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AP Chemistry

Unit 8.1 & 8.2

INTRODUCTION TO ACIDS & BASES

pH and pOH OF STRONG ACIDS & BASES

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Unit 8.1

Enduring Understanding:

The chemistry of acids and bases involves reversible proton-transfer reactions, with equilibrium concentrations being related to the strength of the acids & bases involved

Learning Objective:

Calculate the values of pH and pOH, based on Kw and the concentration of all species present in a neutral solution of water

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Unit 8.2

Enduring Understanding:

The chemistry of acids and bases involves reversible proton-transfer reactions, with equilibrium concentrations being related to the strength of the acids & bases involved

Learning Objective:

Calculate the pH and pOH based on concentration of all species in a solution of a strong acid or a strong base

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Acid-Base Review

...again
Brønsted-Lowry acid: a substance that donates hydrogen ions (H⁺; protons) to a solution
Brønsted-Lowry base: a substance that accepts hydrogen ions
Hydronium ion: H₃O⁺, formed when a hydrogen ion interacts with water
pH: the mathematical manipulation of the concentration of the hydronium (think: H⁺) ions in a solution

Describes how acidic or basic (alkaline) a solution is

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Acid-Base Review

Typically the pH scale is 0-14, < 7 = acidic, > 7 = basic
pH can be less than 0 or higher than 15
Pure water is neutral because [H₃O⁺] = [OH^-]

At 25℃	[H₃O]⁺	pH	[OH^-]	pOH
Acids	> 1 x 10^-7 M	< 7	< 1 x 10^-7 M	> 7
Neutral	1 x 10^-7 M	= 7	1 x 10^-7 M	= 7
Bases	< 1 x 10^-7 M	> 7	> 1 x 10^-7 M	< 7

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pH Calculations

Calculations Tool Kit:

pH = -log[H₃O⁺]
pOH = -log[OH^-]
[H⁺] = antilog(-pH)= 10^-pH
[OH^-] = antilog(-pOH)= 10^-pOH
pH + pOH = 14
Kw = 1 x 10^-14 = [H₃O⁺][OH^-]

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Equilibrium Constant for Water

The equilibrium constant for water, Kw, can be written from the following reaction, the autoionization of water:

Heat + 2H₂O (l) ⇌ H₃O⁺ (aq) + OH^-
Kw = [H₃O⁺][OH^-]

At 25℃, the Kw for water is 1 x 10^-14
So 1 x 10^-14 = [H₃O⁺][OH^-]
Can also simplify this equation using H⁺ instead of H₃O⁺

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)
Kw = [H⁺][OH^-]

Both are accepted by AP, but H₃O⁺ is preferred

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Equilibrium Constant for Water

Kw = [H₃O⁺][OH^-]
We can take the log of each side and find that pKw = pH + pOH
pKw = 14 at 25℃

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Equilibrium Constant for Water

Just like the other K values, Kw is dependent on temperature
As temperature increases:

Kw increases
[H₃O⁺] and [OH^-] increase

Water is still neutral even though the pH value differs with changes in temperature

This is because the concentration of hydronium and hydroxide ions remain equal

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Sig Fig Rule for pH

The sig fig rule for pH (and therefore pOH, pKw, etc) is that only the numbers after the decimal point (mantissa) are significant.
This is because the value before the decimal point (characteristic) is really descriptive of the order of magnitude
**Remember, pH are log values!
Example: pH = 10.7 has 1 sig fig
Example:

Calculate the pH of a solution where [H⁺] = 0.00100 M
pH = 3.000
There are 3 sf after the decimal in 0.00100 so your answer needs 3 sf after the decimal

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Strong Acids & Bases

Strong acids and bases completely dissociate

Irreversible reactions

Six strong acids to know:

HCl
HBr
HI
HNO₃
H₂SO₄
HClO₄
“SO, I Brought NO Clean ClOthes”

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Strong Acids & Bases

Strong acids are strong electrolytes
When added to water they dissociate into their ions to form an aqueous solution
Electrolytes can conduct electricity in aqueous solutions
For monoprotic (1 H⁺) strong acids (like HCl), the concentration of the acid = the concentration (molarity) of the hydronium ions

[H₃O⁺] = [acid]

For polyprotic acids, like H₂SO₄, only the first hydrogen dissociates completely

H₂SO₄ (aq) + H₂O (l) → HSO₄^- (aq) + H₃O⁺
In a reaction with a base, however, both H⁺ will react

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Strong Acids & Bases

Strong acids and bases completely dissociate

Irreversible reactions

The strong bases are Group 1 and Group 2 metals + OH^-

“So Really Strong Bases Can Certainly Look Pleasing”
Sodium, rubidium, strontium, barium, calcium, cesium, lithium, potassium

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Strong Acids & Bases

Strong bases are soluble hydroxides and dissociate completely
They are also strong electrolytes
The [OH^-] can be calculated using the initial concentration (molarity) and the ratio of hydroxide ions in the formula

[OH^-] = [base]

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Practice: I Do

What is the molarity of the H₃O⁺ in the following solutions at 25℃? Determine if they are acids or bases.

pH = 3.689
pOH = 6.410
[OH^-] = 1.56 x 10^-4

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Practice: I Do

What is the molarity of the H₃O⁺ in the following solutions at 25℃? Determine if they are acids or bases.

pH = 3.689

[H⁺] = 10^-3.689 = 2.05 x 10^-4

Acid because pH < 7

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Practice: I Do

What is the molarity of the H₃O⁺ in the following solutions at 25℃? Determine if they are acids or bases.

b. pOH = 6.410

pH = 14 - 6.410 = 7.590

[H⁺] = 10^-7.590 = 2.57 x 10^-8 M

Base because pH > 7

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Practice: I Do

What is the molarity of the H₃O⁺ in the following solutions at 25℃? Determine if they are acids or bases.

c. [OH^-] = 1.56 x 10^-4

[H⁺] = 1 x 10^-14 / 1.56 x 10^-4 = 6.41 x 10^-11 M

pH = -log(6.41 x 10^-11) = 10.193

Base because pH > 7

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Practice: I Do

Find the pH of a mixture of 10.0 mL of 0.00100 M solution of potassium hydroxide, KOH, and 10.0 mL of distilled water

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Practice: I Do

Find the pH of a mixture of 10.0 mL of 0.00100 M solution of potassium hydroxide, KOH, and 10.0 mL of distilled water

[OH^-] = (0.00100 mol/1L)x

0.000500 M

pOH = -log(0.000500) = 3.301

pH = 14 - 3.301 = 10.699