Tutorial on Equilibrium Calculations:
Here are a few examples of a equilibrium problems that may help explain them.
Consider acetic acid HAc with a pKA = -4.75 (or Ka=1.7610-5).
HAc  H +  Ac [H  ][Ac  ]
 1.76 105
[HAc]
1) If we have a solution of 0.1M HAc what is the pH?
Before equilibrium has set in we have only 0.1 M HAc then we let equilibrium take place
and x moles of the HAc dissociates.
HAc
H+
Acstart
0.1 M
0
0
At eq
0.1 – x
x
x
We start with the 0.1M of acetic acid but at equilibrium it dissociates and so you lose x of
it and gain x for both [H+] and [Ac-]. Writing the equilibrium constant you get
[H  ][Ac ] [x][x]
K

 1.76 105
[HAc]
[0.1 x]
To solve this you can ignore the x in the denominator and you get x=[H+]=1.32x10-3 M or
a pH =-log[H+] = 2.88.
K


Now try this
2) You have a solution that you add 0.1 M HAc and 0.2 M NaAc. What is the pH?
Preceding as above we get
HAc
H+
Acstart
0.1 M
0
0.2 M
At eq
0.1 – x
x
0.2+x
Note that the reaction must proceed to the right since we say there is no H+ (actually there
is a little from the dissociation of water but that is negligible in this problem)..
We get K 
[x][0.2  x]
 1.76 105
[0.1 x]
[x][0.2]
 1.76 105
[0.1]
+
-6
solving
we
get
x=[H
]=8.8
x
10
;
pH=5.06.
The
pH
increased
since we added a weak

base NaAc
Now x is small compared to both 0.1 and 0.2 so we get K 

3) Now try a strong acid and a weak base.
You have a solution of HCl and the salt of a weak acid NaAc. [HCl] =.1 M; and NaAc =
.12 M
HAc
H+
Acstart
0.0 M
0.1
0.12 M
At eq
x
0.1-x
0.12-x
[0.1 x][0.12  x]
 1.76 105
[x]
Unfortunately x is not small compared to 0.1 or 0.12 so we can not ignore it in this
equation. It can be solved but is takes a quadratic. Multiplying out and rearranging and
we get

0 1.2 102  (0.22 1.76 105 )x  x 2
if we ignore the 1.76x10-5 compared with 0.22 we get x = 0.1 M!
This is strange since it implies that [H+] = 0.1 – x = 0.0
Now we get: K 

The problem with this answer is due to round off error. If you do it exactly (i.e. don't
ignore 1.76x10-5) you get two roots for x of 0.1201 and 0.0.09991. The first answer
(0.1201 is physically impossible since you do not start with enough Ac to give that much
HAC)
Then
[H+] = 0.1 –x =0.1 – 0.09991 = 0.00009.
However these numbers are really outside of the error in our starting concentrations since
you need 5 digits to do the last subtraction.
So we need to do something else.
Assume that the reaction precedes to the left as far as it can go and you form 0.1 M HAc
and are left with no H+ and only 0.02 M Ac-. An alternative way to look at it is rather
then starting by mixing 0.1 HCL and 0.12 NaAc you add 0.1 M HAc and 0.02 M NaAc
and 0.1 M NaCl. You get the same chemical solution no matter which things you added
originally since they all added up to the same amounts of H, Cl, Na and Ac and
equilibrium allows the various species to form as the equilibrium constant requires.


This second way we have
HAc
H+
Acoriginal start
0M
0.10
0.12 M
rex moved to left 0.01M
0.0
0.02 M
At eq
0.1 - x
x
0.02 + x
Now we get
[x][0.02  x]
K
 1.76 105 in this case x is small compared to 0.02 so we solve and
[0.1 x]
get
[x][0.02]
K
 1.76 105 and x = [H+] =8.8e-5 or pH = 4.06.
[0.1]
So the lesson is sometimes in equilibrium problems it helps to push the equilibrium all
the way one way and them let it come back rather then just using the original starting
species.
A more complicated problem:
Telluric acid is a weak acid, H2TeO4. Here is a problem to help with the HW.
Assume you have a beaker of 0.1M HCl. You are going to add pinches of Na2TeO4 to
the beaker and each pinch you want to know the pH. Each pinch will contain 0.03 M of
Na2TeO2 ( I will just call it Na2T). In the lab of course you would grab a pH meter and
measure the pH but we will calculate it.


The equilibrium is
H 2T  H   HT 
pK1  6.27; K1  5.4 107
HT   H   T 2
pK 2  8.43; K 2  3.7 109
What we will do is calculate the pH at the start. At start [H+] = 0.1M (HCl is strong acid
and completely dissociates). Thus pH = -log([H+])=1
Now add the pinch of Na2T and the salt dissociates so you have
Na2T  2Na  T 2

start
[HT-]
0
[H+]
0.1
[T2-]
0.03
Lets let the reaction go all the way to the left and then come back for equilibrium:
[HT-]
[H+]
[T2-]
start
0
0.1
0.03
shift left 1
0.03
0.07
0
However we still have more H+ that will then protonate the HT[HT2]
[H+]
[HT-]
shift left 1
0
0.07
0.03
shift left 2
0.03
0.04
0
eq
0.03-y
0.04+y
y
Now we get:
K1  5.4 107 
4.1107  y

[H  ][HA  ] (0.04  y)(y) .04 y


[HA2 ]
0.03  y
.03
where we guessed that y was much smaller than either 0.04 or 0.03 and so could ignore it
when compared to both of them. This was then an easy equation to solve. and the pH is
pH  log([ H  ])  log(. 04)  1.4

Note that here since in the second equilibrium we don't use up all the acid it pH can be
calculated without actuualy solving the above equation since H+ is just 0.4 the left over
acid.
Now add 0.03 more Na2T.
start
shift left 1
[HT-]
0
0.06
[H+]
0.1
0.04
However we still have some acid that will then protonate the HT[HT2]
[H+]
shift left 1
0.0
0.04
shift left 2
0.04
0.0
eq
0.04-y
y
[T2-]
0.06
0
[HT-]
0.06
0.02
0.02+y
Now y is the H+ concentration and we get
[H  ][HA  ] (y)(0.02  y) .02y
K1  5.4 107 


[HA2 ]
0.04  y
.04
1.08 106  y
and
pH  log([ H  ])  log(1.08 106 )  6.0


Do this again and what do we get
[HT-]
start
0
shift left
0.09
[H+]
0.1
0.01
However we still have some acid that will then protonate the HT[HT2]
[H+]
shift left 1
0.0
0.01
shift left 2
0.01
0.0
eq
0.01-y
y
Now y is the H+ concentration and we get
[H  ][HA  ] (y)(0.08  y) .08y
7
K1  5.4 10 


[HA2 ]
0.01 y
.01
6.8 108  y
and
pH  log([ H  ])  log( 6.8 108 )  7.2


[T2-]
0.09
0
[HT-]
0.09
0.08
0.08+y
A general approach to equilibrium problems.
The most general approach to equilibrium problems is to write n equations in n
unknowns and solve them. This is not as hard as it sounds since the equations are easy to
write down. They include three classes of equations, conservation of matter, charge
neutrality, and equilibrium constants.
An example is the above problem of a solution made up of NaAc 0.12M and HCl
0.1M. We know that NaAc and HCl completely dissociate. We have by the conservation
of Ac- that
Total Ac = [Ac] + [HAc]
where the Total Ac is just 0.12 M
Charge neutrality requires that the number of negative and positive ions be equal:
+
[H ] + [Na+] = [Ac-] + [OH-] +[Cl-]
where [Na+] = 0.12 M and [Cl-] = 0.1 M.
The equilibrium constant equations are:
[H  ][Ac  ]
KA 
[HAc]
K w  [H  ][OH  ]
or listing all together and plugging in the concentrations that we know:

0.12  [Ac  ]  [HAc]
[H  ]  0.12  [Ac  ]  [OH  ]  0.1
KA 
[H  ][Ac  ]
[HAc]
K w  [H  ][OH  ]

We have 4 equations and 4 unknowns. To solve we approximate the [OH-]
concentration as much less then 0.1 M, so we drop that term in the second equation and
solve for [Ac-].
0.12  [Ac  ]  [HAc]
[Ac  ]  [H  ]  0.12  0.1  [H  ]  .02
KA 
[H  ][Ac  ]
[HAc]
K w  [H  ][OH  ]
using this value in the first and the third equations we get

0.12  [H  ]  .02  [HAc]
KA 
[H  ][H  ]  .02
[HAc]
now plugging the first into the second for [HAc] we get:

KA 
[H  ][H  ]  0.02
0.12  [H ]

This equation is the familiar one from above that we solved by assuming that [H+] is
small compared to 0.02 or 0.12.
