Calculating pH and pOH for WEAK acids and bases

Just when you were thinking things were easy….

Just Listen for a few minutesRemember that Strong acids and Bases are different from Weak Acids.

Every bit of STRONG acids or bases put into water completely split into the [H+] or [OH-] ions.

However for WEAK acids and weak bases, we will need to determine how much of the ions dissociate in water.

Equilibrium & KA weak substance will only partially dissociate and will do so until the reaction reaches equilibrium.

We will solve for the Ka for acids and Kb for bases. (Dissociation Constants)

These values represent the relationship between the concentrations of the products over the reactants when equilibrium is reached.

A quick reminderHow we solve for K is the law of mass action:

Ka = [Y]c [Z]d

[W]a [X]b

The brackets[ ] represent the molar concentrations.

aW + bX ⟺ cY + dZ

A KA Equilibrium Expression:Were going to write the “Blueprint” equation for any acid dissociation:

HA H+ + A-

Any acid H+ ion + Conj. BaseWhat would the KA = ?KA = [H+][A-] [HA]

Reaction Practice: Lets write a weak acid reaction and practice: First lets predict the products: HCOOH + H2O ↔ H+ + HCOO-

* You need to remember from equilibrium calculations, that pure liquids are not included.

HCOOH ↔ H+ + HCOO-

Notice the arrow goes both ways.

Equilibrium Expression Lets write an EQ. expression. What will

KA = ?

HCOOH ↔ H+ + HCOO-

KA = [H+][HCOO-] [HCOOH]

Ice Ice BabyThe tool for solving equilibrium problems is the ICE chart.

It’s broken into three sections:I = Initial concentrations

C = Change in concentrations E = Equilibrium concentrations

Each of these is recorded in Molarityor concentration of the solution.

Ice Chart Template:

Acid H+ Ion Conjugate Base

Initial

Change

Equilibrium

Solving for the KASolving for KA can be done if you know the pH or [H+] of the solution at equilibrium.

A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the KA.

To solve for the KA, we need first need the equilibrium expression and then we can use an ICE table.

HCOOH ↔ H+ + HCOO-

Ice Chart:

HCOOH H+ HCOO-

Initial

Change

Equilibrium

Initially: A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the KA.

Initially we know the concentration of the formic acid = 0.10 M.

But think about it, AT THE VERY BEGINNING, how much of the products have we made?

None! So we complete the Initial Conc.

Ice Chart:

HCOOH H+ HCOO-

Initial 0.10 M 0.0 M

0.0 M

Change

Equilibrium

Change in Concentrations We don’t know the numbers to this part yet,

but we do know something about it. When this acid molecule dissociates, it will

create one hydrogen ion, and one molecule of the conjugate base. Not always the case, but true for this acid!

HCOOH ↔ H+ + HCOO-

We put X in for the change in amounts: - X for decreasing amounts (HCOOH)+X for increasing amounts. (H+) and (HCOO-)

Ice Chart:

HCOOH H+ HCOO-

Initial 0.10 M 0.0 M

0.0 M

Change

- X + X + X

Equilibrium

Equilibrium ConcentrationsThis is pretty easy. Read straight down the column.

Keep the same signs.

Ice Chart:

HCOOH H+ HCOO-

Initial 0.10 M 0.0 M

0.0 M

Change

- X + X + X

Equilibrium

0.10 - X

X X

Final ConcentrationsBut we know something about the final concentration!

A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the KA.

pH can tell us [H+] so we can use that to solve for X in the ICE table and plug it in!

pH = -log[H+] 10-2.38 = 0.0042657 *Dolphins in San Fran4.3 x 10-3 [H+] * This is the [H+] and X!!!

Ice Chart:HCOOH H+ HCOO-

Initial 0.10 M 0.0 M 0.0 M

Change

- 4.3 x10-3

4.3 x10-3

4.3 x10-3

Final 0.10 – 4.3 x10-

3

4.3 x10-3

4.3 x10-3

Finally: solving for the KA Using the equation we need to write the

equilibrium expression: We will plug in concentrations from the “Final” part of the Ice Chart:

KA = [H+][HCOO-] [HCOOH]

KA = [4.3 x10-3][4.3 x10-3] [0.10 – 4.3 x10-3]

KA = 1.9 x 10-4