1.1 - The Ionization of Water and Kw
Module 8
Lesson 1.1 The Ionization of Water and Kw
Key Concepts
Highly purified water is slightly conductive; therefore, we know that some ions are present. Ions are produced when two water molecules collide successfully to produce a hydronium ion and a hydroxide ion.
A water molecule occasionally can accept a proton from another molecule to form a hydronium ion. The products of this reversible reaction are a hydronium ion and a hydroxide ion in a 1:1 ratio. The equilibrium for this reaction at SATP greatly favours the reactants. In other words, the Kc for the ionization of pure water is an extremely small number.
The equilibrium expression for this reversible reaction is
Because liquid water does not change in concentration, it is not included in the equilibrium expression. Notice that the denominator has a line through it.
This particular equilibrium constant is very important in chemistry and is referred to as the ionization constant for water (Kw). At SATP, it has been determined that Kw = 1.0 x 10-14.
\( \mathrm { 1.0 \times 10^{-14} = [H_3O^+(aq)] [OH^- (aq)] } \)
This mathematical relationship applies both to pure water and any solution that is mostly water. Because the hydronium ions and hydroxide ions form in a 1:1 ratio, the concentrations of hydronium ions and hydroxide ions are equal.
In acidic solutions, the concentration of hydronium ions exceeds 10-7 mol/L. This is because an acid is a substance that reacts with water to produce hydronium ions.
In basic solutions, the concentration of hydroxide ions exceeds 10-7 mol/L. This is because hydroxide ions are formed either by the dissociation of an ionic hydroxide or by the reaction of a base entity with water to form hydroxide ions.
\( \mathrm { K_w= [H_3O^+(aq)] [OH^- (aq)] } \)
Because Kw is a constant for dilute aqueous solutions at SATP, the Kw expression can be used to calculate the hydronium ion concentration if the hydroxide ion concentration is known. Conversely, the hydroxide ion concentration can be calculated if the hydronium ion concentration is known.
Watch
Read pages 712 to 716 in your text. Study Communication examples 1 to 4.
Check Your Understanding
Go to your textbook and complete Practice Questions 1 to 4 on page 716 and Section 16.1 Questions 1 and 2 on page 721. Check your answers by clicking the link below.
Page 716 Practice Question 1
\( \mathrm { [OH^-(aq)] = \dfrac{K_w }{ [H_3O^+(aq)] } = \dfrac{1.0 \times 10^{-14} }{ 4.40 \times 10^{-3}~ \frac{mol}{L}} = 2.3 \times 10^{-12}~ \frac{mol}{L} } \)
Page 716 Practice Question 2
\( \mathrm { [H_3O^+(aq)] = \dfrac{Kw}{[OH^-(aq)]} = \dfrac{1.0 \times 10^{-14} }{ 0.299 \frac{mmol}{L} } = 3.3 \times 10^{-11} \frac{mol}{L} } \)
Page 716 Practice Question 3
\( \mathrm { HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq) } \)
Since HCl(aq) is a strong acid, it reacts 100% with water.
Therefore,
\( \mathrm { [H_3O^+(aq)] = [HCl(aq)] } \)
\( \mathrm { [HCl(aq)] = \dfrac{ \left( \dfrac{ 0.37 g }{ 1.01 \frac{g}{mol} + 35.45 \frac{g}{mol} } \right) }{ \left(250~ mL \times \dfrac{1~L}{ 1000 ~mL} \right) } = 0.041 \frac{mol}{L } } \)
\( \mathrm { [H_3O^+(aq)] = 0.041 \frac{mol}{L} } \)
\( \mathrm { [OH^-(aq)] = \dfrac{Kw }{ [H_3O^+(aq)] }= \dfrac{ 1.0 \times 10^{-14} }{ 0.041 \frac{mol}{L} } = 2.5 \times 10^{-13} \frac{mol}{L} } \)
Page 716 Practice Question 4
\( \mathrm { Ca(OH)_2(aq) \rightarrow Ca^{2+}(aq) + 2 OH^-(aq) } \)
\( \mathrm { [OH^-(aq)] = 6.9 \frac{mmol}{L} \times \dfrac{2}{1} = 14 \frac{mmol}{L} ~ or~ 1.4 \times 10^{-2} \frac{mol}{L} } \)
\( \mathrm { [H_3O^+(aq)] = \dfrac{Kw }{[OH^-(aq)]} = \dfrac{1.0 \times 10^{-14}}{ 1.4 \times 10^{-2} \frac{mol}{L}} = 7.2 \times 10^{-13} \frac{mol}{L} } \)
Page 721 Section 16.1 Question 1
- If a solution is neutral, then [H3O+(aq)] = [OH-(aq)]
- If a solution is acidic, then [H3O+(aq)] > [OH-(aq)]
- If a solution is basic, then [H3O+(aq)] < [OH-(aq)]
Page 721 Section 16.1 Question 2
Conductivity: Conductivity of a solution detects the degree of ionization. Strong acids will ionize to a greater extent than do weak acids - therefore, they produce solutions with higher conductivity.
pH: Because strong acids ionize to greater extents than weak acids do, equimolar solutions of strong acids have lower pHs. Note: Solutions must all be of the same molarity!