Consider the titration of 100.0 mL of 0.20 M acetic acid (ka = 1.8x10^-5) by 0.10 M KOH.

Question

Consider the titration of 100.0 mL of 0.20 M acetic acid (ka = 1.8 x 10^-5) by 0.10 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

a) 200.0 mL

b) 250.0 mL

I know A is at the equivalence point but I don't understand the process to finding the answer so please show steps.

the answers are:

a) 8.79

b) 12.15

please show the steps so I can follow along to both a and b. Thank you!

Answer 1

CH3COOH + OH^- ==> CH3COO^- + H2O ... balanced equation

You are correct that when 200 ml of 0.1 M KOH is added to 100.0 ml of 0.20 M CH3COOH, you are at the equivalence point (0.02 moles acid and 0.02 moles base)

At equivalence, no CH3COOH remains, and you have formed 0.02 moles of the conjugate base, CH3COO^-.

The final volume is now 100 ml + 200 ml = 300 ml = 0.300 L

The final [CH3COO^-] = 0.02 mol/0.300 L = 0.0667 M

Now, we must look at the hydrolysis of the CH3COO^-, since that is the only thing in solution.

CH3COO^- + H2O ==> CH3COOH + OH^- (note that CH3COO^- is a base, so we need to use Kb)

KaKb = Kw = 1x10^-14

Kb = 1x10^-14/1.8x10^-5

Kb = 5.56x10^-10

Kb = 5.56x10^-10 = [CH3COOH][OH^-]/[CH3COO^-]

5.56x10^-10 = (x)(x)/0.0667 -x and if we ignore x in the denominator assuming it is small, we have...

x² = 3.71x10^-11

x = 6.08x10^-6 M = [OH^-]

pOH = -log 6.08x10^-6 = 5.22

pH = 14 - pOH = 8.78

After adding 250 ml of 0.1 M KOH, you have an additional 50 ml of OH^- beyond the equivalence or

50 ml x 1 L/1000 ml x 0.1 mol/L = 0.005 moles of excess OH^- (this is the amount of OH^- not neutralized)

The final volume = 100 ml + 250 ml = 350 ml = 0.35 L

The final [OH^-] = 0.005 moles/0.35 L = 0.0143 M

pOH = -log 0.0143 = 1.85

pH = 14 - 1.85 = 12.15

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Additional answers for Mathias N.

100 ml 0.2 M CH3COOH + 0.1 M KOH

For 0 ml KOH added:

CH3COOH ==> CH3COO- + H+

Ka = 1.8x10^-5 = [CH3COO-][H+] / [CH3COOH] = (x)(x) / 0.2

x² = 9x10^-5 and x = [H+] = 9.49x10^-3 M

pH = -log 9.49x10^-3 = 2.02

For 50 ml (0.05 L) 0.1 M KOH added = 0.005 mols KOH

CH3COOH + KOH ==> CH3COOK + H2O

0.02 mol.......0.005 mol............0..........Initial

-0.005........-0.005................+0.005....Change

0.015 mol.....0 mol...............0.005 mol..Equilibrium

Final volume = 100 ml + 50 ml = 150 ml = 0.150 L

[CH3COOH] = 0.015 mol/0.150L = 0.1 M

CH3COOK] = 0.005 mol/0.150 L = 0.033 M

pH = pKa + log (0.033/0.1) = 4.74 -0.48

pH = 4.26

For 100 ml (0.1 L) of 0.1 M KOH added = 0.010 mols KOH

CH3COOH + KOH ==> CH3COOK + H2O

0.02mol.......0.01mol...........0............Initial

-0.01..........-0.01..............+0.01.......Change

0.01..............0...................0.01.......Equilibrium

Final volume = 200 ml = 0.2 L

Final [CH3COOH] = 0.01 mol / 0.2 L = 0.05 M

Final [CH3COOK] = 0.01 mol / 0.2 L = 0.05 M

pH = pKa

pH = 4.74

For 150 ml (0.15 L) of 0.1 M KOH added = 0.015 mls KOH

CH3COOH + KOH ==> CH3COOK + H2O

0.02mol.......0.015mol..........0.............Initial

-0.015.......-0.015...............+0.015......Change

0.005..............0.................0.015.........Equilibrium

Final volume = 250 ml = 0.25 L

Final [CH3COOH] = 0.005 mol / 0.25 L = 0.02 M

Final [CH3COOK] = 0.015 mol / 0.25 L = 0.06 M

pH = pKa + log (0.06/0.02) = 4.74 + 0.48

pH = 5.22