Question

Be sure to answer all parts. A 23.0−mL solution of 0.110 M CH3COOH is titrated with a 0.220 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 10.0 mL (b) 11.5 mL (c) 15.0 mL

Answer 1

a)when 10.0 mL of KOH is added

we have:

Molarity of CH3COOH = 0.11 M

Volume of CH3COOH = 23 mL

Molarity of KOH = 0.22 M

Volume of KOH = 10 mL

mol of CH3COOH = Molarity of CH3COOH * Volume of CH3COOH

mol of CH3COOH = 0.11 M * 23 mL = 2.53 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.22 M * 10 mL = 2.2 mmol

We have:

mol of CH3COOH = 2.53 mmol

mol of KOH = 2.2 mmol

2.2 mmol of both will react

excess CH3COOH remaining = 0.33 mmol

Volume of Solution = 23 + 10 = 33 mL

[CH3COOH] = 0.33 mmol/33 mL = 0.01M

[CH3COO-] = 2.2/33 = 0.0667M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {6.667*10^-2/1*10^-2}

= 5.569

Answer: 5.57

b)when 11.5 mL of KOH is added

we have:

Molarity of CH3COOH = 0.11 M

Volume of CH3COOH = 23 mL

Molarity of KOH = 0.22 M

Volume of KOH = 11.5 mL

mol of CH3COOH = Molarity of CH3COOH * Volume of CH3COOH

mol of CH3COOH = 0.11 M * 23 mL = 2.53 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.22 M * 11.5 mL = 2.53 mmol

We have:

mol of CH3COOH = 2.53 mmol

mol of KOH = 2.53 mmol

2.53 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 2.53 mmol

Volume of Solution = 23 + 11.5 = 34.5 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 2.53 mmol/34.5 mL = 0.0733M

CH3COO- dissociates as

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

0.0733 0 0

0.0733-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*7.333*10^-2) = 6.383*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.383*10^-6 M

[OH-] = x = 6.383*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (6.383*10^-6)

= 5.20

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.20

= 8.80

Answer: 8.80

c)when 15.0 mL of KOH is added

we have:

Molarity of CH3COOH = 0.11 M

Volume of CH3COOH = 23 mL

Molarity of KOH = 0.22 M

Volume of KOH = 15 mL

mol of CH3COOH = Molarity of CH3COOH * Volume of CH3COOH

mol of CH3COOH = 0.11 M * 23 mL = 2.53 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.22 M * 15 mL = 3.3 mmol

We have:

mol of CH3COOH = 2.53 mmol

mol of KOH = 3.3 mmol

2.53 mmol of both will react

excess KOH remaining = 0.77 mmol

Volume of Solution = 23 + 15 = 38 mL

[OH-] = 0.77 mmol/38 mL = 0.0203 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.026*10^-2)

= 1.69

we have below equation to be used:

PH = 14 - pOH

= 14 - 1.69

= 12.31

Answer: 12.31