The Equilibrium ofWeak Acids and Bases

The Equilibrium ofWeak Acids and Bases

• The dissociation of an acidic or basic compound in aqueous solution produces ions that interact with water (REVIEW!)

• Pure water contains a few ions, produced by the dissociation of water molecules:

• At 25°C, only about two water molecules in one billion dissociate.

• In neutral water, at 25°C, the concentration of hydronium ions is the same as the concentration of hydroxide ions: 1.0 × 10−7 mol/L.

• These concentrations must be the same because the dissociation of water produces equal numbers of hydronium and hydroxide ions:

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• [H3O+] is not 1.0 × 10−7 mol/L at other temperatures. (Le Chatelier?)

• The same is true of [OH−] .

The Ion Product Constant for Water

What would the equilibrium constant, Kc, be for the dissociation of water?

The Ion Product Constant for Water

• So few ions form that the concentration of water is essentially constant.

• The product Kc[H2O]2 is equal to the product of the concentrations of hydronium ions and hydroxide ions (work it out!)

• The equilibrium value of the concentration ion product [H3O+][OH−] at 25°C is called the ion product constant for water. It is given the symbol Kw.

The Ion Product Constant for Water

The units are commonly dropped, as in other equilibrium expressions you have encountered.

The Ion Product Constant for Water

pH and pOH• Acidity of an aqueous solution can be stated

quantitatively by the concentration of the hydronium ions that are present.

• pH scale - devised by Danish biochemist named Søren Sørensen to represent acidity (and, by extension, basicity).

• We use logarithms to define pH because [H3O+] is often a very small number.

(Side-note: review of Logarithms)

• Write the exponential equation, 43 = 64 as a logarithm.

• Write 25 as a logarithm.

• Write 5-2 as a logarithm.

pH: What does it mean?

• Since [H3O+] is so small, pH is defined by using the logarithmic equation:

pH = - log10 [H3O+]

Calculate the pH of Water

In neutral water, [H3O+] is:

pH = -log10 [H3O+]

Calculate the pH of Water

In neutral water, [H3O+] is:

pH = -log10 [H3O+] = -log10 (1.0 X 10-7 )

What exponent do we need to apply to the base10 to equal 1.0 X 10-7 ?

Calculate the pH of Water

In neutral water, [H3O+] is:

pH = -log10 [H3O+] = - (-7) = -log10 (1.0 X 10-7 ) = 7

What exponent do we need to apply to the base10 to equal 1.0 X 10-7 ?

Calculating pOH

• You can calculate the pOH (the power of hydroxide ions) of a solution from the [OH−].

• pOH = −log[OH−]

• Kw = [H3O+][OH−] = 1.0 × 10−14 at 25°C

• ∴pH + pOH = 14

Sample Problem, Page 390

pH and pOH

Another Way to Find [H3O+] and [OH−]

• You can calculate [H3O+] or [OH−] by finding the antilog of the pH or pOH.

• [H3O+] = 10−pH

• [OH−] = 10−pOH

Sample Problem, Page 392

The Acid Dissociation Constant

• When a weak acid dissolves in water, it does not completely dissociate.

The Acid Dissociation Constant

• You can represent any weak monoprotic acid with the general formula HA. The equilibrium of a weak monoprotic acid in aqueous solution can be expressed as follows:

The Acid Dissociation Constant• This new constant is called the acid dissociation

constant, Ka, (aka the acid ionization constant. With either name, the symbol is Ka.)

pH and Ka of a Weak Acid• The smaller the value of Ka, the less the

acid ionizes in aqueous solution.

Ka constants

Percent Dissociation

• The percent dissociation of a weak acid is the fraction of acid molecules that dissociate compared with the initial concentration of the acid, expressed as a percent.

• The percent dissociation depends on the value of Ka for the acid, as well as the initial concentration of the weak acid.

Sample Problem

Sample Problem

Polyprotic Acids

• polyprotic acids have more than one hydrogen atom that dissociates.

• Each dissociation has a corresponding acid dissociation

• constant. • Problems that involve polyprotic acids can be

divided into as many sub-problems as there are hydrogen atoms that dissociate.

• The ion concentrations that are calculated for the first dissociation are substituted as initial ion concentrations for the second dissociation, and so on.

HOMEWORK:

PPs 13-18 (Pg. 392)PPs 19-25 (Pg. 400)PPs 25-28 (Pg. 402)

SR, Pg. 403:1-5.