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):

$\ce{pH = p}K_a=6$

  • $\begingroup$ Hey, how did you take [OH−]=10^−8M ? $\endgroup$ – On the way to success Jun 13 '15 at 12:49
  • $\begingroup$ If the pH=6, then [H3O+]=10-6M, and [OH-]=10-8M $\endgroup$ – Yomen Atassi Jun 13 '15 at 13:35
  • $\begingroup$ But do we need that thing. I feels it is unnecessary $\endgroup$ – On the way to success Jun 13 '15 at 14:46
  • $\begingroup$ Of course, it is not necessary here. But please see the answer as I've added another method to find Ka (using the concentration of OH-). So, you have two ways to find Ka $\endgroup$ – Yomen Atassi Jun 13 '15 at 19:22

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