
C -x
E 0.25-x
𝐾𝑎 =
⇄ C5H5N + H+
0
0
+x
+x
x
x
2
𝑥
𝑥2
=
≈
0.25 − 𝑥 0.25
𝐾𝑤
1 𝑥 10−14
=
= 5.88 𝑥 10−6
𝐾𝑏 1.7 𝑥 10−9
1.21 𝑥 10−3
𝐶ℎ𝑒𝑐𝑘
× 100 = 0.48 %
0.25
𝑥 = 1.21 𝑥 10−3 𝑀 = [𝐻 +]
𝑝𝐻 = − log(1.21 𝑥 10−3 ) = 𝟐. 𝟗𝟏
Chemistry 102 Exam 2
Name ______Mr. Perfect_______________________________________ Date ____F 16____________
4. 4. Use the Henderson-Hasselbach to calculate the molar concentration of the weak acid HF
when it is mixed with 0.34 M KF to give a pH equal to 2.57. (10 pts)
KF → K+ + FKa = 7.2 x 10-4
Strong Electrolyte (100 %)
[𝐴−]
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔
[𝐻𝐴]
2.57 = − log(7.2 𝑥 10−4 ) + 𝑙𝑜𝑔
(0.34)
(𝑥)
2.57 - 3.14 = log 0.34 – log x
-0.1 = -log x
x = 1.26
5. Predict if an increase in temperature is favorable for the following reactions: (5 pts)
a. CO2(s) → CO2(g)
∆G = ∆H – T∆S and ∆G < 0 is favorable
+∆S increase in entropy
Increase in temperature is favorable
b. 2CO(g) + O2(g) → 2CO2(g)
-∆S decrease in entropy
Increase in temperature is unfavorable
6. Predict if a precipitate will form from when 750 mL of 4.00 x 10-3 M Ce(NO3)3 is mixed with
300 mL of 2.00 x 10-4 M KIO3. The Ksp for Ce(IO3)3 is 1.9 x 10-10 (15 pts)
Ce(NO3)3(aq) +
3KIO3(aq) →
Ce(IO3)3(s) + 3KNO3(aq)
Ce(IO3)3(s) ⇄ Ce+3(aq) + 3IO3-(aq)
[𝐶𝑒 +3 ] =
(0.750 𝐿)(4.00 𝑥 10−3 𝑀)
= 2.86 𝑥10−3 𝑀
(0.750 𝐿 + 0.300𝐿)
[𝐼𝑂3− ] =
(0.300 𝐿)(2.00 𝑥 10−4 𝑀)
= 5.71 𝑥10−5 𝑀
(0.750 𝐿 + 0.300𝐿)
Qsp = [Ce+3][IO3-]3 = (2.86 x 10-3)(5.71 x 10-5)3 = 5.34 x 10-16
Qsp (5.32 x 10-16) < Ksp (1.9 x 10-10) ; therefore a precipitation will form.
Unsaturated
Chemistry 102 Exam 2
Name ______Mr. Perfect_______________________________________ Date ____F 16____________
7. A 20.0 mL sample of 0.100 M Lactic Acid, HC3H5O3, is titrated with 0.200 M of NaOH.
Calculate the pH after the following additions of base: (15 pts)
Ka = 1.38 x 10-4
a. 10.0 mL of base. (At equivalence point)
0.02 L x 0.1 mol/L = 2.0 x 10-3 molA
0.01 L x 0.2 mol/L = 2.0 x 10-3 molB
A- + H2O ⇄ HA + OHI 0.067
--0
0
C -x
--+x
+x
E 0.067- x
x
x
𝐾𝑏 =
𝐾𝑤 1.00 𝑥 10−14
𝑥2
𝑥2
−11
=
=
7.25
𝑥
10
=
≈
𝐾𝑎
1.38 𝑥 10−4
0.067 − 𝑥 0.067
𝑥 = 1.20 𝑥 10−6 = [𝑂𝐻 −]
pOH = -log(1.20 x10-6) = 5.66
pH = 14 – 5.66 = 8.34
b. 20.0 mL of base. (After equivalence point)
0.020 L x 0.200 M base = 4.0 x 10-3 mol base
0.004 mol base – 0.002 mol acid = 0.002 mol base left over
[𝑂𝐻 −] =
0.002 𝑚𝑜𝑙
= 5.0 𝑥 10−2 𝑀
0.04 𝐿
pOH = -log(5.0 x 10-2) = 1.30
pH = 14 – 1.30 = 12.69
Chemistry 102 Exam 2
Name ______Mr. Perfect_______________________________________ Date ____F 16____________
8. Which indicator of the following indicators would be more appropriate for a weak base-strong
acid titration, methyl orange (pKa = 3.7) or phenolphthalein (pKa = 9.3)? (5 pts)
Weak base - Strong Acid equivalence points are acidic (below pH 7); therefore methyl orange
would be more appropriate for this titration.
9. Complete the following table. (10 pts)
[H+]
3.6 x 10-12
[OH-]
2.8 x 10-3
pH
11.4
Acid or Base?
base
2.7 x 10-6
3.7 x 10-9
5.6
acid
6.3 x 10-7
1.6 x 10-8
6.2
acid
5.8 x 10-8
1.7 x 10-7
7.2
base
10. Calculate the solubility (g/L) of Ag3PO4 (Mw = 419 g/mol). (10 pts)
Ksp = 1.8 x 10-18
Ag3PO4(s) ⇄ 3Ag+(aq) + PO43-(aq)
I
--0
0
C
-s
+3s
+s
E
--3s
s
Ksp = 1.8 x 10-18 = (3s)3s = 27s4
Molar solubility = s = 1.61 x 10-5 M
𝑆𝑜𝑙𝑢𝑏𝑖𝑙𝑖𝑡𝑦 = 1.61 𝑥 10−5
𝑚𝑜𝑙 419𝑔
𝑔
×
= 0.00675 𝑜𝑟 𝟔. 𝟕𝟓 𝒙 𝟏𝟎−𝟑 𝒈/𝑳
𝐿
𝑚𝑜𝑙
𝐿
11. (Extra Credit). Can a buffer solution be prepared from 20 mL of 0.1 M HCl and 20 mL of 0.1
M NaCl? (5 pts)
No, because strong acids and there salts have no equilibrium.
Chemistry 102 Exam 2