IB Chemistry: Calculations involving acids and bases

Click here to go back to Table of Contents

18.1 Calculations involving acids and bases

18.1.1 State the expression for the ionic product constant of water (Kw)

Water ionizes slightly to a hydrogen and hydroxide ion.

            H2O ↔ H+ + OH

Remember the equilibrium constant in unit 7? We’re going to introduce a new equilibrium constant called the ionic product constant of water, Kw, which basically works off the sample concept as Unit 7, but this time we’re going to be a bit more specific.

            Kw = [H+][OH] / [H2O]

As the concentration of water can be considered constant (as it barely ionizes), we can rearrange this constant to become

            Kw = [H+][OH] / 1

            Kw = [H+][OH]

18.1.3 Solve problems involving [H+(aq)], [OH–(aq)], pH and pOH.

E.g A substance has pH of 3.1 at room temperature. Calculate [H+], [OH] and pOH

 [H+] is simply the concentration of hydrogen. To find this, simply use formula learned in core, concentration = 10-pH,:

[H+] = 10-3.1 = 0.000794 moldm-3

 Let’s calculate pOH before finding [OH-].

 pOH + pH = 14 by definition, so pOH = 14 – 3.1 = 10.9

 Now, to calculate [OH] , simply use formula [OH] = 10-pOH

 [OH] = 10-10.9 = 1.25 x 10-11 moldm-3

18.1.4 State the equation for the reaction of any weak acid or weak base with water, and hence deduce the expressions for Ka and Kb.

Let’s assume we have a weak acid, HA, that dissolves in water. The reaction can be described as:

 HA (aq) + H2O (l) ↔ A (aq) + H3O+ (aq)

 Therefore, the equilibrium constant of this acid reaction, Ka, is:

           Ka = [A][H3O+] / [HA][H2O]

 Again, we can ignore the concentration of water.

 Ka = [A][H3O+] / [HA]

 The same process can be applied to a weak base, B,  that dissolves in water.

  B (aq) + H2O (l) ↔ BH+ (aq) + OH (aq)

 The equilibrium constant, Kb,

 Kb = [OH][BH+] / [B]

Again, the concentration of water can be ignored.

 18.1.5 Solve problems involving solutions of weak acids and bases using the expressions:

A few rules that should be learned:

 Ka x Kb  = Kw

 pKa + pKb = pKw

pH + pOH = pKw = 14

 18.1.6 Identify the relative strengths of acidsand bases using values of Ka, Kb, pKa and pKb.

The larger the value of Ka, the more acidic the solution is. This is because as Ka value increases, [H3O+] also increases, and hence the solution is more acidic. However, as Ka increases, pKa decreases.

 The same applies to Kb. As Kb increases, it becomes a stronger base. However, as Kb increases, pKb decreases.

About Acceleratedstudynotes

IB Student studying at Shatin College. Passionate about education and using technology to enhance the distribution of knowledge and how we learn. I'm working on improving on the IGCSE Coordinated science page at the moment.
This entry was posted in Everything. Bookmark the permalink.