Question

A titration is performed by adding .435M KOH to 80ml of .205M HC3H5O2.

A) Calculate the pH before the addition of any KOH?

B) Calculate the pH after the addition of 7.54ml of KOH?

C) Calculate the volume of base needed to reach the equivalence point?

D) Enter the pH of the solution at the equivalence point of the titration?

Answer 1

M1 = 0.435 KOG

V2 = 80 ml

M2 = 0.205 M HC3H5O2

From internet database = Ka = 1.3*10^-5

There is an equilibrium

HAc <-> H+ and Ac-

Ka = [H+][Ac-] / [HAc]

a)

when there is no base:

pH = -log[H+]

Calculate [H+]

[H+] = [Ac-] = x (1 mol of H+ is free when 1 mol of Ac- is also free in equilibrium)

Substitute in Ka

Ka = [H+][Ac-] / [HAc]

1.3*10^-5 = x^2 / [0.205 - x]

Solve for x

x = 0.0016 or -0.0016

choose positive since negative is imposible for a concentration

x = [H+] = 0.0016

pH= -log[H+] = -log(0.0016) = -log(0.0016) = 2.79

B)

when you add 7.54 ml of KOH, you will react some acid

moles of KOH = M1*V1 = 0.435 * 7.54 ml = 3.27mmol

moles of Acid = M2*V2 = 0.205* 80 ml = 16.4 mmol

16.4 mmol - 3.27 mmol = 13.13 mmol of acid are left

Recalculate concentration (total Volume changed = V1+V2 = 80 ml + 7.54 ml = 87.54 ml

M of Acid = moles Acid/ Volume of Acid = 13.13 mmol / 87.54 ml = 0.149 M

Repeat A)

[H+] = [Ac-] = x (1 mol of H+ is free when 1 mol of Ac- is also free in equilibrium)

Substitute in Ka

Ka = [H+][Ac-] / [HAc]

1.3*10^-5 = x^2 / [0.149 - x]

Solve for x

x = 0.00139

x = [H+] = 0.00139

pH= -log[H+] = -log(0.00139) = -log(0.00139) = 2.86

C)

Calculate total volume to reach equivalent point

equivalent point =

moles of OH- = moles of H+

find out how many moles of H+ are in 80 ml of that solution

moles = M*V = 80*0.205 = 16.4 mmol of H+

MV= mmol of OH-

0.435 * V = 16.4 mmol

V = 16.4/0.435 = 37.7 ml of this base is needed

V = 37.7 ml of KOH

D)

Calculate pH at this point

in the equivalent point

there is no acid, no base left

only the "acetate"

Ac- + H2O <-> HAc + OH-

This solution will turn out to be basic (H+ ions are being used for the HAc acid)

We need to calculate OH-

Kw = KaKb

10^-14 = (1.3*10^-5) Kb

Kb = 7.7*10^-10

Kb = [HAc][OH-] / [Ac-]

[OH-] = [HAc] = x

[Ac-] = M - x

Substitute

Kb = [HAc][OH-] / [Ac-]

Kb = x^2 / ( M - x)

NOTE we need to recalculate molarity of acetate

M = moles of Acetate / New volume

M = (0.205*80ml) / (80ml + 37.7 ml) = 0.139

1.3*10^-5 = x^2 / (0.139 - x)

Solving for x

x = 0.001337

that is [OH-] = 0.001337

pOH = -log[OH-] = -log[0.001337) = 2.87

pH = 14- pOH = 14- 2.87 = 11.13

pH= 11.13