Acid and Base Dissociation Constants Calculations
HSC Chemistry Syllabus
 Explore the use of K_{eq} for different types of chemical reactions, including but not limited to:
 calculate and apply the dissociation constant ( K_{a}) and p K_{a} ( p K_{a} = log_{10}( K_{a})) to determine the difference between strong and weak acids (ACSCH098)
Calculations Involving Acid and Base Dissociation Constants
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Calculations for Acid Dissociation Constant K_{a}
 For a generic weak acid HA:
$$HA(aq) + H_2O(l) \rightleftharpoons A^–(aq) + H_3O^+(aq)$$
 When acid molecules dissociate, the concentration of hydrogen ions, [H_{3}O^{+}] equals [A^{–}] at equilibrium.
Commons assumptions

When the dissociation constant, K_{a}, is small, the proportion of acid molecules that undergoes ionisation is small. As such, it can be assumed that the initial concentration of acid, HA, approximately equals its concentration at equilibrium. In other words, the change in [HA] during ionisation is negligible.

It can be assumed, during calculation, that selfionisation of water does not affect [H_{3}O^{+}] of the solution and therefore, it does not contribute to the pH.

If the acid is diprotic or triprotic, it can be usually assumed that the second and third ionisation occur to a smaller extent than the first ionisation.
In other words, if K_{a} of first ionisation is already small, K_{a }of subsequent ionisation reactions can be disregarded. For example, hydrogen sulfide is diprotic and its two ionisation reactions are as follows:
H_{2}S(aq) + H_{2}O(l) ⇌ HS^{–}(aq) + H_{3}O^{+}(aq) K_{a1} = 8.9 × 10^{–8}
HS^{–}(aq) + H_{2}O(l) ⇌ S^{2–}(aq) + H_{3}O^{+}(aq) K_{a2} = 1.3 × 10^{–14}
This assumption does not apply to all weak acids and certainly not strong acids. Some weak acids ionise to a considerable extent, to which its initial concentration differs substantially with that at equilibrium.
Example 1 – Calculating [H^{+}] from K_{a}
The K_{a} of hypochlorous acid, HOCl, is 3.5 × 10^{–8}.
A solution of hypochlorous acid, HOCl, is made by adding 0.0400 moles of HOCl to 500.0 mL of distilled water. What is the hydrogen ion concentration of the solution?
Step 1: Write a balanced equation and an expression for K_{a}
$$HOCl(aq) \rightleftharpoons H^+(aq) + OCl^–(aq)$$
$$K_a = \frac{[H^+][OCl^–]}{[HOCl]}$$
Step 2: Setup an ICE table

[HOCl] 
[H^{+}] 
[OCl^{–}] 
Initial 
$$\frac{0.0400}{0.5000} = 0.0800 \text{ M}$$ 
0 
0 
Change 
– x 
+ x 
+ x 
Equilibrium 
0.0800 – x 
x 
x 
Step 3: Substitute equilibrium concentration expressions (in terms of x) into K_{a} expression:
$$3.5 \times 10^{–8} = \frac{(x)(x)}{0.0800–x}$$
Step 4: By making the assumption: x is neglible as K_{a} is small, simplify the equation to find x:
$$3.5 \times 10^{–8} = \frac{x^2}{0.0800}$$
$$x = [H^+] = \sqrt{0.0800 \times 3.5 \times 10^{8}}$$
$$[H^+] = 5.3 \times 10^{–5} \text{ mol/L (2 s.f.)}$$
Example 2 – Calculating K_{a} from [H^{+}]
A 0.750 mol L^{–1} solution of a particular weak acid, HA, has a [H^{+}] = `3.4 xx 10^(–3)` mol L^{–1} at 25ºC. Calculate its K_{a}.
Step 1: Write a balanced equation and an expression for K_{a}
_{ }
$$HA(aq) + H_2O(l) \rightleftharpoons A^–(aq) + H_3O^+(aq)$$
$$K_a = \frac{[A^–][H_3O^+]}{[HA]}$$
Step 2: Calculate K_{a} from equilibrium expression.
$$K_a = \frac{[3.4 \times 10^{–3}][3.4 \times 10^{–3}]}{[0.750  3.4 \times 10^{–3}]}$$
$$K_a = 1.5 \times 10^{–5} \text{ (2 s.f.)}$$
Calculations for Base Dissociation Constant, K_{b}
 A base by definition, is a substance that accepts hydrogen ions or protons in solution.
 The equilibrium constant for the equilibrium established when a weak base reacts with water is referred to as the base dissociation constant. For example, for a generic base `B`.
$$B(aq) + H_2O(l) \rightleftharpoons HB^+(aq) + OH^–(aq)$$
The base dissociation constant is:
$$K_b = \frac{[OH^–][HB^+]}{[B]}$$
 The base dissociation constant of carbonate ion in water:
$$CO_3^{2–}(aq) + H_2O(l) \rightleftharpoons HCO_3^–(aq) + OH^–(aq)$$
$$K_b = \frac{[OH^–][HCO_3^–]}{[CO_3^{2–}]}$$
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