Ionic Equilibrium in Solutions

Ionic Equilibrium in Solutions

Ksp, Ka and Kb

Much like with a system of equations, a solution is also an equilibrium

NaCl(aq) Na+(aq) + Cl-(aq)

The ions in this solution are constantly dissociating and re-associating

Ionic Equilibrium

What is a saturated solution?◦ A solution which has reached its capacity of a

solute

What is a super-saturated solution?◦ A solution which holds more than its full capacity

of a solute◦ Video demonstration…

Now onto the real stuff…

Ionic Equilibrium

Ka – The acidity constant

Kb – The Alkalinity (base) constant

Ksp – Solubility product constant

Kwater – Water ionization constant

What we will be covering...

An acid is a substance that dissociates in water to produce hydrogen ions (H+)

◦ HCl (aq) -> H+ (aq) + Cl- (aq)

A base is a substance that dissociates in water to produce hydroxide ions (OH-)

◦ NaOH (aq) -> Na+ (aq) + OH- (aq)

Arrhenius theory of acids and bases

Acids are neutralized by a base and vice versa

◦ NaOH + HCl -> NaCl + H2O

Acids and bases can be stronger or weaker

You need more of a weak base to neutralize a strong acid

Neutralization

NaOH◦ Base!

HCl◦ Acid!

H2SO4◦ Acid!

NH3◦ Base!

Acid or base?

An acid is a substance in which a proton (Hydrogen atom, H+) can be removed. An acid is seen as a proton donor. Seeing how a single H+ cannot exist on its own, it can also be shown as a hydronium ion (H3O+)

A base is a substance that can remove a proton from an acid. A base is seen as a proton acceptor.

Brønsted-Lowry theory of acids and bases

In each acid/base reaction, there are 2 conjugate acid-base pairs

Ex: HCl (aq) + H2O (aq) -> H3O+ (aq)+ Cl- (aq)

HCl and Cl- are one pair (A-B)◦ HCl is the acid and Cl- is the base

H2O and H3O+ are the other pair◦ H2O is the base and H3O+ is the acid

Conjugate Acid-Base Pairs

NH3 (aq) + H2O (aq) -> NH4+ (aq) + OH- (aq)

NH3 and NH4+

◦ NH3 is the base and NH4+ is the acid

H2O and OH-

◦ H2O is the acid and OH- is the base

Conjugate Acid-Base Pairs

Looking back at the two conjugate Acid-Base pairs, is H2O an acid or a base?

In the Brønsted-Lowry theory of acids and bases, water can be considered an acid or a base, depending on its role in the reaction.

Water, Acid or Base?

The dissociation of ions in water is an equilibrium

The pH of the solution, which measures its acidity, is determined by where the equilibrium settles

This equilibrium can be quantified using the ionization of water constant Kwater

Ionization constant of Water

This constant Kwater makes it possible to understand the interdependence of Hydronium ions (H3O+) and Hydroxide ions (OH-)

Before we explore this wonderful relationship, let us go over what exactly the pH and pOH are...

Ionization constant of Water

pH is a quantitative value attributed to the acidity of a solution

The lower the pH value, the higher the concentration of the hydronium ions (H3O+), and therefore the stronger the acid.

pOH is a quantitative value attributed to the alkalinity or basicity of a solution

The lower the pOH, the higher the concentration of the hydroxide ions (OH-) and therefore the stronger the base.

What are pH and pOH?

pH and pOH can be expressed as the following mathematical expressions

pH = -log [H3O+]◦ [H3O+] = 10-pH

pOH = -log [OH-]◦ [OH-] = 10-pOH

Mathematical Expressions

Express in the form of pH, the hydronium (H3O+) concentration of 4.7 x 10-11 mol/L in an aqueous solution. Is this solution acidic, neutral or basic?

Data◦ [H3O+] = 4.7 x 10-11

◦ pH = ?

Example #1

pH = - log [H3O+]

pH = - log (4.7 x 10 -11)

pH = 10.33

The solution is basic due to its pH being higher than 7

Solution

Express the pOH of 3.60 in the form of the hydroxide (OH-) concentration

Data◦ pOH = 3.60◦ [OH-] = ?

Example 2

[OH-] = 10 -pOH

[OH-] = 10 -3.60

[OH-] = 2.5 x 10 -4

The concentration of hydroxide (OH-) is 2.5 x 10 -4 mol/L

Solution

What is [H3O+] at pH 7?

◦ 1.00 x 10 -7

What is [OH-] at pOH 7?

◦ 1.00 x 10 -7

Relationship between pH and pOH

The ionization of water follows this simple formula

2 H2O (l) H3O+ (aq) + OH- (aq)

Once this equation has reached equilibrium, we obtain the ionization of water constant Kwater

Ionization Constant of Water

Kw = [H3O+] x [OH-]

If water is neutral, pH 7, then we know the concentrations of the hydronium and hydroxide ions.

The concentration of both ions is 1.00 x 10-7

Therefore...

Kw

Kw = [H3O+] x [OH-]

Kw = 1.00 x 10-7 x 1.00 x 10-7

Kw = 1.00 x 10-14

This is always at 25°C

Ionization constant of water

By carrying out the logarithmic inverse of each side of the equation, the following equivalence can be obtained

-log [H3O+] + -log [OH-] = -log (1.00 x 10-14) -log [H3O+] + -log [OH-] =14 pH + pOH = 14

How this all fits together...

Knowing that Kw is constant in all aqueous solutions, we can use this to determine the concentration H3O+ and OH- ions in any acidic or basic solutions

Example! At 25°C, a hydrochloric acid solution has a

pH of 3.2. What is the concentration of each of the ions in this solution?

Relationship between the pH and [H3O+] and [OH-]

[H3O+] = 10 –pH

[H3O+] = 10 -3.2 = 6.3 x 10-4

Kw = [H3O+] x [OH-] = 1.00 x 10-14

[OH-] = 1.00 x 10-14 / [H3O+] [OH-] = 1.00 x 10-14 / 6.3 x 10-4

[OH-] = 1.58 x 10-11

Solution

Ka and KbAcidity and Basicity Constants

Here we will be quantifying the strength of acids and bases

The stronger the acid or base depends on how it dissociates

The more dissociation, the stronger the acid

Ka and Kb

When an acid comes into contact with water, a certain amount of dissociation takes place

In a strong acid, as much as 100% will dissociate◦ Ex: HCl

In a weak acid, very little will dissociate. As little as 1%◦ Ex: Acetic acid (Vinegar)

Strength of Acids

Acid Dissociation

Ionization percentage can be calculated by dividing the concentration of the H3O+ ions by the concentration of the original acid and multiplied by 100

% = [H3O+] eq / [HA] i * 100

Ensure that all of the concentrations are in the same units

Ionization Percentage

This can only be done using a weak acid, why?

◦ If there is none of the original acid left, you can’t calculate an equilibrium constant

Using the following general equilibrium, we can calculate the Ka◦ HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Calculating the Acidity

Ka = ([H3O+] * [A-]) / [HA]

Since water is a liquid, there is no concentration

We cannot do this with a strong acid because there is none of the original HA acid left and you cannot divide by 0

Would a weak acid have a higher or lower acidity constant?◦ Lower!

Acidity Constant

To find the acidity constant, all of the concentrations must be known

Also, if the Ka is known, then we can use that to predict either the final concentration of the [H3O+], or the initial concentration of the [HA]

It can also be used to calculate the pH

Acidity Constant

The basicity constant can also be calculated along the same lines

Using the following general equilibrium formula

B (aq) + H2O (l) HB+ (aq) + OH- (aq)

Kb = ([HB+] * [OH-]) / [B]

Again, the weaker the base, the smaller the constant

Basicity Constant

KspSolubility Product Constant

A saturated solution that contains non-dissolved solute deposited at the bottom of a container is an example of a system at equilibrium.

The solubility of a substance corresponds to the maximum quantity of a substance that dissolves in a given volume of water

Usually given as g/100ml

Solubility Product Constant

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

Ksp = [Ba2+] * [SO42-]

General formula

XnYm(s) nX+ (aq) + mY-(aq)

Ksp = [X+]n * [Y-]m

Example

The solubility of silver carbonate (Ag2CO3) is 3.6 x 10-3 g/100ml of solvent at 25°C. Calculate the value of the solubility product constant of silver carbonate.

Steps

1- Find the concentration of the Ag2CO3 using M = m / n then the solubility◦ Where M is molar mass, m is mass and n is the amount in

moles of Ag2CO3

2- Calculate the Ksp

Problem

1- M = m/n

n = m/M n = 3.6 x 10-3 g / 275.8 g/mol n = 1.3 x 10-5 mol for 100 ml (0.1 L)

Solubility = 1.3 x 10-5 / 0.1 L Solubility = 1.3 x 10-4 mol/L

Solution

Ag2CO3 2 Ag+ (aq) + CO3 2- (aq) Ksp = [X+]n * [Y-]m

Ksp = [Ag+]2 * [CO32-]

[Ag+] = 2 * [Ag2CO3] = 2 * 1.3 x 10-4 mol/L [Ag+] = 2.6 x 10-4 mol/L [CO3

2-] = [Ag2CO3] = 1.3 x 10-4 mol/L Ksp = [Ag+] * [CO3

2-] Ksp = (2.6 x 10-4)2 mol/L * 1.3 x10-4 mol/L Ksp = 8.8 x 10-12

Solution Continued