Biology Equations - Lecture notes 3-6 PDF

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Lecture notes these notes are very informative these help to understand how they can be used to make your own revision notes!...

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Biology – Equations The descriptions in the previous section can seem a bit intimidating, however, application of this formula is straight-forward. Always remember the Henderson-Hasselbalch equation:

pH = pKa + log10 [A-]

[HA]

Example 1: Consider a buffer solution of an acid with a pKa of 4,3 and an acid concentration that is half the concentration of the conjugate base (do not forget concentrations are in mol/L)..

Although we do not know the actual concentrations of the acid and base we do know the ratio ([A]/[HA] = 2/1 = 2) and we also know the pKa. Therefore:

pH = 4.3 + log10 2

pH = 4.3 + 0.301

pH = 4.601

And that's it. A solution with the above composition will have a pH of 4.601.

Example 2: You have made up a buffer solution containing a final concentration of acetic acid of 0.05 mol/L and sodium acetate at 0.2 mol/L. Acetic acid has a pKa = 4.76 at 25°C. What is the pH of this solution?

The ratio of base to acid ([A-]/[HA]) is 0.2/0.05 which is equal to 4. Therefore:

pH = 4.76 + log10 4

pH = 4.76 + 0.602

pH = 5.362

That's it, another straight-forward calculation.

What if the acid/concentrations are flipped so acetic acid is 0.2 mol/L and sodium acetate is 0.05 mol/L? Now the ratio of base to acid is 0.25. Therefore:

pH = 4.76 + log10 0.25

pH = 4.76 + (-0.602) Note: the log of 0.25 is negative.

pH = 4.158

Straight-forward to calculate and as expected the pH of this solution is more acidic than the previous one (because it contains more acid)....