K sp, K a and K b. Much like with a system of equations, a solution is also an equilibrium ...
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Transcript of K sp, K a and K b. Much like with a system of equations, a solution is also an equilibrium ...
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 Kb
Acidity 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
Ksp
Solubility 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