chem 127 exam 2
October 6, 2004
100 points with all questions being worth 12-14 points,
justify your answers / show your work
1.(12 pts) a. What is the pH of a 0.0023 M solution of HCl ?
100% dissociated so [H3O+]
= 0.0023 M
pH= -log10( 0.0023) = 2.64
(since data is 2 sig figures, we
should keep two digits behind the decimal)
rounding off to pH = 2.6 is a bit
too severe
b. What is the pH of a
0.0023 M solution of NaOH ?
pOH = -log10[OH-] = 2.64
pH = 14.00 - pOH = 11.36
alternatively, [H3O+]
= 1.00 x 10-14 / (0.0023) = 4.35 x 10-12 M
pH= - log10(4.35 x 10-12) = 11.36
c. what is the [H3O+]
of a solution that has pH 4.33 ?
[H3O+] = 10-pH
= 10-4.33 = 4.7 x 10-5 M
the pH has two significant figures
(to right of decomal) so 2 digits in the result
d. Acetic Acid has Ka = 1.80 x 10-5 ; what is pKa for this acid ?
by definition pKa= -log10(Ka) =
-log( 1.80 x 10-5) = 4.74
2. (14
pts) What is the equilibrium hydronium ion concentration, [H3O+]
,
of
a 0.25 M solution of benzoic acid (just
write benzoic acid as HA or HB)
Ka =
6.5 x 10-5
hint-- write the
balanced chemical reaction; write the
expression for Ka; what are the initial concentrations and the changes...
HA + H2O ---> H3O+ + A-
(A is benzoate)
Ka = [H3O+] x
[A-] / [HA]
initially: HA = 0.25 M; H3O+=0 M; A-=0 M
change HA (-x); H3O+=
+; A- (+x)
equil: HA (0.25-x); [H3O+]=
x; [A-]=x
6.5 x 10-5 = (x)*(x)/(0.25-x)
we expect x << 0.25 so this is
almopst 6.5 x 10-5 = (x)*(x)/(0.25)
or x = square root ( 6.5 x 10-5
* 0.25) = 0.040 M = [H3O+]
3.
(12 pts) The gas NO2 reacts
to form the gas N2O4
first, write the balanced equation for the reaction
Kp for the reaction has a
value of 0.15 (atm).
A
gas mixture is at equilibrium;
it contains NO2 at a pressure of 0.73 atm.
What is the pressure of N2O4
?
2 NO2 (g) --> N2O4
Kp = P(N2O4) /
P2(NO2)
0.15 = P(N2O4)
/ (0.73)2
P(N2O4) =
0.07994 = 0.080
4.
(14 points) The solubility of Ag2CrO4 (silver chromate)
is 1.15 x 10-4 mol/liter
it
dissolves to form silver ions, Ag+ and chromate ions CrO42-
what is Ksp for this species?
Ag2CrO4 --> 2 Ag+ + CrO4-2
Ksp = [Ag+]2 [CrO42-]
[Ag+] = 2 x 1.15 x 10-4 M
[CrO42-] =1.15
x 10-4
Ksp = [Ag+]2 [CrO42-]
= (2.30 x10-4)2 x 1.15 x 10-4 = 6.08 x 10-12
5.(12
points) Nitrous is a weak acid : HNO2 + H2O
---> H3O+ + NO2-
a.
How does HNO2 in this equation fit the Bronstead Lowry definition
of an acid
Acid is a proton donor; HNO2 gives up H+ to H2O
b.
Identify the conjugate base of Nitrous Acid
Nitrite ion, NO2-
c.
Would you expect a solution of Sodium
Nitrite to be acidic / basic/ or neutral ? (why?)
basic; we just decided (part b) that nitrite ion is a base
NO2- + H2O
---> HNO2 + OH- (basic)
sodium is a boring spectator ion
here
d.
In Bronstead Lowry language, what roles are H2O and H3O+
playing here?
in equation ,H2O accepts a proton (base) to form H3O+ (acid)
6 . (12 points) Use the
concept of le Chatlier to answer the questions below
(justify your answers)
The
reaction 2 SO2 (g) + O2 (g) ---> 2SO3 (g)
is exothermic.
If
we start with a mixture of SO2, SO3 and O2 at equilibrium, would we increase /
decrease /( or cause no change) the
amount of O2
Note: In grading we insisited on a justification since the
answer is up/down with 50% guessing outcome.
a.
if we raise the temperature of the mixture
for an exothermic reaction, K decreases with temperature
reaction shifts to left
more O2 is formed
b.
if we forced the original mixture into
a smaller volume
reaction goes from 3 gas molecules to 2 gas molecules
reduced volume raises pressure
shift to right (fewer molecules of
gas) relieves the stress
therefore less O2
c.
if we injected additional SO2 gas into the original mixture
relieve stress by removing SO2 ; shift to right
this uses up O2 also
reduces O2 pressure
d.
if we injected additional O2 gas into the original mixture
similar to (c) except we expect only partial shift
therefore much of the added O2
will remain so P(O2) will be hgiher.
7 . (12 points) Silver Chloride has a very low solubility in water. Silver acetate is slightly soluble in water. Explain why Silver acetate is more soluble in 0.1 M nitric acid than in water, but is less soluble in 0.1M HCl than it is in water.
In HNO3, the acetate ion reacts
with H+, forming acetic acid
this reduces the [acetate] ion below
what would form in water only
so more silver acetate dissolves
at equilibrium, [Ag+] is what ever
dissolves and [acetate] is much less
With HCl we need to worry bout the
very low solubility of AgCl
If we had a saturated solution of
silver acetate in water and added the HCl
we'd immediately get a preciptiate
of AgCl
the H+ would tie up
acetate ions, the the [Ag+] is fixed by Ksp of AgCl
(we'd actually convert solid silver
acetate to solid silver chloride)
8.
(12 points) If we mix equal volumes of 0.016 M Pb(NO3)2 and 0.080 M of KI, would we get a precipitate of PbI2 to
form. For lead Iodide, Ksp = 8.5 x 10-9
(Justify
/ explain your answer)
After mixing [Pb2+} = 0.0080 M
and [I-]= 0.040
Q = [Pb2+] [I-]2
= 0.080 x (0.040)2 = 1.25 x 10-5
Q much greater than Ksp so a precipitate definitely forms
(had earlier said no precipitate, but that's in error)
(end of exam)