0.50 cm3 of 0.200 mol dm-3 hydrochloric acid is added to the buffer solution. Calculate the new pH.

Answer 1

Use the Henderson Hasselbalch equation: pH = pKa + [salt]/[acid]

Final volume = 40 ml + 100 ml = 140 ml = 0.140 L

Final [ethanoic acid] = 0.040 L x 0.200 mol/L = 0.008 moles/0.140 L = 0.0571 M

Final [Na ethanoate] = 0.1 L x 0.100 mol/L = 0.01 moles/0.140 L = 0.0714 M

Substituting into the HH equation we have...

pH = 4.76 + log (0.0714/0.0571)

pH = 4.76 + 0.097

pH = 4.86

Now, if you add HCl (hydrochloric acid) it will react with the conjugate base (ethanoate) to produce ethanoic acid. The moles of HCl is calculated as 0.005 L x 0.2 mol/L = 0.001 moles HCl (and we can neglect the contribution of 0.5 ml to the final volume)

CH3COO- + H⁺ ==> CH3COOH

0.01............0.001..........0.008........Initial

-0.001.....-0.001.........+0.001.........Change

0.009..........0..............0.009..........Equilibrium

Since [CH3COOH] = [CH3COO-] = 0.009 moles, the pH will be equal to the pKa

pH = 4.76