University:
High School
Course:
High School Chemistry
Academic year:

2023
Views:

6

Pages:

3
Author:

anon

IB Chemistry: Calculations involving acids and bases Calculations involving acids and bases State the expression for the ionic product constant of water (Kw) Water ionizes slightly to a hydrogen and hydroxide ion. H2O ↔ H+ + OH– Recall from unit 7 the equilibrium constant? The ionic product constant of water, abbreviated Kw, which basically uses the same sample concept as Unit 7, but this time we're going to be a little more specific, is a new equilibrium constant that we're going to introduce. 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–] 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 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. 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 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.