Question

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.a. 0 mLb. 17.5 mLc. 34.5 mLd. 35.0 mLe. 35.5 mLf. 50.0 mL

Answer 1

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 =  0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 =  0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of  A -   in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42