Today, class is devoted to some Basic equilibrium problems,
slightly concealed (some more so than others)
can you translate these back into basic questions ?
what type of problem is it?
how would you attack it? (we might omit the final calculations)
1. The bicarbonate ion can act as a buffer in water
HCO3- + OH- <---> CO32- + H2O
with Kc = 1.80 x 10-4 for this process
if we mixed up a solution that starts out
0.100 M HCO3- and 0.050 M CO32-
what is the equilibrium OH- concentration ?
2. A 1.00 liter flask is filled with an equilibrium mixture of NO2 and N2O4 gases.
The weight of the gas is 2.895 grams;
the pressure is 1.000 atm and the temperature is 85oC.
What is Keq for the equilibrium ?
3. A compound MX2 forms a hydride
MX2 (s) + 2H2O
(g) ---> MX2.2H2O
(S) <note-- incorrectly said G>
Keq for this process is 24.5 (atm-2)
The equilibrium vapor pressure of water is 27.5 torr at this temperature
At what relative humidity will a sample of MX2 begin to gain weight in air?
4. we mix 100 ml of 0. 050 M Pb+2 solution with
400 ml of 0.25 M Cl-
a. will PbCl2 precipitate?
b. if so, how much of the original Pb2+ will remain dissolved
Ksp = 1.2 x 10-5 for PbCl2
5. A solution is made to contain
0.050 N Ag+ 0.025M Cl- 1.00 M NH3
In the absence of Ammonia, the Ag+ and Cl- would certainly precipitate AgCl
Ksp of AgCl = 1.8 x 10-10
Ag+ + 2 NH3 ---> Ag(NH3)2+ Keq = 1.7 x 107
Ammonia
can remove (tie up) much of the silver ion
will AgCl form a precipitate or will all the silver stay in solution?
6. A 0.25 Molar solution of HA is 2.5% dissociated into H+ and A-
what is Ka ?
7. A 0.25 M solution of HB (a weak acid) has a pH of 3.45
What is
Ka for this acid [H+] = 10-pH
Stage 1 -- write Keq
expressions
in a fe cases we actually need to
provide the equation
1. Keq = Kc = [CO32-] / ( [OH-] [HCO3-]
2. see text: N2O4 (g) --> 2 NO2 (g)
Keq = Kp = P2(NO2) / P(N2O4)
3. note the typo.. the hydrated salt is a solid (typo said g)
Kp = 1 /
P2(H2O)
also, I wrote
H2 rather than H2O on the board Friday
4. Since the solubility prodcut
is given (dissolving a salt), we’d write the equation and K as
PbCl2 (s) ----> Pb2+ (aq) +
2 Cl- (aq)
Ksp = Keq = [Pb2+] [Cl-]2
5. Two reactions, two equations and two equilibrium expressions
AgCl (s) ----> Ag+ (aq) + Cl- (aq)
Keq = Ksp = [Ag+] [Cl-]
Ag+ + 2 NH3 ---> Ag(NH3)2+
Keq = [Ag(NH3)2+/ ([Ag+] [ NH3]2 )
6 and 7 are both weak acids in water
HA + H2O --> H3O+ + A-
Ka = Keq = [H3O+] [A-] / [HA]
7. same, except HA and A- are replaced by HB and B-
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Stage 2 -- what exactly are we asked to find (and
therefore what type of a problem is it)
1. what is the equilibrium concentration of OH-
set up inital conditions, change is “x”, express all final conc with “x”
substitute into Keq expression, solve for x
2. What is Keq?
the only way to do this is to know both P(NO2) and P(N2O4) in the mixture
then it is easy-- fill into Keq and evaluate a numerical answer
Hint-- this is, at heart, a PV = n RT system
but n= number of moles of both species
3. What is the relative humidity ....
need to translate that to equilibrium language
a. what is equilibrium P(H2O) with this salt
b. if P(H2O) in air > that value, we get sample absorbing humidity
c. finally, what humiidty matches the equilibrium P(H2O)
4. a. Will PbCl2 precipitate?
initial conditions given, will conc make Q > K?
b. How much Pb+2 remains in solution
let x = loss of Pb2+ due t precipitate (a change)
then 2x = loss of Cl-
substitute, solve for x
answer is Pb2+(original) - x
5. This is more complex
the question: will Ag+ and Cl- all stay in solution?
is Q = [Ag+][Cl-] > Ksp
but [Ag+] is after encountering ammonia
6, 7 What is Ka
we need to know [H3O+], [A-] and [HA] to evaluate Ka
data is disguised, but is given
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Solutions to these problems (somewhat abbreviated now)
1. At equilibrium
Keq = Kc = [CO32-] / ( [OH-] [HCO3-]
Keq = Kc = [0.050-x] / ( [x] [0.10+x]
multiply and move all to the left side
Kc * x2 + x (0.1*Kc +1) -0.050 = 0
quadratic, solve for x after substituting the value for Kc
2. PV=nRT solve for n = number of moles of both gases in the mixture
n= PV/RT where all values are given
mass / n = apparent mol wt = J (a purely artifical symbol)
but J =( f) FW(NO2) + (1-f) FW(N2O4)
f is the fraction of gas that’s NO2; (1-f) is, by default, what’s left
once you find f
P(NO2) = 1.00 atm * f
P(N2O4)= 1.000 *(1-f)
then easy to evaluate K
3. K = 1/ P2(H2O)
or, the equilibrium pressuure is (K)-1/2 = 1/(24.5)1/2 = 0.202 atm = 153 torr
even at 100% RH the P(H2O) in air is < 153 torr
so the hydrated salt loses water, not gains it
(I guess I chose a bad value for Keq here)
4. After mixing [Pb2+] = (100/500)*0.05 = 0.010 M
[Cl-] = (400/500)*0.25 = 0.20 M
Q = (0.010)*(0.20)2 = 4 x 10-4 > Keq = 1.2 x 10-5
so we will get a precipitate
we exceed K by a factor of 10 so we can expect lots of precipitate
part b-- how much (we could solve normally, but there’s a simplification here)
if all the Pb2+ were to precipitate, the Cl- would still be 0.20 - 0.02 = 0.18 M
so, within 10%, the Cl- concentration is essentially 0.20 yet
since we expect much of the Pb to come out, it’s better to assume 0.18
K = (Pb2+)(0.18)2 = 1.2 x 10-5
[Pb2+] = 3.7 x 10-4
notice that the error in Cl- (assuming all Pb2+ precipitated) is really quite small
5. This is simultaneous equilibrium
let’s assume, for the moment, that all material stays in solution
Keq = 1.7 x 107 = [Ag(NH3)2+] / ([Ag+][NH3]2)
one big number and we can’t possibly have big concentrations
so we must get a big ratio of [Ag(NH3)2+] / [Ag+]
so let’s guess that [Ag(NH3)2+] = 0.050 M
this uses up twice that amount of NH3 so [NH3] = 1.0-0.10 = 0.9
solve for free [Ag+]
1.7 x 107 = [0.05] / ([Ag+][0.9]2)
[Ag+] = 4 x 10-9
this means our original assumption is awfully good
precipitate?
Q = (4x10-9)*(0.025) = 1 x10-10 <Ksp so no precipitate
6. 2.5% dissociated means
[H3O+] = [A-] = 0.025 * 0.25M = 6.25 x 10-3
[HA] = 0.975 * 0.25 = 0.243
Ka = 0.243 * 6.25 x 10-3 = 3.7 x 10-4
7. pH of 3.45 means = 10-3.45 = 3.54 x 10-4 M
we must have equal [B-]
we must lose that amount of HB. so [HB]= 0.25 -3.54 x 10-4 = 0.2497 (0.25 is ok)
K = 3.54 x 10-4 * 3.54 x 10-4 / 0.2497 = 5.0 x 10-7