Question

Equal volumes of 200 mM lactic acid and 200 mM potassium lactate are mixed to make a buffer. The pKa of lactic acid is 3.86 (a) Determine the pH of this buffer and its final concentration. [4 marks] (b) Calculate the concentrations of the buffer components after acid or base is added to bring the pH to 4.5. Assume a negligible change in volume. 6 marks]

Answer 1

According to Henderson-Hasselbalch equation :

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

pH = 3.86 + log [200 mM/200 mM] (same volume)

pH = 3.86 + log 1 = 3.86 + 0 = 3.86

pH = pKa = 3.86

Final concentration of lactic acid , M2 = M

Initial concentration, M1 = 200 mM

Initial volume, V1 = 100 ml (Let us consider)

Final volume, V2 = 100 + 100 = 200 ml

M1V1 =M2V2

200 mM x 100 ml = M x 200 ml

Final concentration of lactic acid = 100 mM

Similarly, final concentration of lactate = 100 mM

b) pH = 4.5 and pKa = 3.86

According to Henderson-Hasselbalch equation :

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

4.5 - 3.86 = log [salt]/[acid]

0.64 = log [salt]/[acid]

[salt]/[acid] = 10^0.64 = 4.365

As there is negligible change in volume and initial concentrations of lactic acid and lactate are equal

If acid is added then lactic acid reacts with water to form lactate

HA+ H₂O --> H₃O⁺ + A^-

Hence acid concentration decreases by x (concentration of H⁺ ion added) and salt increases by X

200 mM +x/ 200mM -x = 4.365

200+x = 4.365 (200-x)

200 + x = 873 - 4.365 x

x +4.365 x = 873 -200

5.365 x = 673

x = 673/5.365 = 125.4 mM

Hence Concentration of lactic acid after addition of acid = 200-x = 200-125.4 mM = 74.6 mM

Concentration of lactate after addition of acid= 200 +x = 200+125.4 = 325.4 mM

Similarly if base is added

Lactate ion react with the base and form lactic acid

A^- + H₂O --> HA + OH^-

Hence acid concentration increases by x (concentration of H⁺ ion added) and salt decreases by X

Here concentration of lactate = 200 - x = 74.6 mM

Concentration of lactic acid = 200 +x = 325.4 mM