8
$\begingroup$

I completed a titration of Ammonium Acetate buffer solution and to it I added $\pu{2M}$ Hydrochloric Acid.

I measured the initial $\ce{pH}$ of the buffer solution before any acid was added and I used methyl orange as an indicator which has a $\mathrm{p}K_\mathrm{a}$ value of 3. My independent variable was the temperature of the buffer.

I was just wondering how I would calculate the buffer capacity. Is it $\Delta\ce{pH}$/Volume?

$\endgroup$
8
$\begingroup$

In addition to Martin's answer, there is at least an old recommendation by E. B. Sandell and T. S. West in Pure Appl. Chem., 1969, 18, 427-436 (DOI), which states:

Buffer capacity or buffer index. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or strong base required to change the pH by one unit when added to one litre of the specified buffer solution.

$\endgroup$
  • $\begingroup$ oh okay thanks! Any suggestions how I should calculate that? $\endgroup$ – Sushanth Saha May 22 '15 at 10:34
6
$\begingroup$

As far as I can remember from my tutorials, the formula was $$\beta =\frac{n(\ce{H+})}{\Delta\ce{pH}},$$ meaning: How many protons have to be added to produce a change in the pH of one unit. But there might be different definitions. The IUPAC does not provide an official one.

$\endgroup$
  • $\begingroup$ oh okay thanks! Any suggestions how I should calculate that? $\endgroup$ – Sushanth Saha May 22 '15 at 10:34
  • $\begingroup$ @Sushanth Take a sample of your buffer and add a known amount of acid to it and see how the pH changes. Ideally this should be done with an pH electrode. $\endgroup$ – Martin - マーチン May 22 '15 at 10:45
4
$\begingroup$

Perhaps a derivation of the definition I was taught. Consider the Henderson-Hasselbalch equation: $$pH=pK_a+\log\bigg(\frac{[A^-]}{[HA]}\bigg). $$ Given the definition provided by Klaus above and the fact that a weak acid's $pK_a$ is unique, the only term that could possible alter the $pH$ by an amount of $\pm1$ is $\log\big(\frac{[A^-]}{[HA]}\big).$ We decide to choose the case of the $pH$ increasing by $+1$ (it does not matter which direction---either more acidic or more basic---the reaction proceeds, we simply want the magnitude of the numerical change to be 1). If the $pH$ increases by 1, then the solution becomes more basic. Consequently, the term $\log\big(\frac{[A^-]}{[HA]}\big)$ changes by $\log\big(\frac{[A^-]+\beta}{[HA]-\beta}\big)$, where $\beta$ denotes the buffer capacity. We know this term must equal $+1$ because a perfect buffer has equimolar solutions of $[HA]$ and $[A^-]$. We obtain the following equation: $$1=\log\bigg(\frac{[A^-]+\beta}{[HA]-\beta}\bigg). $$ Solving for $\beta$ yields the buffer capacity equation $$\beta=\frac{10[HA]-[A^-]}{11}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.