18.2 Calculations involving Acids and Bases

What does this mean?

In Topic 8 we learned that:

A strong acid dissociates ______________________

A weak acid _________________ dissociates.

We also learned to distinguish between 1M solutions of HCl & CH₃COOH by….

1. Passing an electric current through the solutions

Observed difference ___________________________________________

Reason _____________________________________________________

2. Adding a Carbonate.

Observed difference ___________________________________________

Reason _____________________________________________________

3. Measuring a temperature change with an excess of powdered Zinc

Observed difference ___________________________________________

Reason _____________________________________________________

Q. If we reacted both solutions with an excess of Zinc they would both produce the same volume of Hydrogen. Why?

________________________________________________________________

____________________________________________________________

That depends on exactly how well they dissociate.

The dissociation is in equilibrium, so we use an Equilibrium Constant

You should know how to write a K_c for the below equilibrium

What would the K_c be for the equilibrium of Ethanoic Acid in water?

CH₃COOH_(aq) + H₂O_(l) ⇌ CH₃COO^-_(aq) + H₃O ⁺_(aq)

K_c = ______________________

This is obviously not sensible because the concentration of water in water is so much higher than any of the other concentrations that it would make the K_c almost useless.

So, for the purposes of the Acid Dissociation Constant (Ka) we pretend that the dissociation is as below

CH₃COOH_(aq) ⇌ CH₃COO^-_(aq) + H ⁺_(aq)

K_a = ______________________

We could think of any weak acid as HA a molecule made up from an H⁺ ion and the rest of the molecule is A^-

HA ⇌ H⁺ + A^-

If a weak acid is particularly poor at dissociating then [H⁺] will be HIGH/LOW

If a weak acid is particularly poor at dissociating then [HA] will be HIGH/LOW

If a weak acid is particularly poor at dissociating then K_a will be HIGH/LOW

So, the higher the K_a the STRONGER/WEAKER the Acid

So, the higher the K_a the HIGHER/LOWER the pH.

We could think of any weak base as B - a molecule that can accept H⁺ ions from to become BH⁺ and produce an OH^- ion

B + H₂O ⇌ BH⁺ + OH^-

Again, the H₂O is ignored for this expression

If a weak base is particularly poor at dissociating then [OH^-] will be HIGH/LOW

If a weak base is particularly poor at dissociating then [B] will be HIGH/LOW

If a weak base is particularly poor at dissociating then K_b will be HIGH/LOW

So, the higher the K_b the STRONGER/WEAKER the Base

Find the K_a for a Weak Acid with a concentration of 0.001 moldm^-3. If the pH is 6.1

STEP ONE: Work out [H⁺] = 10^–pH = 7.94 x 10 ^-7 moldm^-3

STEP TWO: Write the dissociation

HA ⇌ A^- + H⁺

STEP THREE: Write in the initial concentration before dissociation

HA ⇌ A^- + H⁺

Initial 0.001 0 0

STEP FOUR: Write the final concentrations after dissociation

HA ⇌ A^- + H⁺

Initial 0.001 -7.94 x 10 ^-7 7.94 x 10 ^-7 7.94 x 10 ^-7

= 0.001 !!!!!

K_a = [H⁺][A^-]/[HA]

= (7.94 x 10 ^-7) x (7.94 x 10 ^-7) / 0.001 = 6.3 x 10^-10

1. Find the K_a of Ethanoic acid if a 1.4 moldm^-3 solution has a pH of 2.3

STEP ONE: Work out [H⁺] = 10^–pH = __________ moldm^-3

STEP TWO: Write the dissociation

HA ⇌ A^- + H⁺

STEP THREE: Write in the initial concentration

HA ⇌ A^- + H⁺

Initial 1.4 0 0

STEP FOUR: Write the final concentrations

HA ⇌ A^- + H⁺

Initial 1.4 -______ __________ __________

= _________

K_a = [H⁺][A^-]/[HA]

= _______________ ?

2. Find the K_a of Butanoic acid if a 0.5 moldm^-3 solution has a pH of 2.6

3. Find the K_a of Carbonic acid if a 2.0 moldm^-3 solution has a pH of 6.0

4. Find the K_a of Citric acid if a 0.8 moldm^-3 solution has a pH of 3.2

5. Find the K_a of Methanoic acid if a 1.8 moldm^-3 solution has a pH of 3.5

EXAMPLE

Find the K_b of a base B with concentration 2.3 mol/dm³ if the pH is 11.1

pH + pOH =14

so pOH = 2.9

pOH = -log₁₀ [OH^-] so [OH] = 10^-2.9 = 1.25 x10 ^-3

B + H₂O => BH⁺ + OH^-

Initial 2 x 10^-3 0 0

Final 2.3 – 1.25 x10 ^-3 1.25 x10 ^-3 1.25 x10 ^-3

K_b = [BH⁺][OH^-]/[B] = (1.25 x10 ^-3 )( 1.25 x10 ^-3)/ 2.3 = 6.8 x 10 ^-7

QUESTIONS

1. Find the K_b of a base B with concentration 1.3 mol/dm³ if the pH is 12.0

2. Find the K_b of a base B with concentration 0.3 mol/dm³ if the pH is 10.2

3. Find the K_b of a base B with concentration 0.03 mol/dm³ if the pH is 9.1

Hopefully, you recall that the K_w describes the ionization of water.

H₂O ⇌ H⁺ + OH^-

K_w = [H⁺][OH^-]

1. At 25^oC, Kw = 1x10^-14, what is the pH of a 1.2 moldm^-3 solution of NaOH?

2. At 25^oC, Kw = 1x10^-14, what is the pH of a 1.2 moldm^-3 solution of Ca(OH)₂?

Hopefully, you’ll also remember the idea of conjugate acids and bases.

Weak acids, like Methanoic acid, have relatively strong conjugate bases.

HCOOH ⇌ HCOO^- + H⁺

The Acid has a K_a = [HCOO^-][H+] / [HCOOH]

Its Conjugate Base has K_b = [HCOOH][OH^-] / [HCOO^-]

K_a x K_b = [HCOO^-][H⁺] / [HCOOH] x [HCOOH][OH^-] / [HCOO^-]

K_a x K_b = [H⁺] [HCOOH] [OH^-] / [HCOOH]

K_a x K_b = [H⁺] [OH^-] = K_w

Propanoic acid has a K_a = 1.4 x 10^-5 at 25^oC. What is the K_b of its conjugate base, the Propanoate ion?

CH₃CH₂COOH + H₂O ⇌ CH₃CH₂COO^- + H₃O⁺

H₂O ⇌ H⁺ + OH^-

K_w = [H⁺][OH^-]

K_w describes the dissociation of water.

Q1. Is this an Exo- or an Endothermic process? How do you know?

________________________________________________________________

________________________________________________________________

________________________________________________________________

Q2. If the temperature of water increases is the dissociation more or less common?

________________________________________________________________

________________________________________________________________

Q3. If the temperature of water increases what will happen to K_w?

________________________________________________________________

________________________________________________________________

Q4. At 35 ^oC K_w = 2.09 x 10^-14. What is [H⁺] of water at 35 ^oC?

________________________________________________________________

________________________________________________________________

Q5. At 35 ^oC K_w = 2.09 x 10^-14. What is pH of water at 35 ^oC? Explain why it is still neutral at this pH

________________________________________________________________

________________________________________________________________

In Chapter 8 we said that changes in [H⁺] are less easy to understand than changes in pH

The same is true about the tiny numbers of K_a and K_b.

So, pKa = -log₁₀(K_a)

And, pKb = -log₁₀(K_b)

And just like Increasing acidity reduces pH, reducing pK_a reduces the strength of acids, and reducing pK_b reduces the strength of bases.

Fill in the missing values in the table below

1. Find the pK_a of Lactic Acid given that at equilibrium [Lactic Acid] = 0.01 moldm^-3, [Lactate] = 0.087 moldm^-3, and pH = 4.8

2. A pH meter was calibrated and used to test a 0.10 molar solution of a weak acid. It gave a pH reading of 4.2, calculate the Hydrogen ion concentration and the value of the acid dissociation constant K_a and pK_a.