Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3CH2COOH

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M CH₃CH₂COOH(aq) with 0.2000 M LiOH(aq) after 7 mL of the base have been added. K_a of propanoic acid = 1.3 x 10^-5.

The answer is 5.25

Answer 1

In this titration, you are creating a buffer as you add the OH^- to the CH3CH2COOH...

CH3CH2COOH + OH^- ==> CH3CH2COO^- + H2O

So you now not only have the weak acid, CH3CH2COOH, but you also have the salt of that weak acid, CH3CH2COO^- and this makes a buffer.

20.00 ml x 1 L/1000 ml x 0.1000 mol/L = 0.002 moles CH3CH2COOH initially present

7 ml x 1 L/1000 ml x 0.2000 mol/L = 0.0014 moles OH^- added

Total final volume = 20 ml + 7 ml = 27 ml = 0.027 L

CH3CH2COOH + OH^- ===> CH3CH2COO^-

0.002.....................0.0014...................0................Initial

-0.0014...................-0.0014...............+0.0014........Change

0.0006......................0......................0.0014...........Equilibrium

Final [CH3CH2COOH] = 0.0006 mol/0.027 L = 0.0222 M

Final [CH3CH2COO-] = 0.0014 mol/0.027 L = 0.0519 M

Using the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid] where pKa = -log Ka = -log 1.3x10^-5 and pKa = 4.89

pH = 4.89 + log (0.0519/0.0222)

pH = 4.89 + 0.37

pH = 5.25