Calculate the pH during the titration of 20.00 ML of 0.1000M with 0.2000M solution after the following edition of acid. The Ka of CH3COOH is 1.8 x10^-5 0ml 5.000ml 10.00ml 15ml

Question

Calculate the pH during the titration of 20.00 ML of 0.1000M CH3COOH with 0.2000M NaOH solution after the following addition of acid. The Ka of CH3COOH is 1.8 x10^-5 :

0ml

5.000ml

10.00ml

15.00 ml

Answer 1

Probably easiest to use the Henderson Hasselbalch equation for this problem:

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

So, we need to first find the [salt] and [acid] and then plug this value into the equation to solve for pH.

Before addition of any NaOH, we simply have CH3COOH ==> CH3COO- + H+

Ka = [CH3COO-][H+] / [CH3COOH] = (x)(x) / 0.1 = 1.8x10-5

x = 1.34x10-3 M = [H+]

pH = -log [H+] = 2.87

For the addition of 5 ml of 0.2 M NaOH:

mols CH3COOH initially present = 20.00 ml x 1 L/1000 ml x 0.1 mol/L = 0.002 mols CH3COOH

mols NaOH added = 5.000 ml x 1 L /1000 ml x 0.2000 mol/L = 0.001

CH3COOH + NaOH ===> CH3COONa + H2O

0.002...............0.001...................0.......................Initial

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

0.001.................0.......................0.001.................Equilibrium

pH = pKa + log [0.001/0.001]

pH = pK

pH = 4.82

For the addition of 10 ml of 0.2 M NaOH: 10 ml x 1 L/1000 ml x 0.2 mol/L = 0.002 mol NaOH added

CH3COOH + NaOH ===> CH3COONa + H2O

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

-0.002..........-0.002................+0.002...............Change

0......................0.........................0.002............Equilibrium

At this point, we no longer have a buffer since all the CH3COOH has been neutralized. To find pH, we look at the hydrolysis of CH3COO- as follows:

CH3COO- + H2O ==> CH3COOH- + OH-

Kb = 1x10^-14/Ka = 1x10^-14 / 1.8x10^-5 = 5.56x10^-10

Kb = [CH3COO-][OH-] / [CH3COO-]

[CH3COO-] = 0.002 mol/0.03 L = 0.3707 M

5.56x10^-10 = (x)(x) / 0.3707

x = 1.44x10^-5 = [OH-]

pOH = -log [OH-] = 4.84

pH = 9.16

Do the same procedure for 15 ml of NaOH. No CH3COOH remains, and the entire solution is essentially OH-.