Assume that $\ce{pKa}$ is 4.76

Mix $\pu{50 mL}$ of $\pu{0.1M}$ $\ce{HA}$ and $\pu{50 mL}$ $\pu{0.1M}$ $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ and $\ce{A-}$, measure the $\ce{pH}$.

Then, mix $\ce{10 mL}$ of solution (i) and $\ce{90 mL}$ of water. Calculate $\ce{[HA]}$ and $\ce{[A-]}$, and measure the $\ce{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two.

Equation is: $$\ce{pH} = \ce{pKa}+\log{\left(\frac{\ce{[A-]}} {\ce{[HA]}}\right)}$$

For the first one, I got, $$4.76 + \log{\left(\frac{0.1}{0.1}\right) = 4.76}$$ But I don't know if that's correct. How would I approach the subsequent problem when water is added?

  • $\begingroup$ I have edited your post with MathJax. You can learn how to use MathJax to write chemical equations here: chemistry.meta.stackexchange.com/questions/86/… $\endgroup$ Commented Apr 27, 2017 at 12:11
  • $\begingroup$ Moreover I changed A to (i), so that we don't confuse with A anion. $\endgroup$ Commented Apr 27, 2017 at 12:12
  • $\begingroup$ The solution (i) is a weak acid and it's conjugate base type of buffer, the $pK_a$ value matches that of acetic acid. Although the $pH$ of a buffer remains unchanged upon reasonable dilution, $10 \pu{mL}$ of solution (i) and $90 \pu{mL}$ of water will have quite an impact. Have you tried writing the ionic equilibrium equation of the dissociation of weak acid $\ce{HA}$? You can probably apply common-ion effect for the concentration of $\ce{H+}$ (due to water). $\endgroup$ Commented Apr 27, 2017 at 19:23

2 Answers 2


@Tyberius: I think you're neglecting Berry Holmes' comment above. Dilution is not always inconsequential to the pH, especially when concentrations get quite low. As my analytical chemistry teacher once put it, if you squeeze a lemon into a lake...

@notiarogers: Tyberius already showed you that you miscalculated the formal concentrations of $\ce{HA}$ and $\ce{A^-}$. Don't skip any steps: $50 \space mL$ of $0.1 \space M$ solution contain $50 \cdot 0.1 = 5 \space mmol$ of substance; the final volume is $100 \space mL$, therefore the final concentration is $5 \space mmol / 100 \space mL = 0.05 M$.

The formula used above (for the calculation of the pH of an aqueous buffer solution made from a weak acid + its salt with a strong base) is only an approximation. Its main assumption is that the concentrations of acid $\ce{HA}$ and anion $\ce{A^-}$ are equal to the nominal ones (0.05 M, as explained above). In reality, when you dissolve $\ce{HA}$ in water, some of it already dissociates (forming some $\ce{A^-}$), and when you dissolve $\ce{NaA}$ in water, it dissociates to $\ce{Na^+}$ and $\ce{A^-}$, and some $\ce{A^-}$ already reverts to $\ce{HA}$ (which, BTW, is why the former solution is slightly acidic and the latter is slightly basic). So things are quite a bit more complicated than it seems: even your initial solutions don't contain 'only' $\ce{HA}$ or 'only' $\ce{A^-}$.

Now, it may be that whoever is asking you to do this calculation wants you to stick to such approximation, and in fairness the final concentration after dilution is probably still large enough, unless you want a very precise result.

In case they want you to do a more complicated calculation, you can write an equation for each equilibrium, one for each mass balance and one for the charge balance. Your equation is the dissociation equilibrium of $\ce{HA}$: $$\ce{K_A=\frac {[H^+] \cdot [A^-]}{[HA]}}$$ You can add the dissociation equilibrium of water: $$\ce{K_w=[H^+] \cdot [OH^-]}$$The mass balance of all acid-related species is:$$\ce{[HA] + [A^-] = C_{HA} + C_{NaA}}$$ where $\ce{C_HA}$ is the nominal concentration of $\ce{HA}$, i.e. 0.05 M as calculated above, and $\ce{C_{NaA}}$ is the nominal concentration of $\ce{NaA}$, i.e. also 0.05 M in this case.

The charge balance is: $$\ce{[H^+] + [Na^+] = [OH^-] + [A^-]}$$ And as $\ce{[NaA]}$ is a strong electrolyte, we know that: $$\ce{[Na^+] = C_{NaA}}$$

If you put all these equations together and eliminate all unknowns except $\ce{[H^+]}$, you get a 3rd degree equation in $\ce{[H^+]}$ and known constants alone, which you can solve (probably best numerically) to find the pH.

The exact solution would look very different from the approximate one you were using, which was essentially:$$\ce{[H^+] = K_A \cdot \frac{C_{HA}}{C_{NaA}}}$$

I can tell you that the pH after your 10-fold dilution is only slightly higher than 4.76 (it's 4.763). You would have to dilute 100-fold to start seeing some difference (4.79); and frankly that looks to me like a minor difference for most practical applications.

Note: we are neglecting the difference between activity and concentration; but I guess we have seen enough maths for one day...


Equal amounts of the acid and its corresponding base are mixed. The concentration of the acid will be equal to its corresponding base, i.e. $0.05 \text{ M}$. When the concentration of the acid is equal to the concentration of its corresponding base, $pH = pk_a$. Thus the $\text{pH} = 4.76$.

Dilution of the solution will decrease the concentration of the acid, as well as, of its corresponding base. The concentrations will be $[\text{acid}] = [\text{base}] = 10*0.05/100 = 0.005 \text{ M}$. The ratio between the acid and the base will not change. Thus, the pH will still be at $4.76$.


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