Chemistry 114
Hour Exam IV
Name:____________
1. (10 points) What is the pH and pOH of a 0.005M solution of KOH?
.005M KOH = .005M OHpOH=-log(.005)
pOH=2.30
pH=14-pOH
pH=14-2.3 = 11.70
2. (10 points) What is the pH and pOH of a 0.005M solution of NH3 ?
(I couldn’t find a K b for this in the back of the book - All I could find was the KA of NH4+= 5.6x1010
)
NH3 you should recognize as a weak base
NH3 + H2OW NH4+ + OHKb = [NH4+][OH-]/[NH3]
But no K b is given so you will have to calculate it
Kb = K w/Ka = 1x10-14/5.6x10-10 = 1.786x10-5
1.786x10-5 =X 2/(.005-X)
assuming .005-X ..005
X=[OH-]=2.99x10-4
pOH= -log[2.99x10-4 ]=3.52
pH=14-3.52= 10.48
If you are a stickler for details and solve this equation exactly
X=[OH-]=2.90x10-4
pOH= -log[2.90x10-4 ]=3.54
pH=14-3.52= 10.46
3. (10 points) Are the following salts or oxides acidic(A), neutral (N) or basic (B)?
LiCl
N (Li is cation of strong base, Cl is anion of strong acid)
NaF
B (F- is conjugate base of weak acid HF)
CH3NH3Cl A (CH3NH3+ is the conjugate acid of a weak base)
Al(NO3)3 A (Al+3 is a highly charged cation)
SO2
A (SO2 is a covalent oxide)
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4. I have 20 ml of a 0.1M solution of Formic Acid (K a=1.8x10-4 ). To this solution I will add 10
ml of 0.1M Sodium Formate (The sodium salt of formic acid)
A.(3 points) Will the pH of the solution go up or down?
[H+].9, pH8
B. (7 points) What is the final pH of the solution?
Mixture of an acid and its conjugate base, use Henderson Hasselbach
pH=pKa + log (A -/HA)
pKa = -log(1.8x10-4 ) = 3.74
mMole A - = M@V = .1(10) = 1
mMole HA = M@V = .1(20) = 2
Since you don’t need concentration, just relative number of moles for the HH
equation we can plug and chug at this point
pH= 3.74 + log (½) = 3.44
5. (10 points) Last week some biologists asked me to determine the bicarbonate levels in a
water sample. The easiest procedure for doing this is to titrate the bicarbonate with base.
The endpoint of this titration has a pH of 4. What is the best indicator for me to use in this
titration?
Cresol Red pKa=1.75
Erythrosin B pKa=2
Methyl Orange pKa= 3.5
Phenolphthalein pKa=8
Methyl Orange pKa= 3.5
You want the pKa of the indicator to match the pH at the equivalence point as closely as
possible
6. (10 points) CuI has a K sp of 5.0x10-12. What is the solubility of this salt in water?
(Express solubility in moles of CuI that will dissolve in 1l of water)
Solubility means the amount of CuI in solution. Actually you won’t find any CuI in
solution because it ionizes to Cu+ and I-, thus CuIsolution = either Cu+(aq) or I-(aq) since you get
1 mole of Cu+ or I- for every mole of CuI that dissolves and ionizes
CuI(s)WCu+(aq) + I-(aq) K sp = [Cu+][I-]; X = [Cu+]=[I-] =solubility of CuI
5.0x10-12 =X 2
X = sqrt(5.0x10-12) = 2.24x10-6 M
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7. (10 points) Predict the sign of S in the following changes:
H2O(l)6H2O(s)
- (l6s becomes more ordered)
AgCl(s)6Ag+(aq) + Cl-(aq) + (solid to 2 aqueous ions, less ordered
N2(g) + 3H2(g)62NH3(g) - (4 gas molecules to 2 gas molecules, more ordered)
H2(g) + 1/2O2(g)6H2O(l) - (1 & ½ gas molecules to 1 liquids, more ordered)
CH3COOH(aq)6 H+(aq) + CH3COO-(aq) +(1 molecule to 2 molecules, less ordered)
8. (10 points) From information in the following table, calculate )Go for the reaction C 2H6(g)
+ 7/2 O2(g)63H2O(g) + 2CO2(g) (assume T=25oC)
C2H6(g)
O2(g)
H2O(g)
CO2(g)
)Hf0 (kJ/mol) )Gf0(kJ/mol)
-84.7
-32.9
0
0
-232
-239
-393.5
-394
So(J/K @mol)
229.5
225
189
214
From )G0 values
=[3(-239)+2(-393.5)]-[1(-32.9)+7/2(0)] = -1471.1kJ
From )H and )S
)H = [3(-232)+2(-393.5)]-[1(-84.7)+7/2(0)] = -1398.3kJ
)S = [3(189)+2(214)]-[1(229.5)-7/2(225)] = -22J/K
)G=)H-T)S = -1398.3kJ -298(-22)J
= -1398.3 + 6.6kJ = -1391.7 kJ
The answers from the 2 calculations should be the same, but there were a couple of errors in
the table so the answers don’t check with each other.
9. Back in chapter 14 you found that water undergoes the autoproteolysis reaction
H2O(l)W H+(aq) + OH- (aq) (Assume T=25oC)
A. (3 points) What is the K for this reaction? Kw=1x10-14
B. (7 points) What is the )Go for this reaction?
)G0 = -RTlnK
-8.3145J/k@mol(298K)ln1x10-14
= 79,900J, 79.9kJ
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10. I want to make a battery that uses the following two reactions:
Al+3(aq) + 3e-6Al(s) E0= -1.66V
Mg+2(aq) + 2e- 6Mg(s) E 0 = -2.37 V
Diagram this cell, placing the reactions on the correct sides of the cell, showing the proper
flow of electrons. Finally calculate the potential of this cell if both Al+3 and Mg+2 are present
in 1M concentrations.
Ecell = E r-E l = -1.66-(-2.37) = .71V
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