Chapter 15 Acids and Bases• First defined by Svante Arrhenius

1903 Nobel Prize winner who proposed that:

Acids - produce H⁺ ions in aqueous solution.

HCl(g) H⁺(aq) + Cl⁻ (aq) H2O

Arrhenius also proposed a useful definition for bases.

Bases - produce OH⁻ ions in aqueous solution.

NaOH(s) Na⁺(aq) + OH⁻ (aq) H2O

Acids and bases using according to the Bronsted-Lowry model.

• Acid - proton donor

• Base – proton acceptor

• According to this model water, due to its polar nature, aids in removing the proton.

In solution, the following reaction occurs

HA (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + A⁻ (aq)

• In this example: HA donates a proton to water which would make HA the acid and water accepts the proton which make it a base.

• The conjugate acid is H₃O⁺ the hydronium ion.• The conjugate base is A⁻, which is everything

that remains after the acid donates its proton

Using a specific acid

• HBr (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + Br⁻ (aq) • Acid base conj. acid conj. Base

• H₂SO₄ (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HSO₄⁻ (aq) • Acid Base conj. Acid conj. Base

• H₂S (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HS⁻ (aq) • Acid base conj. acid conj. Base

• With Bronsted-Lowry acid base theory some themes persist.• Water is the base, the conjugate acid is H₃O⁺ and the conjugate

base is simply the acid without a proton (H⁺ ion)

Differences between the 2 theories

• With Arrhenius, acidic aqueous solutions contain the H⁺ ion.

• With Bronsted-Lowry H₂O is included in the product, so a Hydronium ion (H₃O⁺) is found in acidic solutions.

The Hydronium ion, H₃O⁺

• In water, H⁺ combines with water to form the hydronium ion

H⁺ + H₂O → H₃O⁺

• For our purposes H⁺ and H₃O⁺ can be used interchangeably.

Including H₂O as a reactant

• The ionization of an acid would look slightly different if water is included as a reactant.

• Let’s use HBr as an example

• HBr + H₂O → H₃O⁺ + Br⁻

• Presence of either H⁺ or H₃O⁺ make a solution acidic the difference is only how we choose to write the ionization.

Water is amphoteric

• Amphoteric - a substance that can behave as either an acid or a base.

• Equation for the ionization of water

H₂0₍ι₎ + H₂0₍ι₎ → H₃O⁺(acid) + OH⁻(base)Or

H₂O → H⁺(acid) + OH⁻(base)

[ # ]

• Placing a number in brackets means that that number then represents the a concentration in “moles per liter” aka Molarity

• [OH⁻] = 1.6• Means that the hydroxide ion concentration is

1.6 M

Comparing [H⁺] to [OH⁻]

• In neutral solution [H⁺] = [OH⁻]

• In acidic solution [H⁺] > [OH⁻]

• In basic solution [H⁺] < [OH⁻]

Neutralization Reaction

• This is usually a double replacement reaction with an acid as a reactant and a base as the other reactant. The products are salt and water.

• Ex.

• HCl (aq) + NaOH (aq) → NaCl (aq) + HOH₍ι₎• Acid base salt water

Other salts are possible

H₂CO₃ + Mg(OH)₂ → MgCO₃ + HOHAcid base salt water

HNO₃ + LiOH → LiNO₃ + HOHAcid base salt water

The pH scale

This is a convenient way to express relatively small [H⁺] Generally, pH results should range from 0-14.

To calculate pH you will use the log function on your calculator.

pH values of common materials

Calculating pH

•pH = - log [H⁺]

Remember, [H⁺] is the hydrogen ion concentration expressed in mol/L

a.k.a. MOLARITY

Calculating pH

An aqueous solution has a [H⁺] =2.97x10⁻⁶ what is the pH of the solution?

pH = - log [H⁺]pH = - log [2.97x10⁻⁶] 3 sig figs

pH = 5.527 3 sig figs

Calculating [H⁺] from pH

• This operation involves the inv log function or the antilog so we will have an activity sheet designed to help you to enter this correctly in your individual calculator.

• A solution has a pH of 4.58 what is the [H⁺]?

• [H⁺] = 2.6 x 10⁻⁵ M

An HCl solution has a [H⁺] of 1.4 x 10⁻⁴ M, what is the pH?

pH = - log [H⁺]pH = - log [1.4 x 10⁻⁴]pH = 3.85

Calculating pOH

• The calculation is identical to the pH calculation except it involves the [OH⁻].

• pOH = - log [OH⁻]

Calculating pOH

What is the pOH for a solution whose [OH⁻] = 1.89 x10⁻⁶ ?

pOH = - log [OH⁻]pOH = - log [1.89 x 10⁻⁶]

pOH = 5.724

What is the pOH of a solution with an [OH⁻] = 3.33 x 10⁻⁴ ?

pOH = - log [OH⁻] pOH = - log 3. 33 x 10⁻⁴ pOH = 3.478

Calculating [OH⁺] from pH

• This operation involves the inv log function or the antilog so we will have an activity sheet designed to help you to enter this correctly in your individual calculator.

• A solution has a pOH of 5.55 what is the [OH⁺]?

• [OH⁺] = 2.8 x 10⁻⁶ M

Calculating: pOH from pH, pOH from pH

Due to experiment/observation we know that

pH + pOH = 14.00

So we can convert between pH + pOH with a simple subtraction problem.

pH and pOH conversions

The pH of a solution is 2.29, what is the pOH?pOH = 14.00 – pHpOH = 14.00 – 2.29pOH = 11.71The pOH of a solution is 6.65, what is the pH?pH = 14.00 – pOHpH = 14.00 - 6.65pH = 7.35

Reactions of Acids with Metals

Many acids will react with the more reactive metals to produce the salt of that acid and hydrogen gas.

Ex.

2Al + 6HCl → 2AlCl₃ + 3H₂Solid metal aqueous acid salt hydrogen gas

Mg + H₂SO₄ → MgSO₄ + H₂Solid metal aqueous acid salt hydrogen gas

Acidity of Basicity on pH Scale

Summary:pH = 7 Neutral

pH < 7 Acidic

pH > 7 Basic

Ion Product Constant

In aqueous solution

[OH⁻] [H⁺] = 1.00 x 10⁻¹⁴

Ex.If an aqueous solution has a [H⁺] = 1.83 x 10⁻⁵ what is the [OH⁻] ?

[OH⁻] = 1.00 x 10⁻¹⁴ / 1.83 x 10⁻⁵[OH⁻] = 5.46 x 10⁻¹⁰

If an aqueous solution has an [OH⁻] of 5.43 x 10⁻⁴, what is the [H⁺] ?

[H⁺] = 1.00 x 10⁻¹⁴ / [OH⁻]

[H⁺] = 1.00 x 10⁻¹⁴ / 5.43 x 10⁻⁴

[H⁺] = 1.84 x 10⁻¹¹