Buffer Solution Chemistry Math Question 3

Question

If equal volumes of 0.11 M acetic acid and 0.06 M sodium hydroxide are mixed, what is the pH of the resulting solution? The pK_a of acetic acid is 4.75

Answer 1

HC2H3O2 + NaOH ==> C2H3O2Na + H2O

This is a neutralization or acid/base reaction.

Since there are equal volumes of acid and base, but the concentrations are different, there will be an excess of acid left after reaction (0.11 M acid vs 0.06 M base) and all the base will be used up. There will also be the formation of the salt of the acid (C2H3O2Na). To illustrate, let us assume 100 ml (0.1 L) of each is use:

moles acid = 0.1 L x 0.11 mol/L = 0.011 moles acid

moles base = 0.1 L x 0.06 mol/L = 0.006 moles

After reaction you have 200 ml (0.2 L) and the following moles:

moles acid = 0.011 - 0.006 = 0.005 moles acid final

moles base = 0.006 - 0.006 = 0 moles base

moles salt (C2H3O2Na) = 0.006 moles salt

If you prefer to use an ICE table, it looks like this:

HC2H3O2 + NaOH ===> C2H3O2Na + H2O

0.011...........0.006................0.......................0.........Initial

-0.006.........0.006.................+0.006........................Change

0.005.............0.....................0.006..........................Equilibrium

So, at this point, we have formed a BUFFER solution with a weak acid (acetic acid) and the salt (sodium acetate). To find the pH, we can use the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid]

pH = ?

pKa = 4.75

[salt] = 0.006 moles/0.2L = 0.03 M

[acid] = 0.005/0.2L = 0.025 M

pH = 4.75 + log (0.03/0.025) = 4.75 + 0.079

pH = 4.83