# Derivation and application of the Henderson-Hasselbalch equation

1. Write down an expression for the dissociation constant, $K_{\mathrm{a}}$, of a very weak monobasic acid $\ce{HA}$ in an aqueous solution in terms of the concentrations of $\ce{H+ (aq), A- (aq) and HA (aq)}$ in the solution.

2. Hence, show that

$$\mathrm{p}K_{\mathrm{a}} = \mathrm{p}\ce{H} - \mathrm{log}_{10} \frac{[\ce{A- (aq)}]}{[\ce{HA (aq)}]} \text{ where } \mathrm{p}K_{\mathrm{a}} = - \mathrm{log}_{10} K_{\mathrm{a}}$$

1. At a particular temperature, $\pu{2.00e-3 mol}$ of the acid $\ce{HA}$ was dissolved in water and the solution diluted until the volume was $\pu{75.00 cm^3}$. When $\pu{25.00 cm^3}$ of a $\pu{0.004 mol dm^{-3}}\,\ce{NaOH}$ solution was added to that acid solution, the $\mathrm{p}\ce{H}$ of the resulting solution was found to be 6.0. Calculate the dissociation constant, $K_{\mathrm{a}}$, of the acid $\ce{HA}$ at that temperature.

Below you can see what I have tried with my current answers.

• For 1, I got $$K_{\mathrm{a}} = \frac{[\ce{H+ (aq)}][\ce{A- (aq)}]}{[\ce{HA (aq)}]}.$$

• For 2, I got the same answer.

But I am stuck on 3. These are the things I already know:

• Moles of NaOH added: $(25/10000) * 0.004 = \pu{0.001 mol}$
• Moles of acid HA present initially: $\pu{0.002 mol}$

How can I proceed further?

The reaction taking place is the quantitative reaction: $$\ce{AH + OH- -> A- + H2O}$$
The initial concentrations are: $$\ce{[OH- ]_0}=\frac{0.001}{0.1}=0.01 \mathrm{M}$$ $$\ce{[AH ]_0}=\frac{0.002}{0.1}=0.02 \mathrm{M}$$ The equilibrium concentrations: $$\ce{[AH ]}=\frac{0.002-0.001}{0.1}=0.01 \mathrm{M}$$ $$\ce{[OH- ]}=10^{-8} \mathrm{M}$$ $$\ce{[A- ]}=0.02-0.01=0.01 \mathrm{M}$$ Using the equation of question (ii):
$\ce{pH = p}K_a=6$