UNIT 4 – EQUILIBRIUM
-
Chemical Systems in Balance
o When there is almost a complete conversion of reactants to products or when the
reaction proceeds until the limiting reactant is gone, the reaction goes to completion
 However many reactions do not proceed to completion
o
A reversible reaction is a chemical reaction that proceeds in both forward and reverse
 Ex. N2 (g) + 3H2 (g) → 2NH3 (g)
 If reaction occurs in a closed system the initial concentration of N and H
is high and NH3 is zero
o As reaction proceeds in forward direction the concentration of
N and H decreases, and concentration of NH3 increases
 After a certain time the reactants and products reach a
specific concentration in which they will not change
 Reverse reaction is occurring simultaneously

o
o
o
In a chemical equilibrium reactions are proceeding in both directions but at the same
rate, so there is no net change in concentration of reactants or products
 Ex. A tug of war
A chemical system at equilibrium has the macroscopic property of constant
concentrations of the reactant and products
 A macroscopic property is an observable or measurable property such as
concentration, colour, temperature, pressure, and pH
When a system is changing at the molecular level, but its macroscopic properties remain
constant the system is said to be in dynamic equilibrium

o
When the system is balanced in the forward and reverse reaction it is
said to be in chemical equilibrium
Keep in mind when reaching an equilibrium each substance stabilizes at
a different concentration
Some equilibrium systems consist of 1 substance in 2 physical states
 i.e. When water changes state at constant temperature
 ex. Consider water in an open system:
o Over time it will evaporate
o At least some of the water molecules at the surface have
enough K.E. and are overcoming the intermolecular attractions
of the other water molecules


o
o
o
These fast moving water molecules vaporize
 Nearby water molecules move into the gap left
by the vaporized water
ex. Consider water in a closed system:
o Same steps occur to vaporize water at the surface
o As more water vaporizes the pressure of the vapour above the
liquid increases
o Some vaporized water collides with liquid water
 If the K.E. of the vaporized water is too low they cannot
escape the attractive forces of the liquid water
 They condense and form liquid
o Therefore, rate of vaporization will
equal rate of condensation, and system
reaches equilibrium
 Although it appears static it is in
a dynamic equilibrium
A chemical system in equilibrium in which all of the components are in the same
physical state is known as homogeneous equilibrium
 Ex. N2 (g) + 3H2 (g) ↔ 2NH3 (g)
A heterogeneous equilibrium is a chemical system in which the components are in
different physical states
 Ex. H2O (l) ↔ H2O (g)
 CaCO3 (s) ↔ CaO (s) + CO2 (g)
Law of chemical equilibrium (or law of mass action)
 In a chemical system at equilibrium, there is a constant ratio between the
concentrations of the products and the concentrations of the reactants
o Ratefwd = Raterev
 The equilibrium constant keq is equal to the ratio of equilibrium concentrations
at a particular temperature for a particular chemical system
 Each molar concentration value is divided by a reference state and each is a
unit-less number; therefore keq is unit-less
o aA + bB ↔ cC + dD
 keq = [C]c [D]d
[A]a [B]b
o a, b, c, d are coefficients from balanced equation
 The value of keq tells you the relative concentration of products compared with
the concentration of reactants at equilibrium
 keq = [products]/[reactants]
o

If keq ˃ 1 product formation is favoured
 Equilibrium lies to the right
 Reactions with keq greater than 1010 are said to
be proceeding to completion
o If keq = 1 approximately equal concentrations of reactants and
products at equilibrium
o If keq ˂ 1 reactant formation is favoured
 Equilibrium lies to the left
 Reactions with keq less than 10-10 are said to
have not occurred
Ex. Write the equilibrium expression for 2SO2 (g) + O2 (g) ↔ 2SO3 (g)


-
If reaction is heterogeneous solid and liquid concentrations are constant
and therefore cancel out
Ex. Write the equilibrium expression for 3Fe (s) + 4H2O (g) ↔ Fe3O4 (s) + 4H2 (g)
The Effect of External Changes on Equilibrium
o Le Chatelier’s Principle
 If an external stress (a change in concentration, pressure, volume, or
temperature) is applied to a chemical system at equilibrium, the rates of the
forward and reverse reactions are temporarily unequal because the stress
affects the reaction rates
 However, equilibrium is eventually restored in the chemical system
o
Effect of concentration changes on equilibrium:
 Increase concentration of products causes a shift to reactant formation
 Decrease concentration of products causes a shift to product formation
 Increase concentration of reactants causes a shift to product formation
 Decrease concentration of reactants causes a shift to reactant formation
 keq is not affected by changes in the product or reaction concentrations
because the ratio of product to reactants is the same
o
Ex. Our blood contains carbonic acid that is in equilibrium with
carbon dioxide and water. We control the concentration of acid
by removing CO2 when we exhale
 Removing CO2 causes a shift in the equilibrium system
giving a shift in the reaction rates
 i.e. decrease in concentration of reactant,
therefore, a safe level for our bodies
o
Effect of volume and pressure changes on equilibrium:
 Have effect on equilibrium systems that contain gases only
o V α 1/p
 If p increases (V decreases) reaction shifts so that total number of
particles decreases, which decreases the p of the system
 If p decreases (V increases) reaction shifts so that the total number of
particles increases, which increases the p of the system
 An inert gas added to a chemical system at equilibrium does not cause a
shift in either direction
 keq is not affected by a change in p or V because the
ratio of products to reactants is the same
o Ex. Used in the process of freeze-drying foods for preservation
 First step is to freeze the food so the water in it
becomes ice, then the container is pressurized, when p
decreases the water sublimes (solid to gas), equilibrium
shifts to the right in favour of water vapour, water
vapour is then removed to reinforce the equilibrium
shift
o
Effect of temperature changes on equilibrium:
 A temperature increase (addition of thermal energy) favours the endothermic
(heat-absorbing) reaction, whereas a temperature decrease favours the
exothermic reaction
 keq increases in an endothermic system and decreases in an exothermic
system
o
Effect of a catalyst on equilibrium:
 Presence of a catalyst does not shift equilibrium in either direction
 Ex. In which direction does the reaction, PCl5 (g) + heat ↔ PCl3 (g) + Cl2
(g), shift as a result of each of the following changes?
o Adding additional PCl5 (g)
o Removing Cl2 (g)
o Decrease temperature
o
o
-
Increase pressure by adding helium gas (an inert gas)
Using a catalyst
Calculating Equilibrium Constants
o Calculating keq
 Ex. A 5.0 L flask at constant temperature contains N2 (g), Cl2 (g), and NCl3 (g).
The chemical equilibrium system is represented by: N2 (g) + 3Cl2 (g) ↔ 2NCl3
(g). It contains 0.0070 mol of N2, 0.0022 mol of Cl2 and 0.95 mol NCl3. Calculate
the equilibrium constant for this reaction.

o
Ex. N2 (g) and H2 (g) are mixed in a closed 3500 ml flask. They react and form
ammonia. At equilibrium the mixture contains 0.25 mol of NH3 and 0.08 mol of
H2. The equilibrium constant for this reaction is keq = 5.81 x 105. What amount
of N2 is present at equilibrium?
Equilibrium constant, kp
 As long as each gas behaves as an ideal gas law can be used to derive the
equilibrium constant using gas partial pressures
 Recall: PV = nRT
o Note that, n/V = p/RT which means the molar concentration is
equivalent to p/RT
 Ex. CO (g) + 3H2 (g) ↔ CH4 (g) + H2O (g)
o kp = PCH4 PH2O
PCO P3H2
 Units for partial pressures are given in
atmosphere, atms (pascals, Pa, or torr)
c
 kp = P C PdD
PaA PbB
 Relation between keq and kp
 kp = keq (RT)Δn
o Where Δn is the sum of the coefficients of gas products minus
the sum of the coefficients of gas reactants
-
Using ICE Tables to Determine Equilibrium Information
o If the system info is organized in a table it is easier to see the relationships among the
components and if info is missing
o I refers to initial molar concentration, C refers to amount of change in reactant and
product, E refers to the molar concentrations of reactants and products at equilibrium
 Ex. You want to experimentally determine the equilibrium constant for the
decomposition of HI (g) at 453 C. You fill an evacuated 2.0 L flask with 0.200
mol of HI. You let the HI decompose into H2 and I2 until it reaches equilibrium.
You determine that the concentration of HI is 0.078 mol/L at equilibrium but
you do not know the concentration of H2 or I2.
 Step 1
o
Write out the equilibrium equation and make a column below
each reactant and product. Then label the rows with Initial,
Change, and Equilibrium
2HI (g)
[HI]
↔
H2 (g)
[H2]
+
I2 (g)
[I2]
I
C
E

I
C
E

Step 2
o Add the known values to the table
 HI = 0.2 mol/2 L = 0.1 mol/L
2HI (g)
[HI]
0.1
↔
H2 (g)
[H2]
0
+
I2 (g)
[I2]
0
Step 3
o Let x represent change in 1 concentration and then express the
other concentration in terms of the change to x
o Use a plus sign to show that a substance forms in the reaction
or is increasing
o A minus sign is used to show that a substance is consumed in
the reaction or is decreasing
 Recall: coefficients in balanced equation show molar
relationships of components in the reaction
I
C
E

I
C
E

o
2HI (g)
[HI]
0.1
-2x
↔
H2 (g)
[H2]
0
+x
+
I2 (g)
[I2]
0
+x
S
Step 4
o The equilibrium concentration of each component equals the
initial concentration plus or minus the change in concentration
2HI (g)
↔
H2 (g)
+
I2 (g)
[HI]
[H2]
[I2]
0.1
0
0
-2x
+x
+x
x
x
S 0.1 – 2x
Step 5
o You also measured the equilibrium concentration of HI to be
0.078 mol/L
 0.1 mol/L – 2x = 0.078 mol/L
 2x/2 = (0.1 - 0.078)/2
 x = 0.011 mol/L

Step 6
o Therefore [H2] = [I2] = x = 0.011 mol/L

Step 7
o Solve for equilibrium constant
 keq = [H2][I2]
[HI]2
 keq = 0.011 (0.011)
(0.078)2
 keq = 0.020
Ex. CO (g) + H2O (g) ↔ H2 (g) + CO2 (g), At 700 K, the equilibrium constant is 0.83.
Suppose that you start with 1.0 mol CO (g) and 1 mol of H2O (g) in a 5.0 L container.
What amount of each substance will be present in the container when the gases are in
equilibrium?


Can also use quadratic equation to solve
o 1 solution should be impossible or improbable
 Ex. A negative value or a concentration higher than the
sum of the initial concentrations
When you have an expression that requires the quadratic equation to
solve, you can sometimes make an approximation that will simplify the
equation
o If the keq for a reaction is very small, then a very small
concentration of product is present at equilibrium
o The concentration of reactant at equilibrium is almost the same
as the initial concentration


o
An approximation can be made if the initial
concentrations of reactants are at least 1000 times
greater than keq
It is also possible to measure the equilibrium concentrations using colour
 When a reaction involves a coloured substance, the change in colour
intensity can be measured and used to determine the equilibrium
constant for the reaction
o Because the reaction involves a colour change it is possible to
determine the concentration by measuring the intensity of the
colour
The reaction quotients, Qeq and Qp
 Before a reaction reaches equilibrium, the ratio is not constant
 A reaction quotient has the same formula as keq or kp but the chemical system
may or may not be at equilibrium
 If Qeq < keq (or Qp < kp) the ratio of products to reactants is less than
keq
o To reach equilibrium more products must form and reactants
must be consumed
 The reaction shifts to the right to reach equilibrium
 If Qeq = keq the system is at equilibrium
 If Qeq > keq the ratio of products to reactants is greater than keq
o Products must decompose into reactants to reach equilibrium
 Reaction shifts to the left to reach equilibrium

-
Ex. N2 (g) + 3H2 (g) ↔ 2NH3 (g), At 500 C the value of keq for this reaction is
0.40. The following concentrations of gases are present in the container: [N2] =
0.1 mol/L, [H2] = 0.30 mol/L, and [NH3] = 0.20 mol/L. Is this mixture of gases at
equilibrium? If not, in which direction will the reaction shift to reach
equilibrium?
Acid-Base Equilibrium
o Many acid-base reactions occur in water
 HCl + NaOH → H2O + NaCl
 Net ionic: H+ + OH- → H2O
o When HCl breaks apart in water (dissociation) the positively charged H ion from the acid
is attracted to the negatively charged electrons on the surrounding water molecules
 Recall: That 1 oxygen in water can be bonding with up to 6 hydrogen’s
o The attraction is so strong that the H ion forms a covalent bond with the water, forming
H3O+ (hydronium ion)
 The formation of this charged particle is called ionization
 Therefore we assume that all acids form H3O+ and all bases form OH- in
water
o Arrhenius Principle
o
o
o
o
However Bronsted-Lowry theory defines acids and bases according to the donation and
acceptance of a proton
 Acids and bases do not need to be in aqueous solution and their activity in
water is not part of the definition
A Bronsted-Lowry acid is a proton donor or any substance that donates a H ion
 Must contain H in acid formula, ex. HCl, HNO3
A Bronsted-Lowry base is a proton acceptor or any substance that accepts a H ion
 A base must have a lone pair of electrons to bind with H ion
 Ex. NaOH, NH3, HF
According to Bronsted-Lowry any substance that behaves as an acid can do so only if
another substance behaves as a base at the same time (or vice versa)
 Ex. HCl + H2O → H3O+ + Cl 2 molecules or ions that differ because of the transfer of a proton are called a
conjugate acid-base pair
 Conjugate base of an acid is the particle that remains when a proton is
removed from the acid
 Conjugate acid of a base is the particle formed when the base receives
the proton from the acid
o
o
o
Ex.
o
Ex. Identify the conjugate acid-base pairing:
 NH3 + H2O → NH4+ + OH-
A molecule or ion that can accept or donate a proton can act as an acid or a base
 It is said to be amphiprotic
 i.e. H2O, HSO4Due to the acid-base reactions use of water and it being amphiprotic we represent
water in an auto-ionization reaction (or self-ionization)
 H2O ↔ H+ + OH H2O + H2O ↔ H3O+ + OH keq = [H3O+][OH-]
o Therefore, kw = [H3O+][OH-]
 Where kw is the ion product constant of water
 kw = 1.0 x 10-14 mol2/L2 (at 25 C)
o i.e. [H3O+] = 1.0 X 10-7 mol/L; [OH-] = 1.0 x 10-7 mol/L



If concentration of hydronium increases then concentration of hydroxide
decreases (and vice versa)
Recall that a change in concentration will result in a shift of equilibrium
 Le Chatelier’s Principle
If additional acid is added to hydronium concentration increases, if additional
base is added hydroxide concentration increases
o Therefore,
 Acidic solution [H3O+] > [OH-]
 Neutral solution [H3O+] = [OH-]
 Basic solution [H3O+] < [OH-]

Ex. The [OH-] is household window cleaner is 0.002 mol/L at 25 C. What
is [H3O+]?
o
o
To determine acidic values
 pH = - log [H3O+]
 If we want concentration
o [H3O+] = 10-pH
To determine basic values
 pOH = - log [OH-]
 if we want concentration
o [OH-] = 10-pOH


-
Relationship:
o pkw = pH + pOH = 14
Ex. A solution of [H3O+] is 0.50 mol/L. Calculate pH, pOH, and [OH-].
Acid-Base Strength
o Strong acids dissociate completely into ions in water
 HA (g or l) + H2O (l) → H3O+ + AAcid
base
con. Acid con. base
 Once the acid dissolves in water, there are no intact acid molecules left in the
solution
 The concentration of hydronium is approximately equal to the initial
concentration of the acid
o
A weak acid dissociated slightly into ions when dissolved in water
 Most molecules of the acid remain intact and in equilibrium with only a few
hydronium ions that form
 Initial concentration of weak acid is approximately equal to concentration of
weak acid at equilibrium
 i.e. very few hydronium and A- formed
 When a weak acid is added to water, dynamic equilibrium is established in the
solution
 Reaction is very slow with weak acid
o
Strong bases are similar to strong acids in that they ionize completely in water
 Ex. Mainly form with hydroxide and group 1 and 2 metals
 NaOH, KOH, Mg(OH)2, Ba(OH)2
 NaOH → Na+ + OH-

Water is written above the arrow because it is required as a solvent,
and is not considered a reactant
o
Weak bases are similar to weak acids due to ionizing slightly when dissolved in water
o
In general, acid-base equilibrium reactions shift in the direction in which a strong acid
and strong base form a weaker acid and weaker base
 Ex. H2S + NH3 ↔ HS- + NH4+
 Reaction shifts right because H2S is a stronger acid than NH4+ and NH3 is
a stronger base than HS-
o
A special equilibrium constant is used for weak acids because not all of the acid ionizes
in the water
 Called acid-dissociation constant (or acid-ionization constant), ka
 Consider HA that donates 1 proton and is a monoprotic acid
 HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)
 keq = [H3O+][ A-]
[H2O][ HA]
 but water is constant and therefore part of keq
o keq = ka = [H3O+][ A-]
[HA]
 At a constant temperature for the ionization of a weak acid at equilibrium, the
stronger the acid is, the higher the [H3O+] and the larger the value of ka will be


2 assumptions are made:
 1. [H3O+] from auto-ionization of water is negligible because it is much
smaller than [H3O+] formed from ionization of weak acid
 2. A weak acid dissociated into ions to a small extent that in calculations
to determine its equilibrium concentration, its change in concentration
is ignored
o i.e. concentration of acid at equilibrium is approximately equal
to initial concentration of acid


A percent dissociation is used to determine that amount of weak acid
that ionizes
o % dissociation = [HA]dissociated x 100
[HA]initial
Ex. An industrial chemist prepares a 0.050 mol/L solution of nitrous acid
which is represented by the equation below. What is the pH of the
solution?
HNO2 (aq) + H2O (l) ↔ NO2- + H3O+ (aq)
Some acids are polyprotic acids which have more than 1 ionizable proton
 One proton ionizes at a time
o Ex. H3PO4 (aq) + H2O (l) ↔ H2PO4- (aq) + H3O+ (aq)
 ka1 = [H2PO4-][ H3O+]
[H3PO4]
 The conjugate base now becomes the acid
 H2PO4- + H2O ↔ HPO4-2 + H3O+
o ka2 = [HPO4-2][ H3O+]
[H2PO4-]
 The conjugate base now becomes the acid
o HPO4-2 + H2O ↔ PO4-3 + H3O+
 ka3 = [PO4-3][ H3O+]
[HPO4-2]
o The ka values decrease as more protons are removed
 Recall smaller ka demonstrate weak acids
-
Base Ionization
o Consider a generic weak base dissolving in water
 B (aq) + H2O (l) ↔ BH+ (aq) + OH- (aq)
o The base does not dissociate, but it does become ionized
 Gives base ionization constant, kb
 i.e. kb = [BH+][ OH-]
[B]

o
Ex. One of the uses of aniline, C6H5NH2 (l) is the manufacture of dye. It is soluble
in water and acts as a weak base. When a solution containing 5 g/L of aniline is
prepared the pH was determined to be 8.68. Calculate kb.
Relationship between ka, kb, and kw
 CH3COOH + H2O ↔ H3O+
+ CH3COOAcid
base
con. Acid
con. Base

The conjugate base then reacts with water
 CH3COO- + H2O ↔ CH3COOH
Acid
base
con. Acid

-
+
OHcon. Base
Therefore, ka = [H3O+] [CH3COO-]
kb = [OH-] [CH3COOH]
[CH3COOH]
[CH3COO-]
o kw = kakb
o kw = [H3O+] [CH3COO-][OH-] [CH3COOH]
[CH3COOH] [CH3COO-]
o kw = [H3O+][ OH-]
 Therefore, the overall equilibrium constant for an acid
and its conjugate base is kw = kakb
Salts, Buffers, Titrations, and Solubility
o Salt hydrolysis
 When an acid and base react, water and salt are formed
 Referred to as neutralization reaction
 Salts are electrolytes that completely ionize in water and often affect pH of a
solution


o
When it dissociates in water a anion and cation form in a process called salt
hydrolysis
Types:
 Neutral
o Salt consisting of anion of a strong acid or a cation of a strong
base reacts with water, no reaction takes place
 Therefore the acid/base completely dissociates and
ionizes
 Ex. HNO3 + H2O → NO3- + H3O+
 Acidic
o When salt consisting of the anion of a strong acid and the cation
of a weak base ionize in water, the solution is acidic because the
cation acts as a weak acid
 Ex. NH4Cl → NH4+ + Cl NH4+ + H2O ↔ NH3 + H3O+
o From [H3O+], therefore acidic
o Metal ions also form acidic solutions; ex. Al+3, Cr+3, Fe+3, Bi+3, and
Be+2
 Ex. [Al(H2O)6]+3 + H2O ↔ [Al(OH)(H2O)5]+2 + [H3O+]
 Basic
o When a salt consisting of the anion of a weak acid and cation of
a strong base ionizes in water the solution is basic
 CH3COONa → Na+ + CH3COO CH3COO- + H2O ↔ CH3COOH + OH-
Buffered solutions
 A buffer solution resists change in pH when a limited amount of acid or base is
added to the solution
 Must contain a large quantity of acid to react with any base added to the
solution (vice versa)
 The acid and base components in the solution cannot react in a neutralization
reaction
 Contain a mixture of weak acid and conjugate base (and vice versa)
 Sometimes in buffer solutions a common-ion effect can be observed
 Which is the shift in equilibrium position caused by the addition of a
compound that has an ion in common with 1 of the dissolved
substances
o Ex. CH3COONa → CH3COO- + Na+
 When acid is added to buffer:
 CH3COO- + H+ → CH3COOH
 When base is added to buffer:

o
 CH3COOH + OH- → CH3COO- + H2O
Buffering capacity depends on how much acid/base the solution can absorb
without a significant change in pH
Acid-Base titration
 We use indicators to measure the endpoint which has a specific pH when it will
change colour



Indicators are weak acids; the acid is 1 colour and the conjugate base is another
 HInd + H2O → H3O+ + IndAcid
con. Base
An equivalence point is reached when the amount (in moles) of the substance
added is stiochiometrically equivalent to the amount (in moles) present in the
original test solution
 Or the point at which the acid and base have completely reacted
3 types of titration reactions:
 Strong Acid-Strong Base
o The pH of the solution increases slowly at first
o The low pH indicates a high concentration of hydronium ion
from the strong acid
 Base is added and near equivalence point a pH curve
demonstrates an almost vertical increase

o

pH will be approximately 7 when at equivalence
point
Ex. A 40.00 mL solution of 0.10 M HCl is titrated with 0.10 M
NaOH. Calculate pH of solution for:
 A. pH of original solution of strong acid
 B. pH of solution before equivalence point; after adding
20 mL NaOH
 C. pH of solution at equivalence point; after adding 40
mL of NaOH
 D. pH of solution after equivalence point; after adding
50 mL NaOH
Weak Acid-Strong Base
o The partial ionization of the weak acid has to be considered in
calculations
o
o
o
Initial pH is higher for a weak acid and there is less hydronium
ions
Has a buffer region
 As base is added, more conjugate base forms; creating
acid/conjugate base buffer
pH at equivalence point will be greater than 7
o
o

Weak Base-Strong Acid
o
o
o
o
The weak acid anion acts as a weak base and accepts a proton
from water
 Yields OH ions that increase pH of solution
Ex. A chemist titrated 25 mL of a 0.1 mol/L solution of acetic
acid with 0.1 mol/L solution of NaOH. Calculate pH of resulting
solution after addition of 10 Ml and 25 mL of NaOH.
Same shape as weak acid-strong base but inverted
 pH decreases throughout the process and initial pH is
basic because it sits in the flask
After buffer region pH drops almost vertically
 i.e. base and conjugate acid
The solution is acidic at the equivalence point
Equilibrium for solubility
 Slightly soluble ionic compounds reach equilibrium with only small amounts of
solute dissolved in solvent
 Ex. AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
o Expressed by a reaction quotient
 Qeq = [Ag+][ Cl-]
[AgCl]
 The solid value is part of the constant
o Therefore, we get ion-product
expression, Qsp
 Qsp = Qeq[AgCl]



 Qsp = [Ag+][Cl-]
At equilibrium Qsp becomes constant and this constant value is called the
solubility-product constant, ksp
 Therefore, ksp = [Ag+][Cl-]
Can use ksp and Qsp to determine if a precipitate will form
o Qsp < ksp solution is unsaturated and no precipitate
o Qsp = ksp solution is saturated and no change occurs
o Qsp > ksp a precipitate forms until the solution is saturated
Ex. The solubility of silver carbonate is 1.3 x 10-4 mol/L. Calculate ksp.
Ag2CO3 (s) ↔ 2Ag+ + CO3-2

Ex. Solubility of CaSO4 is 0.67 g/L. What is the volume of ksp?
CaSO4 (s) ↔ Ca+2 + SO4-2

Ex. If exactly 200 mL of 0.004 mol/L BaCl2 is mixed with exactly 600 mL of 0.008
mol/L K2SO4. The particles ionize and the only possible precipitate that can form
is BaSO4. Will a precipitate form?