Equilibrium in a “Nutshell”
Here is a series of questions that will guide you through the different types of problems you will encounter when dealing with
equilibrium. A worked example is given along with two similar problems (with answers). Have fun!
A. Creating an equilibrium expression from a chemical equation.
Worked example:
H2 (g) + I2 (g)  2 HI (g)
Remember to use only aqueous or gaseous parts of the equation. Solids and pure liquids don’t
count.
Answer: Keq = products = [HI]2
reactants
[H2][I2]
This equilibrium expression is a Kp because all the parts are gaseous.
Questions:
Write the equilibrium expressions for the following chemical equations.
1) S(s) + O2 (g)  SO2 (g) (Kp = PSO2 /PO2)
2) HCl (aq) + NaOH(aq)  H2O(l) + NaCl(aq)
(Keq = [NaCl] / [HCl][NaOH])
B. Finding the equilibrium constant (Keq) from given equilibrium concentrations.
Worked example:
The dissociation of AgCl(s) is represented by the following equation: AgCl(s)  Ag+(aq) + Cl-(aq). If the equilibrium concentration of Ag+
is 1.0 x 10-5 and the Cl- concentration is 1.8 x 10-5, what is the equilibrium constant?
Answer: Ksp = [Ag+][Cl-] = (1.0 x 10-5)(1.8 x 10-5) = 1.8 x 10-10
Note: You may be given mol and the volume of the container. In that case you need to convert that to a concentration.
Questions:
1) Using the equation: 2HI(g)  H2 (g) + I2 (g) equilibrium conditions are found to be: [HI] = 0.27 mol/L and [H2] = [I2] = 0.86
mol/L. Calculate the equilibrium constant. (Keq = 10)
2) Using the equation S(s) + O2 (g)  SO2 (g), equilibrium pressures are found to be [O2] = 0.54 atm and [SO2] = 0.35 atm. Calculate
the equilibrium constant. (Kp = 0.65)
C. Given concentrations and Keq, predict which direction the eqm will shift.
Worked example:
Consider the system PCl5 (g)  PCl3 (g) + Cl2 (g). The equilibrium constant is 4.0 x 10-2. If the concentrations of Cl2 and PCl3 are both
0.30 mol/L, while the concentration of PCl5 is 3.0 mol/L, is the system at equilibrium? If the system is not at eqm, which way will it
proceed?
Answer: Keq = [PCl3][Cl2] / [PCl5] = 4.0 x 10-2
Substitute your values in to see if the Kc value (or rxn quotient – Q) comes out to greater than or less than the Keq (or equal to the Keq if
the system is at eqm) . If it is less than the Keq, then there are more reactants than products and the eqm shifts to the right. And viceversa.
In this case the Q = 3.0 x 10-2 so the system is not at equilibrium and it shifts to the right.
Question:
1) The Kp for the formation of NO(g) using the equation N2(g) + O2(g)  2 NO(g) is 4.0 x 10-4 . If the partial pressure of N2 is 0.50 atm,
O2 is 0.25 atm, and NO is 4.2 x 10-3 atm, decide whether the system is at equilibrium. If not, predict which way it will shift. (right)
D. Given Keq and initial concentrations, calculate equilibrium concentrations.
Worked example:
1.0 mol of each of the gases HI, H2, and I2 are placed in a 2.0 L flask. Given that Keq = 10 for this reaction and that 2HI(g)  H2 (g) + I2
(g), calculate the equilibrium concentrations.
Answer: Find the concentration of each species and create an ICEBOX.
Substitute into the equilibrium expression and solve for x.
[HI]
[H2]
[I2]
Keq = [H2][I2]
10 = (0.5 + x)(0.5 + x)
x = 0.148 mol/L
Initial
0.5
0.5
0.5
[HI]2
(0.5 – 2x)2
Change
-2x
+x
+x
Equilibrium
0.5 – 2x
0.5 + x
0.5 + x
Substitute back into the equilibrium part of the ICEBOX to find the
equilibrium concentrations.
[HI] = 0.5 – 2x = 0.204 mol/L
[H2] = [I2] = 0.5 + x = 0.648 mol/L
You can check your answer if you substitute these values back into the equilibrium expression and see if the answer comes back out to
the Keq. Note: you may need to use the quadratic formula to solve for x.
Questions:
1) The equilibrium constant for the reaction H2 (g) + I2 (g)  2 HI(g) was determined to be 55.17. If you place 1.00 mol of each of H2
and I2 in a 0.500 L flask, what ore the equilibrium concentrations of H2, I2, and HI? (0.42, 0.42, and 3.16 mol/L respectively)
2) Given that CO2 reacts with carbon via the reaction: C(s) + CO2 (g)  2 CO(g) Kp = 1.90 and that at equilibrium the total pressure
in the reaction vessel is 2.00 atm, calculate PCO2 and PCO at equilibrium. Note: Ptotal = PCO2 + PCO (PCO2 = 0.78 atm, PCO = 1.22 atm)
E. Finding equilibrium concentrations after an old equilibrium is changed.
Worked example:
n-butane is in equilibrium with isobutane: n-butane  isobutane Keq = 2.5 . Assume equilibrium has been established in a 1.00 L
flask where [n] = 0.500 mol/L and [iso] = 1.25 mol/L. Then 1.50 mol of n-butane is added to the flask. What are the new equilibrium
conditions?
Answer:
The initial concentrations are immediately after the extra gas is added.
Make the ICEBOX:
Substitute into the equilibrium expression and solve for x
[n-butane]
[isobutane]
Keq = [iso]/[n] x = 1.07 mol/L [n-butane] = 0.93 mol/L [isobutane] = 2.32
Initial
0.500 + 1.50
1.25
mol/L
Change
-x
+x
Questions:
Equilibrium 0.500 + 1.50 - x
1.25 + x
1) Using the same example as above, at equilibrium [n] = 0.20 mol/L and [iso] =
0.50 mol/L. What will the new equilibrium conditions be if 2.00 mol/L of isobutane is added to the original mixture? ([n] = 0.77 mol/L
and [iso] = 1.93 mol/L)
2) Using the equation: 2HI(g)  H2 (g) + I2 (g) Keq = 10 equilibrium conditions are found to be: [HI] = 0.27 mol/L and [H2] = [I2]
= 0.86 mol/L. If 0.2 mol/L of I2 is added, what will be the new concentrations after equilibrium is reestablished?
(x = 0.01377, [HI] = 0.298 mol/L, [H2] = 0.846 mol/L, [I2] = 1.046 mol/L)
F. Finding the molar solubility given the Ksp.
Worked example:
The Ksp of magnesium fluoride, MgF2, is 6.5 x 10-9. Calculate the molar solubility of MgF2.
First write the balanced equation for the dissociation of MgF2:
MgF2(s)  Mg2+(aq) + 2 F-(aq)
Then write the equilibrium expression: Ksp = [Mg2+][F-]2
Using the coefficients in the balanced equation, create variables for the concentrations of the ions.
[Mg2+] = x, [F-] = 2x (from the balanced equation)
Sub into the Ksp expression and solve for x: 6.5 x 10-9 = x(2x)2
6.5 x 10-9 = 4x3
x = 1.18 x 10-3 mol/L
x is the number of mol of MgF2 that will dissolve in 1 litre of solution. This is the molar solubility.
Questions:
1) The Ksp for silver bromate, AgBrO3, is 5.8 x 10-5. Calculate the molar solubility. (7.6 x 10-3 mol/L)
2) The Ksp for barium sulphate, BaSO4, is 1.1 x 10-10. Calculate the molar solubility. (1.0 x 10-5 mol/L)
G. Finding the Ksp given the molar solubility.
Worked example:
The solubility of lead (II) iodate in pure water is 2.24 x 10-2 g/L at 25C. Calculate the value of Ksp at this temperature.
First write the balanced equation for the dissociation: Pb(IO3)2 (s)  Pb2+(aq) + 2 IO3-(aq)
Then find out how many mol/L are soluble 2.24 x 10-2 g/L x 1 mol Pb(IO3)2 /557 g = 4.02 x 10-5 mol/L
Using the balanced equation, determine the concentrations of the ions [Pb2=] = 4.02 x 10-5 mol/L, [IO3-] = 2 x 4.02 x 10-5 = 8.04 x
10-5 mol/L
Write the Ksp expression and sub in your values: Ksp = [Pb2+][ IO3-]2 = (4.02 x 10-5)( 8.04 x 10-5 )2 = 2.60 x 10-13
Questions:
1) Lithium Fluoride dissolves in water to the extent of 0.27 g per 100mL at 18C. Calculate the Ksp at 18C. (Ksp = 0.0108)
2) Potassium perchlorate, KClO4, is soluble in water to the extent of 0.75 g per 100 mL at 0C. Calculate the Ksp of potassium
perchlorate at 0C. (2.9 x 10-3)
H. The Common Ion Effect
Worked example:
Calculate the solubility of copper (II) iodate (Ksp = 7.4 x 10-8) in an aqueous solution that is 0.20 mol/L in copper (II) perchlorate.
First write the balanced equation for the dissociation of both salts: Cu(IO3)2 (s)  Cu2+(aq) + 2 IO3- (aq)
Cu(ClO4)2 (s)  Cu2+ (aq) + 2 ClO4- (aq) Note that the perchlorate salt completely dissociates (one way arrow). It is a source of copper
ions.
Create an ICE Box for the solution, note that there is already some Cu2+ in there.
[Cu2+]
[IO3-]
Initial
0.20
0
Change
+x
+ 2x
Equilibrium
0.20 + x
2x
Both ions increase since Cu(IO3)2 is dissolving from the solid.
sub the values into the Ksp expression: Ksp = [Cu2+][ IO3-]2
7.4 x 10-8 = (0.20 + x)(2x)2 and solve for x. In this case we end up with a cubic expression – that’s too hard so we assume that x is
very small since the Ksp is small. We can approximate 0.20 + x to be ~0.20.
Substituting: 7.4 x 10-8 = (0.20)(2x)2 x = 3.04 x 10-4 mol/L
Thus the solubility of Cu(IO3)2 in a solution of 0.20 mol/L Cu(ClO4)2 is 3.04 x 10-4 mol/L
Questions:
1) Calculate the solubility of silver sulphate (in g/L) in a 1.00 mol/L AgNO3 solution. The Ksp for Ag2SO4 is 1.7 x 10-5. (5.30 x 10-3 g/L)
2) The Ksp of BaF2 is 1.7 x 10-6. Calculate its solubility in an aqueous solution that is 0.25 mol/L in NaF. (2.7 x 10-5 mol/L)
I. Determining if a Precipitate Will Occur
Worked example:
14.5ml of 3.1x10-3 mol/L AgNO3 is mixed with 27.2ml of 4.8x10-8 mol/L AlCl3. If the Ksp of AgCl is 1.8x10-10, will a precipitate form?
As soon as you see “will a precipitate form”, you’ll need to find the concentrations of the ions in the equilibrium expression, and sub
them into the equilibrium expression (Q) and compare the product to the Ksp. If the Q is less than the Ksp, then there will be no
precipitate; if Q=Ksp, then a precipitate is just starting to form; and if Q>Ksp, then there will be a precipitate.
AgCl  Ag+ + Cl- Find the [Ag+] and the [Cl-]. All the [Ag+] comes from the AgNO3 and all the [Cl-] comes from the AlCl3. Find the
new concentrations based on the total volume (41.7 ml).
AgNO3  Ag+ + NO3- 3.1x10-3 mol/L AgNO3 = 3.1x10-3 mol/L Ag+ (1:1 ratio)
M1V1 = M2V2  (3.1x10-3)(14.5) = M2 (41.7)
 M2 = 1.07x10-3 mol/L = [Ag+]
+3
-8
AlCl3  Al + 3Cl 4.8x10 mol/L AlCl3 = 1.44x10-7 mol/L Cl- (1:3 ratio)
M1V1 = M2V2  (1.44 x10-7)(27.2) = M2 (41.7)
 M2 = 9.39x10-8 mol/L = [Cl-]
Ksp = [Ag+][Cl-]
Q = (1.07x10-3)(9.39x10-8)
Q = 1.00x10-10
Since Q<Ksp, then there will be no precipitate.
Questions:
1) The [Ni+2] ion in a solution is 1.5x10-6 mol/L. If enough Na2CO3 is added to make the solution 6.0x10-4 mol/L in the carbonate ion,
will a precipitate of NiCO3 form? The Ksp of NiCO3 is 6.6x10-9. (NO)
2) If the [Ba+2] is 1.0x10-3 mol/L, what concentration of SO4-2 is necessary to just begin precipitating BaSO4 ? The Ksp of BaSO4 is
1.1x10-10. (1.1x10-7 mol/L)
J. Acids and Bases – Finding pH and pOH of a solution given [H+] or [OH-]
Formulae:
pH = -log[H+]
Kw = Ka Kb
pOH = -log[OH-]
Kw = [H+][OH-]
pH + POH = 14
[H+] = 10-pH
Kw = 1.0 x 10-14
[OH-] = 10-pOH
Questions:
1) A solution has an [OH-] = 1.0 x 10-3 mol/L. Calculate the pH, pOH, and [H+] of the solution. (pH = 11.00 pOH = 3.00 [H+] = 1.0
x 10-11)
2) A solution has a [H+] of 2.75 x 10-5. Calculate the pH and the [OH-] of the solution. (pH = 4.561 [OH-] = 3.64 x 10-10 mol/L)
K. Finding the pH of a solution given the initial concentration and the Ka
Worked example:
The Ka for nitrous acid, HNO2, is 4.47 x 10-4 . Calculate the pH of a 0.50 mol/L solution of HNO2.
First set up the dissociation: HNO2  H+ + NO2Set up the ICE Box:
Sub your values into the equilibrium expression: Ka = [H+][NO2-]/[HNO2]
[HNO2]
[H+]
[NO2-]
4.47 x 10-4 = (x)(x)/(0.50 – x) and solve for x. To make this easier, we
Initial
0.50
0
0
can make an assumption that 0.50 – x will be ~0.50.
Change
-x
+x
+x
Thus 4.47 x 10-4 = (x)(x)/(0.50) and x = 0.0149 mol/L.
Check using the 5% rule: 0.0149/.50 x 100 = 3% so the assumption is
Equilibrium
0.50 - x
x
x
OK.
If you’re really picky you can use the quadratic: x = 0.0147 mol/L (pretty darn close) Since x is the same as the [H+], use the pH
equation to find pH:
pH = -log[H+] = -log(0.0149) = 1.83
Questions:
1) The Ka for benzoic acid is 6.46 x 10-5 . Calculate the pH of a solution of 0.025 mol/L benzoic acid. (pH = 2.91)
2) Sulfamic acid, HO3SNH2, is a stabilizer for chlorine in swimming pools. Calculate the pH of a 0.050 mol/L solution of sulfamic acid
given the Ka = 0.10. (pH = 1.4)
3) Ammonia forms a base through the reaction NH3 (aq) + H2O(l)  NH4+(aq) + OH-(aq) . If the Kb of ammonia is 1.75 x 10-5, calculate
the pH of a solution of 0.05 mol/L ammonia. (pH = 10.97)
L. Find the pH of a slightly soluble solid base.
Worked Example.
Calculate the pH of a saturated solution of Copper (II) hydroxide. Ksp = 2.2 x 10-20
Write the dissociation: Cu(OH)2 (s)  Cu2+(aq) + 2 OH-(aq)
Sub into the Ksp expression and solve for x: 2.2 x 10-20 = (x)(2x)2
x = 1.765 x 10-7
Since [OH-] = 2x = 3.53 x 10-7 we can use pOH to find pH.
pH = 7.55
Questions:
1) What is the pH of a saturated solution of Ca(OH)2? Ksp = 5.5 x 10-6 (pH = 12.35)
2) What is the pH of a saturated solution of Cd(OH)2? Ksp = 2.5 x 10-14 (pH = 9.57)
M. Finding the pH of a weak base given initial concentration and Kb (or Ka)
Worked example:
Find the pH of a 0.20 mol/L solution of ammonia. Kb = 1.8 x 10-5 Remember Kw=KaKb if given Ka
First set up the dissociation: NH3 + H2O  NH4+ + OHThis example is now just like section J. Set up the ICE Box but don’t include the H2O (It’s a liquid).
Set up the equilibrium expression: Kb = [NH4+][ OH-]/[NH3]
Use an ICE box and solve for x: in this case x=0.0019
Do the 5% check and find that our assumption of 0.2-x ~ 0.2
x = [OH-] = 1.9 x 10-3 mol/L
From this we can find the pOH = 2.72 and then pH = 14 – pOH = 11.28
Questions:
1) Ethylamine, a weak base, has a Kb of 5.6 x 10-4. What is the pH of a 0.10 mol/L solution? (pH = 11.86)
2) Codeine is a weak base with a pKb of 5.79. What is the pH of a 0.020 mol/L solution of codeine? (pH=10.26)
N. Buffers – Using the Henderson-Hasselbalch equation
Worked example:
What is the pH of a solution of 0.11 mol/L sodium acetate and 0.090 mol/L acetic acid? The Ka of acetic acid is 1.8 x 10-5.
First, convert the Ka to a pKa: pKa = -log (Ka) = -log (1.8 x 10-5) = 4.74
Use the Henderson Hasselbach equation: pH = pKa + log (base/acid)
In this case pH = 4.74 + log (0.11/0.090) = 4.82
Note: even though we often use the concentrations of base and acid for this calculation, we can actually just use the number of mol of
each since they are in the same solution. If they are in the same solution, they both have the same volume – thus the volume in the
log (base/acid) part cancels out. In question 1, you can actually calculate the number of mol of base and from that, the mass.
Questions:
1) How many grams of sodium acetate, NaC2H3O2, would have to be added to 1L of 0.15 mol/L acetic acid (pK a = 4.74) to make the
solution a buffer of pH 5.00? (22 g of NaC2H3O2)
2) What ratio of molar concentrations of NH4Cl and NH3 would you use to make a buffer of pH 9.25 if the Kb of ammonia is 1.8 x 10-5
Hint: Find Ka from Kb using Kw=KaKb (Answer = 1:1)
O. Buffers – Adding strong acid or strong base to a buffer
Worked example:
A 1 L solution containing 0.100 mol of HC2H3O2 and 0.100 mol of C2H3O2- forms a buffer solution, pH = 4.74. What is the pH after
0.020 mol of NaOH are added?
The strong OH- will react completely with the weak acid HC2H3O2 and create the anion: HC2H3O2 (aq) + OH- (aq)  H2O (l) + C2H3O2- (aq)
For every mol of weak acid used up in the reaction, the same number of mol of weak anion will be created. The opposite happens
when you add an acid. For every mol of base used up in the reaction, the same number of mol of acid will be created.
We can plug in the values into the Henderson-Hasselbach equation:
pH = pKa + log (base+x /acid-x) Remember: x is the mol of STRONG base (NaOH).
= 4.74 + log [(0.100+0.020)/(0.100-0.020)]
= 4.92
Questions:
1) What is the pH of a buffer that is made with 0.100 mol HAc and 0.120 mol Ac- after 0.020 mol of HCl is added? The pKa of this
buffer is 4.74 (pH = 4.66)
2) What is the pH of a buffer made from 0.200 mol NH3 and 0.250 mol NH4Cl after 0.050 mol NaOH is added? The pKa of this buffer is
9.255. Hint: remember NH3 is the base and NH4+ is the acid. (pH = 9.352)
P. Finding the pH at the equivalence point
Strong acid- Strong base - pH is 7 at the equivalence point.
Worked example: Weak acid – Strong base
What is the pH at the equivalence point when 10ml of 0.1 mol/L HA is titrated with 0.1 mol/L NaOH? Ka = 1.8 x 10-5
a) Figure out how much NaOH is needed – since the concentrations are the same, we need 10 mL. If they aren’t the same, use C 1V1
= C2V2 to find the volume needed.
b) All of the HA is converted to A-. Find the mol HA that you started with (0.1 mol/L x 0.01 L = 0.001 mol) and then find the new
concentration of the A- with the total volume (0.001 mol/0.02 L = 0.05 mol/L A -).
c) Now you have a weak base: A- + H2O  HA + OH- use KaKb = Kw to find the Kb from the Ka (1.8 x 10-5 Kb = 1 x 10-14
Kb =
5.56 x 10-10)
d) Set up an ice box and find the OH- at equilibrium – from that you can find pH.
Kb = [HA][OH-]/[A-]
5.56 x 10-10 = x2 / 0.05 (using the assumption 0.05-x ~ 0.05)
[A-]
[HA]
[OH-]
x = 5.27 x 10-6 so [OH-] = 5.27 x 10-6 and pOH = 5.28 so pH = 8.72 (5% check OK=
Initial
0.05
0
0
0.01%)
Change
-x
+x
+x
Equilibrium
0.05 - x
x
x