ACIDS AND BASES

Ionization of Water

H

O

H

H+

O

H

-

· Describe the equilibrium between hydronium and hydroxide ions in water.

· Define the ion product of water (Kw).

· Describe the concentrations of hydronium and hydoxide ions .

· Calculate the concentration of hydronium or hydroxide ions.

Water is amphoteric

HA + H2O(l) H3O+(aq) + A¯(aq)

or

B + H2O(l) BH+(aq) + OH¯(aq)

Water dissociates into ions - self-ionization.When water particles collide - ions form.

H2O(l) H+(aq) + OH¯(aq)

H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)

or

ion product for water, Kw

KW = [H3O+][OH¯]

KW = [H+][OH¯]

H2O(l) H+(aq) + OH¯(aq)

In room temperature water, the concentration of hydrogen ions and hydroxide ions, are each

1.0 x 10 -7 mol/LWater is neutral - these two must be a 1:1 ratio

Any solution in which the [H+] and [OH-] are 1.0 x 10-7 M is described as a neutral solution.

Like all eqbm constants, the value of Kw varies with temperature.

Kw = [H+][OH-] = 1.00 x 10-14

Like any reversible reaction, Le Chatelier's applies:

Add a SA – increase [H+] decrease [OH-]

Add a SB – increase [OH-] decrease [H+]

This means that H+ and OH¯ are BOTH present in

any solution - whether they are acidic or basic.

OH- H2O H+ +

If 2.5 moles of hydrochloric acid is dissolved in 5.0 L of water, what is the concentration of the hydroxide ions?

Since HCl is a strong acid (100% dissociation):[H3O+] = 0.50 mol/L.

5.0 L= 0.50 mol/L

2.5 molnV

M=

HCl (s) H+(aq) + Cl¯(aq)

[OH-] = 2.0 x 10-14 M

Kw =

[H3O+]

[OH-]

1.0 x 10-14 =

[0.50

[OH-]+ 1.0 x 10-7]]

0.40 g of NaOH is dissolved in water to make a solution with a volume of 1.0 L. What is the hydronium ion concentration in this solution?

NaOH = 40.0 g/mol

0.40 g1 L 40.0 g

1 mol= 0.01 mol/L

NaOH (s) Na+(aq) + OH¯(aq)

[NaOH] = [OH–] = 0.010 mol/L

Substitute into the equilibrium law and solve for hydronium ion concentration:

[H+] = 1.0 x 10-12 M

Kw =

[OH-]

[H3O+]

1.0 x 10-14 =

[0.010

[H3O+]+ 1.0 x 10-7]]

H2O(l) H+(aq) + OH¯(aq)

KW = [H+][OH¯]

= 1.00 x 10-14

Add a SA – increase [H+] decrease [OH-]

Add a SB – increase [OH-] decrease [H+]