CHE230 Environmental Chemistry 2010
Problem Set 3 Solutions
1)
( a)
1(b)
2.
a)
We know the concentrations of 3 species:
1) CO2(g) ⇌ CO2(aq)
K1 = 3.4x10-2
2) CO2(aq) + H2O ⇌ H2CO3(aq)
K2 = 2x10-3
3) H2CO3(aq) ⇌ H+(aq) + HCO3-(aq)
K3 = 2.25x10-4
4) HCO3- (aq) ⇌ H+(aq) + CO32-(aq)
pKa= 10.33
5) CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)
Ksp = 6x10-9
The partial pressure of
depends on Eq 1, so we need to know
is affected by Eq 2, but we don’t know
. Again,
depends on Eq 3, and this time we know the concentrations of the products, so we can
combine 2 and 3 to write the equilibrium such that all terms are known:
2) CO2(aq) + H2O ⇌ H2CO3(aq)
3) H2CO3(aq) ⇌ H+(aq) + HCO3-(aq)
= 2x10-3
= 2.25x10-4
6) CO2(aq) + H2O ⇌ H+(aq) + HCO3-(aq)
You can verify that K6 = K2K3 by writing out these expressions in full.
Since we have the numerical value of K6, we can now
at 1 atm (1.01325 bar) pressure gives
b)
At saturation, the solubility product is reached:
We will need to find
and can do so using Eq 4, assuming equilibrium:
We find that
This is almost exactly
. We can say that the solution is saturated.
3)
a)
We are interested in
and
. Since these species don’t occur together
in Eqs 1-5, we will have to figure out how they relate. Both limestone and CO2 in the air
affect these species.
The concentration of Ca2+ is given:
A solution in equilibrium with its solid must necessarily be saturated, so we can use
to find that
Because carbonate is in equilibrium with bicarbonate:
, we can find
by considering the equilibria in which it is
involved. First, limestone produces bicarbonate by (Eq 4 and Eq 5):
And second, CO2 produces bicarbonate by (Eq 2 and Eq 3):
We can’t use either of these directly since the pH is unknown, but after noticing that H+ is
cancelled out by these reactions we can write them together as:
We can avoid considering the equilibrium constant (especially since {CO2(aq)} isn’t
known) by observing that the molar ratio of Ca2+ to HCO3- here is 1:2. Assuming that all
Ca2+ and HCO3- in the solution comes from CO2 and CaCO3, we can therefore write
b)
We can find
c)
using Eq. 4:
d)
Combining consecutive equilibria:
1) CO2(g) ⇌ CO2(aq)
= 3.4x10-2
2) CO2(aq) + H2O ⇌ H2CO3(aq)
= 2x10-3
3) H2CO3(aq) ⇌ H+(aq) + HCO3-(aq)
= 2.25x10-4
7) CO2(aq) + H2O ⇌ H+(aq) + HCO3-(aq)
You can write out
to see that they give
as written. Water activity {H2O} is
approximated to 1 because water is nearly pure in dilute solution. Now we can find
by
e)
Ca2+(aq) + 2 HCO3- (aq) ⇌ CaCO3(s) + CO2(g) + H2O(l)
This equation is written in terms of
K=2.8x106
because this is the most abundant CO3
ion in the water. The other ion is CO3- (aq), but in part (a) we found that HCO3- (aq) was
1000 more concentrated than HCO3- (aq).1
Because we wrote the equation this way, we get a simple equilibrium constant:
When the water is heated two HCO3- ions react for each Ca2+ ion. Assuming that all Ca2+
and HCO3- has come from limestone or CO2, their concentrations will obey the molar
ratio given in the expression, so [Ca2+] = 2[HCO3-] and we can write
1
See http://www.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html for a nice graph of
this.
Assuming P=1 atm inside the heater,
Therefore
And change in
In 1000L this is equivalent to
So
f)
4.
a)
b)
c)
CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)
Ksp = 4.6x10-9 (15oC)
. Therefore the water is supersaturated
d)
CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)
therefore seawater is not supersaturated.
Ksp = 4.6x10-9 (15oC)