L3- Equibrium Principles

Page 1

Equilibrium expression- Kc

Example equation. Aa+ Bb ----> Cc+ Dd

ABCD - represent reactants and products

abcd- represent the exponents

Kc= [C]c [D]d / [A]a [B]b

Note: Solids are not written in equilibrium equations, liquid H2O is not written as it I in such small quantities it is negligible

EXAMPLE

N2 (g)+ 3H2(g) <> 2NH3 (g) the equilibrium expression will be...

[ NH3] 2 / [N2] [H2]3

Page 2

Solubility product- Ks

Equilibrium equation used for saturated salts in solutions of sparingly soluble salts

turte solutions contain the maximum amount of dissolved ions and should be thought of as having some amount of undissolved amount of ionic solid in the system

Ks is derived in a similar way to Kc

The equilibrium expression is Ks= [A+][B-]

Formula types include .. AB2, A2B and AB

Example of an A2B style equation: A2B <---> 2A+ (aq) + B2- (aq). Ks= [A+]2[B2-] 2:1 ratio

Example of an AB2 style equation : AB2+ <---> A2+ (aq) + 2B- (aq) Ks= [A2+] [B-]2. 1:2 ratio

Example of an AB style equation : AB <---> A2+ (aq) + B2- (aq). Ks=[A2+][B2-]. 1:1 ratio

Because Ks is calculated from expressions with reactants at very low concentrations this means that sparingly soluble ionic compounds will have very small Ks values.

Solubility is usually given as moles per litre of water (molL-1).

Solubility is calculated Experimentally, and may then be used to calculate the Ks of the compound.

You may be asked to Find Solubility Product (Ks) using a given solubility OR Find the solubility given the solubility product (Ks) value

Formula you need to know

c=n/v ... c= concentration of solution mols per litre. , n= amount of substance being dissolved (moles), v=volume of solution in litres

n=m/M ... n= amount of a substance being dissolved (in moles) , m= mass (in grams) , M= Molar mass (in gmol-1)

Page 3

Calculating Solubility Product (Ks) (AB Type, 1:1 Ratio)

Question: the solubility of Barium Sulfate ( BaSO4-) is 0.00260 gL-1 in pure water at 25 * c. Calculate the solubility product (ks) of Barium Sulfate

M= (BaSO4)= 233 g mol=1

STEP 1- Convert Solubility from gL- to molL-1

n=m/M

n= 0.00260 /233

n= 1.1158 x10-5mol

STEP 2- Convert solubility from gL- to molL-1

c=n/v

c= 1.1158 x 10-5 mol/1

c=1.1158 x10-5 mol

STEP 3- Write out the equation and figure out the expression (AB)

BaSO4(s) <> Ba2+ (aq) + SO42- (aq)

[Ba2+][SO42-] [A2+][B2-] therefore this is AB type question

STEP 4- put the numbers into the expression and calculate Ks ( has not units)

Ks = [1.12 x 10-5] x [1.12x10-5]

Ks = 1.25 x 10 -10

SECOND EXAMPLE QUESTION for Finding solubility product Ks (AB Type, 1:1 ratio)

Solubility of Calcium Sulfate is 2.40gL-. Calculate the solubility product of calcium sulfate. M (CaSO4)= 136mol-1

n=m/M. n= 2.40/136. n= 0.0176

c=n/v. c= .0176/1. c=0.0176

CaSO4- <> Ca2+(aq) + SO4 2- (aq)

ks= [Ca2+][SO42-]. ([A2+][B2-] therefore this is AB type 1:1 ratio )

Ks= (0.0176) x (0.0176)

Ks= 3.0976 x 10-4

Page 4

Calculating Solubility product Ks- (AB2 and A2B types, 2:1 ratio and 1:2 ratios) (2B or not 2B)

In BOTH cases 1 mole of solid produces 3 moles of ions.

if the solubility f the ionic compound is s mol -1, then there will be 3s moles of ions in solution .

The solubility product expression will be s= [A]2.[B] = (2s)2.s or [A].[B]2 = s.(2s) = 4s3

In a question that already provides you with the solubility in mol L-1

Eg: The solubility of Copper hydroxide is 2.29 x 10-7 pure water. Find the solubility of product of Copper hydroxide...

You can use the simple method of Ks=4s3 to find the product.

Ks= 4s3

Ks=4 x (2.29x10^-7) ^3

Ks= 4.80 x 10 ^-20

EXAMPLE QUESTION ( AB2 type 1:2 Ratio)

The solubility of barium flouride BaF is 0.160 g per 100ml of pure water. Calculate the solubility product (Ks) os Barium Flouride.

M (BaF2)=175g mol-2

n=m/M

n= 0.160/175

n= 9.142 x10-4

c=n/v

c= 9.142x10-4/0.100

c= 9.142x10-3

BaF2 <---> Ba2+ (aq) + 2F-

Ks= [Ba2+][F-]2

Ks= 4 X (9.142x10-3) 3

Ks= 3.06 x 10-6

SECOND EXAMPLE QUESTION

The solubility of barium hydroxide is 18.5gL-1 pure water. Calculate the solubility product of barium hydroxide. M(Ba(OH)2)= 171g mol-1

n=n/M

n= 18.5/171

n=0.108

c=n/V

c= 0.108/1

c=0.108

Ba(OH)2 <---> Ba2+ (aq) +2OH-(aq)

[Ba2+][OH-]2 ( [A2+][B-]2 therefore this is AB2 Type ratio 1:2)

Ks= 4x (0.108) ^3

Ks= 5.03x10-3

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Precipitation reactions

When the sparingly soluble ionic solid CaF2 is placed into water, the solid starts to dissolve and its ions go into solution according to the equation...

CaF2(aq) -----> Ca2+ (aq) + 2F- (aq)

As time passes the concentration of dissociated ions increase so the rate of the reverse precipitation reaction increases according to the equation.

Ca2+ (aq) + 2F- (aq) ----> CaF2 (s)

At some point the rate of the solid is dissolving becomes equal to the rate at which the ions are precipitating and dynamic equilibrium's is reached.

The concentration of the ions Ca2+ and F- remains constant. We can represent the solution y an equilibrium equation

Ca2+ (aq) + 2F- (aq) <-----> CaF2 (s)

Equation: K = [Ca2+][F-]2/ [CaF2]

The value of [CaF2] is a constant because CaF2 is a solid , this means that its value can be incorporated into the value for K to give what is called the solubility product expression

Page 6

Calculating Solubility FROM the Solubility product Ks (AB Type 1:1 Ratio)

Using the value of Ks at any certain temperature we can calculate the solubility of the salt and work out its concentration of ions in the saturated solution.

This process is essentially the opposite of calculating Ks.

Equation is S2=Ks

EXAMPLE QUESTION

Q1) The solubility product (Ks) of Silver iodine is 8.52x10-17 in pure water. Calculate the solubility of Silver Iodide in mol L-1

STEP 1- figure out the Ks Expression for the equation

AgI(s) <---> Ag2+ (aq) + I- (aq)

Ks = [Ag2+][I-]

STEP 2- Because of the 1:1 ratio of AB, Ag=I, therefore we use Ks=S2

S2=Ks

S2= 8.52 x 10-17

S= √ (8.52 x 10-17) Remember to put the Ks in brackets or it will not work.

S= 9.23 x 10-9 mol L-1

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Calculating Solubility FROM the Solubility Product (Ks) (A2B and AB2 Types 2:1 ratio and 1:2 ratios)

Ks= [A]2.[B] or [A].[B2] = (2s)2.s or s.(2s)2 = 4s3

Rearranging the Ks=4s3 equation in order to find solubility gives : s= 3√ Ks/4

Question (AB2 type 1:2 ratio)

The solubility product (Ks) of Lead (II) Chloride (PbCl2) in pure water is 1.70 x 10-5

STEP 1 -Write the equation for the dissolution of lead chloride in water

PbCl2 <---> Pb 2+ (aq) + 2Cl - (aq)

b) Calculate the [Pb2+] in a saturated solution of lead chloride

s= 3√ Ks/4

s= 3√ ( 1.70 x 106 ^ -5 /4)

s=0.0162 mol L-1

Question 2- Multi questions (A2B 2:1 ratio)

The solubility product (Ks) of Lithium Carbonate (Li2CO3) in pure water is 2.51x 10-2

Equation- Li2CO3 <---> 2Li+(aq) + CO2- (aq)

Calculate the [CO32-] in a saturated solution of lithium carbonate

S= 3√Ks/4

S= 3√ (2.51x10-2 / 4)

S= 0.1844 mol L-1

Calculate [Li+] in a saturated solution of lithium carbonate, as this is an A2B example with a 2:1 ratio, the solubility of Lithium is 2 times the solubility of CO32-.

2Li+ = 0.1844 x 2

2Li+ = 0.368 mol L-1

State the solubility in molL-1 of lithium carbonate

s Li2CO3= 0.184 mol L-1

Show using a calculation that your answer to the above gives a solubility product (Ks) of 2.51x10^-2 for lithium

Ks= 4(Ks)^3

Ks=4x(0.1844) ^3

Ks= 2.51 x 10^-2

Page 8

Dilution Factors

If a solution f one type is added to another the concentration will decrease proportionally to the volumes and molarity present.

Formula to work out the new concentration of a solution : [New]= volume used/total volume x original]

Example:

6.42 mL of 0.770 molL-1 potassium iodide solution, (KI) is mixed with 95.8 mL of pure water. Determine the new concentration of the potassium iodide solution.

STEP 1

Formula - [new]= volume used / total volume x original.

To find the 'total volume' add 6.42 mL and 95.8mL together giving you 160 mL

[new]= ( 6.42/ 160 ) x 0.770 = 0. 309 mol L-1

Page 9

Common Ion Effect - Effect of concentration (Le Chatliers)

The solubility of an ionic compound is reduced if it is added to a solution already containing one of its ions ( a common ion).

if the solution being added contains either one of the ions in the original ions already present in the solution then this will reduce the solubility of the salt ice the presence of the common ion will moe the equilibrium toward the side of the precipitate salt.

Because Ks is a constant, the increase in concentration of one ion means that the concentration of the other must fall to maintain the value of Ks

Consider Le Chatliers Principle when solving these problems concerning the common ion effect.

Eg: AB(s) <-----> A+(aq) + B- (aq)

if the concentration of either A+ or B- (the products) is increased, the position of the equilibrium will shift to the LEFT, so if [A+] increases then [B-] will decrease and vice versa

EXAMPLE QUESTION

A small amount of the sparingly soluble ionic compound , Barium Sulphate (BaSO3), was added to 50mL of pure water at 25*C.

Equilibrium between the solid and its ions was allowed to form.

This solution and the undissolved solid were added to 50 mL of 0.100 mol L-1 Barium Iodate (Ba (IO3)2) solution.

Ks (BaSO3) = 5.00 x 10-10. Ks( Ba(IO3)2)= 1.5 x 10-9

The common ion between these 2 solutions is Barium (Ba) as they both contain this.

When these two solutions are mixed, the concentration of the sulphate and iodate ions will decrease, and the concentration of solid barium sulphite in the system will increase.

Explain the effect that adding Ba(IO3)2) solution has had on the equilibrium that has been established between solid BaSO3 and its ions in solution (calculations are not required) .

In terms of Le Chatliers:

BaSO <-----> Ba2+ (aq) + SO3 2- (aq)

Le Chatliers principle states that constraints placed on a system at equilibrium will force the equilibrium to shift in order to minimise the effects of these constraints. In this case the constraint is concentration. When the concentration of Barium (Ba2+) increases the position of the equilibrium will shift to the left hand side to decrease the concentration of Ba2+ being produced and also decreasing the concentration of SO32-. This means that the BaSO4 becomes less soluble.

In terms of Ks, the Ks is a constant. The common ion effect causes the Ks to be initially overtaken, causing precipitation until the new concentrations satisfy the value of Ks again

Common Ion Calculations

Note: It is important to note that calculation involving common ions Make major assumptions

"The common ion added to the solution containing the sparingly soluble compound will have a concentration very much higher than the concentration of the common ion in equilibrium solution. In calculations it is therefore acceptable to assume that all of the common ions come from the solution other than the one containing the sparingly soluble compound"

50.0mL of a 0.500 molL-1 solution of sodium chloride was added 50.0mL of a saturated solution of silver chloride. Calculate the [Ag +] after the two solutions were mixed and equilibrium allowed to re-establish. Ks (AgCl)= 1.77 x 10-10

Step 1- Assume all Cl- ions come from the sodium chloride solution, Include a dilution factor- 50.0 ml @ 0.500 made up to 100mL.

[Cl-] = 0.250 mol L-1

Ks= [Ag+][Cl-] = s2

Step 2- Rearrange the solubility product expression to save for the ion whose concentration is required.

Step 3- Substitute values and calculate (include unit)

Example 2

Dissolving of AgCl in pure water.

Equation: AgCl (s)<------> Ag2+ (aq) + Cl- (aq)

Ks (AgCl)= [Ag+][Cl-]

AgCl will be less soluble in sea water than in pure water because of the presence of the common Cl- ion dissolved in the seawater.

The value of Ks cannot change. So the [Cl-] Means that there is a lower [Ag+] at equilibrium. As the equilibrium shifts to minimise the effects .

This reduces the amount of solid AgCl that can dissolve

Example calculation - What is the solubility of AgCl in a -0.0025 molL-1 solution of NaCl?

Ks(AgCl) = [Ag+(aq)] x [Cl-(aq)]= 1.6x10 -10

Ks(AgCl) = [Ag+(aq)] x [Cl-(aq)]

1.6 x 10-10 = s x 0.0025 (rearrange formula to find s)

s= 1.6 x 10-10 / 0.0025

s= 6.4 x 10 -8 molL-1

Page 10

The Ionic product (Qs)

Type AB- Qs=[A+][B-] . Type A2B- Qs= [A+][2-] . Type AB2- Qs- [A2+][B-]2'

while the solubility product applies to saturated solutions at specific temperatures, the ionic product can be calculated for any concentration of solution

Example Question -

Q) write the ionic product expression and then calculate the ionic product of the following solutions.

a) 0.00500 mol L-1 copper iodide (Cul)

[Cu+][I-] AB 1:1 ratio

[0.00500] x [0.00500]

= 2.5 x 10 - 5

b) 0.105 mol L-1 of Silver Chromate (Ag2CrO4)

[Ag+]2 x [CrO24-] 1:2 A2B

=[0.210]^2 [0.105]

= 4.63 x 10 -3

c) 0.00755 mol L-1 Strontium Flouride (SrF2)

[ Sr 2+][F-] 2 AB2 . 1;2 ratio

Qs= [ 0.00755][0.0151]2

Qs= 1.721 x 10 -6

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Predicting Precipitation

if the concentration of ions in a solution exceeds the solubility product (Ks) then a precipitate will form. To determine wether Ks for the sparingly soluble ionic compound has been exceeded, first calculate the ionic product (Qs) from the data given, and compare it to the Ks value for the compound.

Qs < Ks . No precipitate ( solution is UN saturated)

Qs = Ks No precipitate ( Solution IS Saturated)

Qs > Ks . A precipitate will form (solution has exceeded saturation point)

Example question

Determine whether a precipitate of lead chloride PbCl will form when 200 mL of a 0.5 mol L-1 solution of sodium chloride NaCl is added to a solution made by completely dissolving 1.50 g of lead (II) nitrate Pb(NO)2) in 250mL of pure water.

Ks= (PbCl2)= 1.70 x 10-5 . M(Pb(NO)2)= 331 gmol-1

Step 1- Calculate the conc of the Pb(NO3)2 in the total volume and therefore [Pb2+]

n=m/M

1.50/331

n = 4.53 x 10-3

c=n/V

Add 200 mL and 250 mL to get the total volume of the substance.

4.53 x 10-3 / 0.450

c=1.01 x 10 -2 mol L-1

Step 2 - [Cl-] from NaCl using the dilution factor

[Pb2+]= 1.01 x 10-2

[Cl-]= from sodium chloride (NaCl) = 0.500 x 200/ 450

0.222 mol L-1

Step 3- Calculate Qs

Qs Lead Chloride (PbCl2)

Qs= [Pb2+] [Cl-]2

Qs= 1.01 x 10-2 x 0.222 ^2

Qs= 4.97 x 10 -4

Step 4 - Determine wether Qs exceeds Ks for PbCl2

Qs (4.97 x 10-4) > 1.0 x 10 - 5)

Qs is greater than (>) Ks so precipitate WILL occur

Page 12

Effects of pH on Solubility-

Acid Conditions

The presence of H3O+ Hydronium ions can greatly increase the solubility of sparingly soluble compounds

IF they contain an ion that reacts with acids.

Hydroxides (OH-), carbonates (CO32-), flourides (F-) and ethanoates (CH3COO-) all become more soluble in acidic solutions BECAUSE their anions all react with H3O+.

The increase in solubility is explained by Le Chatlier's principle.

Magnesium hydroxide is more soluble in 0.100 molL-1 hydrochloric acid than in pure water. This is because the [presence of hydronium ions in the acid increase the solubility of Magnesium hydroxide as the hydronium ions in the HCl will react with the hydroxide ions of the MgOH to produce water.

Based on Le Chatlier's principle, as the products are removed from the the equilibrium it will shift to the right to attempt to replace the "lost " products and hence decrease the amount of reactants (Magnesium Hydroxide) to attempt to maintain equilibrium.

Alkaline Conditions

Basic conditions will increase the solubility of sparingly soluble ionic compounds IF the metal cation is able to form complex ions with the basic anion.

Example-

the equilibrium between solid copper II hydroxide and its ions in solution:

Cu(OH)2 (s) <------> Cu 2+ (aq) + 2OH- (aq)

the reaction between copper (II) ions and ammonia :

Cu2+ (aq) + 4NH3(aq) <-----> [Cu(NH3)4]2+ (aq)

The solubility of Copper Hydroxide is greatly increased in ammonia solution as copper ions are removed from the equilibrium as they form complex ions (Tetraammine Copper). Equilibrium will shift to the right to replace the lost copper ions thereby reducing the amount of the reactant (solid copper hydroxide)

Example 2-

Copper Carbonate is sparingly soluble in pure water (Ks(CuCO3)= 1.462 x 10-4).

In 1.00 mol L-1 ammonia solution the solubility of copper carbonate is greatly increased.

Explain this increase in solubility using equilibrium principles . Calculations aren't needed.

The solubility of Copper Carbonate is greatly increased by the presence of Ammonia solution. compared to it sparing solubility in water.

This is because the Copper ions are being removed from the solution in the formation of the complex ion tetraammine copper.

As the ions are being removed, the equilibrium will shift to the right (as per Le Chatlier's principle) to replace the Copper ions being 'lost'.

This also subsequently causes a reduction in the reactant Solid Copper Carbonate.

Page 13

Complexes formed with Hydroxide Ions

the presence of hydroxide ions (OH-) will increase the solubility of sparingly soluble ionic compounds if they are able to form complex ions with the metal cation. These include hydroxides of zinc and lead that form tetrahydroxy complex ions.

Examples- write chemical equations for

the tetrahydroxyl complex ions from zinc hydroxide (Zn(OH)2) in presence of excess hydroxide ions.

Zn2+ (aq) + 2OH-(aq) <------> Zn(OH2) (s)

the tetrahydroxyl complex ions from lead hydroxide ( Pb(OH)2) in presence of excess hydroxide ions.

Pb2+ (aq) + 2OH- <--------> Pb(OH)2 (s)

Page 14

Effect of PH on solubility

Acidic and Basic Conditions

Q) The solubility of a sparingly soluble salt calcium carbonate can be increased when the pH of the solution is below 4.

Explain using equilibrium principles and relevant equations

the solubility of calcium carbonate salt increases when the pH of the solution is below 4.

This is because when the solution is more acidic, the hydronium ions in the calcium carbonate react with the carbonate ions in the solution to form Hydrogen Carbonate (HCO3-).

Based on Le Chatlier's principles, in more acidic conditions the equilibrium will shift to the right as more product (Calcium Carbonate) are removed from the equilibrium.

The position of the equilibrium will move right decreasing the amount of reactants in the system.

CaCO3 (s) <---------> Ca 2+ (aq)+ CO32- (aq)

CO32- (aq) + H3O+ (aq) -----------> HCO3- (aq) + H2O (l)

Page 15

Species in Solution

Conductivity - substances ability to conduct electrical current

Conjugate acid - A species that forms by accepting a proton( H+)

Conjugate base- a species that forms by donating a proton (H+)

Dissociation- The process where ionic compounds or molecules split into smaller particles such as ions.

Strong acid or base - Acid or base that FULLY dissociates in water

Weak acid or weak base - an acid or base that PARTIALLY dissociates in water

Acids and Bases

Brosted Lowry theory- when an acid and a base react with each other the Acid will form it's conjugate base and the base will form its conjugate acid by exchanging a proton.

Bronsted -Lowry acid is a species that 'donates' a proton (H+) to a water molecule , when in aqueous solution. This forms hydronium ion (H3O+)

A Bronsted-Lowry base is a species that 'accepts' a proton to become a conjugate acid.

Weak acids and bases will react to form solutions in equilibrium

Acid and Base <------> Conjugate Base + Conjugate Acid

Eg: NH4+ (aq) + H2O (l) <-------> NH3 (aq) + H3O+ (aq)

NH4 is the acid and NH3 is its conjugate base. As it is the Acid NH4 will donate a proton (H+) to the H2O and it will form its conjugate acid H3O and the NH4+ will form its conjugate base NH3.

Page 16

Amphiprotic Nature of Hydrogen Carbonate Ions

Species that can donate AND accept protons are called 'amphiprotic'.

Amphiprotic substances - Water molecules, hydrogen carbonate, hydrogen sulfate ions and amino acids. etc

Eg:

HCO3- (aq) + H2O (l) <-----> H2CO3 (aq) + OH- (aq) - (Hydrogen Carbonate accepts an H+)

HCO3- (aq) + H2O (l) <--------> CO32- (aq) + H3O+ (aq) - (Hydrogen Carbonate donates an H+)

"The HS- is amphiprotic "- discuss this statement and define what an acid and base are in terms of proton donation, include equations.

- An Acid is a species that donates a proton to a water molecule , e.g HS- (aq) + H2O <----> S2- (aq) + H3O+ (aq). A base is a species that accepts a proton e.g HS- (aq) +H3O (aq) <-----> H2S (aq) + H2O (l). Because the HS- ion can both donate and accept a proton it is termed on amphiprotic species.

Page 17

pH

pH is calculate by finding the neg log of the conc of hydrogen

pH = -log [H3O+]

For strong monoprotic acids (acids that completely dissociate into its ions in aqueous solution) where the conc. of the acid will equal [H3O+]

conc. of H3O+ are very small and written as exponents

eg calculate the pH of a solution where the conc. of [H3O+] = 4.00 x 10 -5 mol L-1

pH = -log [H3O+]

pH = -log [ 4.00 x 10 -5 ]

pH = 4.39 , round up to 4.40

A pH value can be converted to a value for [H3O+] by 'undoing' the pH calculation. The equation to determine the H3O + conc when given pH is...

[H3O+]= 10-ph

Eg: Fing the conc of [H3O+] in a solution with a pH of 1.50

[H3O+] =^-pH

= ^-1.50

= 0.0316 mol L-1

Page 18

Ka and pKa- Weak Acids

Ka is the acid dissociation constant.

For a strong acid Ka will be very large and for a weak acid Ka will be very small.

Weak acids do not fully dissociate in water and exist in equilibrium with their ions

HA (aq) + H2O(l) <------> H3O + (aq) + A- (aq)

Equilibrium expression for this reaction is :

Ka= [H3O+] [A-]/[HA]

Ka is the acid dissociation constantFor a strong acid Ka will be very large and for a weak acid Ka will be very small.

Chemists will usually convert Ka to pKa by sing the formula

pKa= -Log Ka

Or pKa can be converted back into Ka by using formula

Ka = 10^-pKa

Example: the reaction of Ethanoic Acid (CH3OOH) with water

Equation: CH3OOH(aq)+ H2) (l) <--------> CH3OO- (aq) + H3O+ (aq)

Ka= [CH3OO-][ H3O+] / [CH3OOH] H2O not included as it is negligible

Page 19

Ka and Conjugate Base Strength

Strong acids have weak conjugate bases. For example in a solution of Hydrochloric acid the conjugate base (Cl-) has a weak affinity for electrons.

Conjugate bases of weak acids will have very strong bases with high affinity for protons

Ka is the strength of dissociation specifically for a weak acid.

Strong acids have high Ka values and low pKa values

Strong acids have weak conjugate bases

As Acid strength decreases so does Ka, therefore pKa increase

Weaker acid will have lower Ka and higher pKa

The conjugate base of a weak acid accepts more ions to form the acid in equilibrium with it's ions

Very weak acids have very strong conjugate bases [H3O+] will be very low compared to [HA] so Ka will be y low ad pKa will be very high

STRONG ACIDS HAVE WEAK CONJUGATE BASES

WEAK ACIDS HAVE STRONG CONJUGATE BASES

Page 20

pH calculations for weak acids

Calculating the pH of a solution given the Ka (or pKa ) of the weak acid, and its concentration.

Formulae: [H3O+]= √ Ka [HA]

[HA] = Acid concentration Note: it is ka that occurs in the formula not pKa. Ka may need to be converted from a given pKa value.

Example:

Calculate the pH of a 0.100 mol L-1 solution of Ethanoic acid.

Ka= (CH3OOH)= 1.74 x 10-5

[H3O+]= √ Ka [HA]

[H3O+] = √ (1.74 x 10-5 x 0.100)

[H3O+]= 1.32x10-3

pH= -log [H3O]+

pH= -log (1.32x10-3)

pH= 2.88 (3 sig figs)

Page 21

Weak Acid Concentration Calculations from pH

calculating the concentration of a weak acid from the given Ka and pH of a solution

essentially the rearrangement of the formula to find Conc. instead

Formula: [HA] = [H3O+]2 / Ka

Note: if given the pH and not the [H3O+] concentration, then use

[H3O+]= 10 ^-ph

Example Question (page 275) :

Calculate the conc of Hydrochloric acid if it has a pH of 4.23

Ka(HCN) = 4.90 x 10-10

[H3O+] = 10 ^-ph

= 10 ^-4.23

= 5.89 x10 ^-5

[HA]= [H3O+]^2/ Ka

= (5.88 x10^-5)2 / (4.90 x 10 -10)

[HCN}= 7.05 mol L-1

Example including finding Ka from pKa.

Calculate the conc. of lactic acid if it has a . pH of 2.87 and a pKa-(HLac) = 3.86

[H3O+]= 10^-pH

[H3O+] = 10^-2.87

[H3O+]= 1.34x10-3

[HA] = [H3O+]^2 / Ka

Ka= 10^-pKa

Ka= 10^-3.86

Ka= 1.38x10-4

[HLac] = [H3O+]2 / Ka

[HLac]= (1.34 x 10^-3)2 /(1.38 x 10 ^-4)

Ans [HLac] = 0.0130 mol L-1

Page 22

Kw- The Ionic product of Water

Water molecules dissociate into ions

2H2O(L) <-----> H3O+(aq) + OH- (aq)

The Eq constant for this reaction is: Kc= [H3O+] [OH-] / [H2O]^2

As the concentration of water is so large compared to the very small amount that dissociates, it is left out of the expression. The expression for the ionic product of water is therefore

Kw = [H3O+][OH-]

As [H3O+] = [OH-], pure water is neutral.

At 25 degc [ H3O+]=[OH-] = 1 x 10^-7

the dissociation of water molecules in an endothermic process , so Kw will vary as temperature changes .

Example Question : State whether the value of Kw increases or decreases with an increase in temperature. Justify your answer.

As the temperature increases equilibrium shifts in the endothermic direction (forwards direction) to decrease the temperature. The forward direction is endothermic so the concentration of H3O+ and OH- increases.

2- Calculate the value of Kw at 25 deg c. Show working

Kw = [H3O+][OH-]

Kw= [1x10^-7][1x10^-7]

Kw= 1 x 10^-14

You can rearrange the ionic product to water to solve for [H3O+] and [OH-]

Eg: [H3O+] = 1x^-14/ [OH-] OR [OH-] = 1x 10^-14 / [H2O+]

Calculate the hydroxide ion concentration in a solution t doesn't have a pH of 7 but with a pH of:

2.05:

[H3O+] = 1 x 10 ^-pH

[H3O+]= 1 x 10 ^-2.05

[H3O+]= 8.9 x 10 ^-3 mol L-1

[OH-] = 1 x 10^-14 / 8.91 x 10- 3 = 1.12 x 10^-12

12.4:

[H3O+] = 1x 10^-ph

[H3O+]= 1x 10 ^-12.4

[H3O+] = 3.98x 10-^13

[OH-] = (1x10^-14)/(3.98 x 10 ^-13)

[OH-]= 0.025 mol L-1

Page 23

Ka and the pH of weak bases

Important formulae for calculating the ph of weak bases

Kb= Kw/ Ka

[OH-] = √Kb. [A-] [A-] is the concentration of a weak base

Example Question- Calculate the pH of a 0.110 mol L-1 solution of CH3NH2, given pKa (CH3NH3+) = 10.6

Step1- recognise that the CH3NH3+ and CH3NH2 are conjugate acid/base pairs, so Kb cn be calculated from Ka.

CH3NH2 is the conjugate base of CH3NH+so...

Kb (CH3NH) = 1x10^-14 / 10^-10.6 = 3.98 x 10 ^-4

Step 2 - Calculate [OH-] using Kb as you would [H3O+] using Ka for a weak acid.

[OH-]= √ .98 x 10-4x 0.110 = 6.62 x 10 ^ -3

Step 3 - Calculate [H3O+] using [OH-] and the ionic product of water

[H3O+] = 1 x10 ^-14 / 6.62 x 10 -3

= 1.51 x 10 ^-12

Step 4 - convert [H3O+] to pH

pH= 11.8

Page 24

pH of the salts of Weak Acids - page 279

The salt of a weak acid (eg the product of the reaction between sodium and Ethanoic acid) will form a basic solution due to the presence of the ethanoate ion.

While the sodium ion Na+ shows neither basic or acidic properties the ethanoate ion reacts with water to form a basic solution :

CH3COO - (aq) + H2O (l) <------> CH3COOH(aq) +OH- (aq)

Acid Metal Cations

Many positive metal ions form acidic solutions. mostly those that carry 3+ charges.

Aqueous solutions of metal cations create complex ions with water molecules ligands. (a metal cation with a number of other molecules or ions rounding it)

Electrons on the oxygen atom are drawn towards the positively charged metal cation, drawing the electrons in the O-H bond in the water molecule toward the oxygen atom, further polarising the atom.

Higher the positive charge on the metal cation the stronger the effect on polarisation of the OH bond.

pH of Acidic Salts - 280

Acidic salts form a solution with pH's less than 7 due to the presence of either cation or anion acting as a proton donor

Cations/ anions include bases such as Ammonium NH4+methyl ammonium CH3NH +, ethyl ammonium C2H5NH3+, Acid anions Hydrogen sulfate HSO4-, Dihydrogen sulfate H2PO4-

Note the acidic and basic ions can behave as both acids and bases (amphipirotic)

If Ka>Kb the solution will be acidic. If Kb>Ka the solution is basic.

Page 25

Conductivity of Solutions

a solutions conductivity is determined by the conc. of the charged ions.

Ionic salts which completely dissociate (strong) result in solutions with higher electrical conductivity when compared to salts with lower solubility.(weak)

the same is true of strong acids and strong bases which fully dissociate into water and therefore have more charged ions in their solutions than weak acids or bases that are just as concentrated but dissociate far less.

Example:

0.100 molL-1 of strong acid (HA) completely dissociates in water making many charged ions available. It is a good electrolyte (carries a current well)

a 0.100 mol L-1 of generic weak acid (HB) will only partially dissociate, despite having the same concentration. There are less charged ions available and it is hence a poor electrolyte (carries a current poorly.

Page 26

Buffer Solutions

A solution that undergoes a minimal change in pH when small amounts of base or acid are added.

When small bits of hydronium or hydroxide ions are added to the solutions, they will react with species in the buffer solution and will not contribute to noticeable change in pH that you would get if it were in pure water.

Acidic buffers are created y mixing weak acid with its conjugate base

Basic (alkaline) buffer solutions may be made using a mix of weak base and its conjugate acid.

At equimolar ratios of the base and its conjugate acid pH= pKa = 14- pKa

The weak acid will react with the small amounts of base added forming water and ethanoate ions.

Ethanoate ions will react with small portions of Hydronium ions added to form water and the weak acid.

Example : Ethanoic / Sodium Ethanoate buffer solution works in resisting a change in pH when small amounts of acid or base are added.

Small amount of base added:

CH3COOH (aq) + OH- <-----> CHCOO- (aq) + H2O (l)

Small amount of acid added :

C3COO (aq) + H3O+ (aq) <--------> CH3COOH (aq) + H2O (l)

If the acid & conjugate base are at the same conc. (Equimolar), then pH= pKa of the acid

Eg- The CH3COOH/CH3COO- buffers pH will be 4 .76 , as this is the pKa of Ethanoic Acid.

By varying the ratio of acid to conjugate base the pH of the buffer solution can be varied within a narrow range of this value.

If a buffer solution with a much lower or higher pH is required , then it will be better to choose an acid (and its conjugate base) that has a pKa closer to the desired value.

Basic (alkaline) buffer solutions may be made using a mix of weak base and its conjugate acid.

At equimolar ratios of the base and its conjugate acid pH= pKa = 14- pKa

Page 27

Calculating pH of Buffers

General equation for weak acid in equilibrium with its conj base :

HA (weak) + H2O <----------> A- (conj base) + H3O+

Eq Equilibrium expression for this is:

Ka = [A-][ H3O+] / [HA]

Because [A+]and [H3O+] are not at equal concentration we do not simplify to [H2O +]2

Rearranging the Eq expression fo find [H3O+]

[H3O+] = Ka [HA] / [A-]

Henderson equation can also be used to calculate pH

Hendersons Equation

Henderson equation can also be used to calculate pH

Done by converting both sides to logarithms'

First use n=cxv to find the n of the acid and base

then use c= n/v to find their concentrations

then put this along with the given Ka value into the Hendersons equation

pH= pKa +log [Base]/ [Acid]

Example Question :

Calculate the pH of a buffer solution made by mixing 100mL of a 0.150 molL-1 ethanoic acid with 100 mL of 0.100 mol L-1 sodium ethanoate . Ka = 1.73 x 10 ^-5

[H3O+]= Ka [HA] /[A-]

Step 1 - Calculate the amount of both the acid and the base

n (acid) = c x v

n= 0.150 x 0.1

n= 0.015 mol

n (base) = c x v

n= 0.100 x 0.1

n= 0.010

Step 2- Use n to find the concentration of both the acid and the base (use the combines value which in this case is 100 or 0.1 + 100 or 0.1 = 0.2

[Acid] c = n/v

= 0.015 / 0.2

= 0.075 mol L-1

[Base] c = n/v

= 0.010/ 0.2

= 0.05 mol L-1

Step 3- Calculate the pKa from the Ka value and substitute all 3 values into the Henderson equation

pKa= -Log Ka

pKa= -Log 1.73 x 10 ^-5

pKa = 4.76

pH= 4.76 + log (0.05 / 0.075)

pH= 4.76+ -0.176

= 4.58

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Calculations for Alkaline or Basic Buffers

Also uses the Henderson equation to calculate the pH

Requires some conversions from Kb to Ka if Kb is given but not Ka

Use pKa= 14-pKb to find the pKa value

Example Question:

Calculate the pH of a buffer made by mixing 200mL of 0.150 mol L-1 ammonia solution, NH3, with 200 mL of 0.100 molL-1 ammonia chloride solution NH4Cl. Kb= (NH3)= 1.78 x 10 ^-5

Step 1- Calculate the amount of both the acid and base added to the solution

(NH4) n=cxv

(NH4) n= 0.200 x 0.100

n= 0.02

(NH3) n= c x v

(NH3) n= 0.200 x 0.150

n= 0.0300

Step 2- Calculate the concentration of the Acid and the base using the Combined volume of 0.400 mL

[NH3] c = n/V

c= 0.0300 / 0.400mL

c = 0.0750

[NH4] c = n/V

c= 0.0200/ 0.400

c= 0.0500

Step 3 -Calculate the pka from the pkb value and substitute into Hendersons eq

pKb = -log Kb

pKb= Log 1.78 x 10 ^-5

pKb= 4.749

pKa = 14 - 4.749

pKa= 9.251

pH= pKa+log ([base/[acid])

pH= 9.251+ log /0.0750/0.0500)

pH 9.251+0.176

pH= 9.427 round to 9.43

Page 29

Making Buffers to a special pH

From Hendersons eq we can see that 'log([base][acid]) gives a value that when added to the pKa gives the pH of the buffer. Therefore its the ratio of the concentration of base to acid that is important not not the specific amount of each.

Example calculation:

Calculate the mass of CH3COONa required when added to 500 mL of a 0.100 mol L-1 CH3COOH solution to give a buffer solution with a pH of 5.00.

assume there is no charge in volume

M(CH3COONa)= 82.0 g mol-1

pKa= (CH3COOH)= 4.76

Step 1 - write out Hendersons equation and sub in known values

pH= pKa + log ([base]/[acid])

5.00= 4.76 = log ([base]/0.100])

5.00-4.76

0.24= log (base/0.100)

Step 2 - rearrange and solve the concentration for the unknown

10^0.24= [base]/0.100

1.74=[base]/[0.100]

[base]= 1.74 x 0.100

base c = 0.174

Step 3- use molar mass calculations to calculate the amount and the mass of CH3COONa needed to achieve the desired concentration.

n= c xv = 0.174 x 0.500 = 0.0869

n= 0.0869

m= n x M

m= 0.0869 x 82.0

m= 7.12g

Page 30

Buffer effectiveness

if the weak acid an its conjugate base are equal in concentrations then the buffer solution will be just as effective at resisting a pH change h excess acid or base being added. This will also mean that the pH of the solution is equal to the pKa value.

Example question- calculate the pH of a buffer made from ammonia and ammonium chloride both with a concentration of 1.50 mol L-1 if pKa (NH4+)= 9.24

pH= pKa + log (base / acid)

pH = 9.24 + log (0.150/0.150) = 9.24

The concentrations are equal therefore ph= pKa so pH= 9.24

pH< pKa

concentration of the acid is greater than the base

if the pH is less that the pKa the buffer will be more effective in overcoming the addition of a base

pH=pKa

if the weak acid an conjugate base are in equal concentrations,

pH and pKa are equal the buffer will be equally as effective for acids and bases

pH> pKa

concentration of the acid is less than the base

if pH is greater than the pKa the buffer is more effective at overcoming additional acid

Page 31

Titration Curves - acids and bases

When acid and base react neutralisation occurs and a salt and water is formed

the amount of base (n) needed to neutralise an acid depends on the concentration and the volume and is dependent on the strength of the acid . n= c x V

concentration of an acid solution is found via titration with a base of accurately known concentration

Given the titration data you can figure out the concentration of the acid.

An acid base indicator is used and the titration is stopped at the point were the indicator changes colour to show that it has changed from an acid to a base. This is known as the titration end point

The equivalence point is where the amount of added base is equal to the initial amount of acid in the flask.

The end point should also be the equivalence point IF the right indicator has been used.

the pH of the equivalence point will determines the choice of indicator used

Titration curve is a plot showing the characteristic points of he titration plotted into a graph

Important values on a titration curve include

Initial pH of the solution in the flask

the pH of the equivalence point - halfway up the steepest area of the curve

pH after the equivalence point- after ALL of the solution has reacted and any excess from the burette is added.

For titrations involving weak acid/base these are also important

Buffer zone

Midpoint of the buffer zone- pint on the curve where half the volume needed to reach equivalence point has been added

Page 32

Page 33

Strong Acid Strong Base titration

If the start point is 1 it starts with an acid

if the start point is 13 it has started with a base

Equivalence points for these are 7 . this is because the product of these is a neutral salt ie - NaCl. Na and Cl ions wont react in water and H3O and OH will be equal therefore the ph is 7.

You will need to be able to sketch this

TO figure out

-initial pH - pH= log [H3O+]

-pH at equivalence point always = 7

- pH after equivalence point

step 1 - [OH-] = [Base] x volume past E.P/ total volume

volume past EP is = to the volume of base added after Equivalence point was reached

total volume = volume of the initial HCl in flask +all NaOH that was added

Step 2 - [H3O+]= Kw / [OH-]

Step 3 - pH = -log [H3O+]

or ph= 14 - pOH can also be used

Calculating volume of added base at equivalence point

V( base @EP)= n (acid)/c(base)

n = cxV

Strong Acid Weak Base

Weak Acid Weak Base

Weak Acid Strong Base

Page 34