titration, pH, and volume

Question

Consider the titration of 100.0 mL of 0.200 M CH₃NH₂ by 0.100 M HCl.

(K_b for CH₃NH₂ = 4.4×10^-4)

Part 1

Calculate the pH after 0.0 mL of HCl added.

pH = 

Part 2

Calculate the pH after 35.0 mL of HCl added.

pH = 

Part 3

Calculate the pH after 60.0 mL of HCl added.

pH = 

Part 4

Calculate the pH at the equivalence point.

pH = 

Part 5

Calculate the pH after 300.0 mL of HCl added.

pH = 

Part 6

At what volume of HCl added, does the pH = 10.64?

 mL

Answer 1

First, determine initial moles of base (CH₃NH₂) present. In the following calculations, the reaction being considered is...

CH₃NH₂ + HCl ==> CH₃NH₃⁺ + Cl^- and Cl^- is a spectator ion so need not be considered.

Under appropriate conditions, this forms a buffer of a weak base (CH₃NH₂) and the conjugate acid

(CH₃NH₃⁺)

Initial moles CH₃NH₂ = 100 ml x 1 L / 1000 ml x 0.200 mol / L = 0.0200 mols

Part 1: no HCl added

CH₃NH₂ + H₂O ==> CH₃NH₃⁺ + OH^-

Kb = 4.4x10^-4 = [CH₃NH₃⁺][ OH^-] / [CH₃NH₂] = (x)(x) /0.2 - x (assume x is small and ignore in denom.)

x² = 8.80x10^-5

x = [OH-] = 9.38x10^-3 M (note: this is less than 5% of 0.2 so above assumption is valid)

pOH = -log 9.38x10^-3 = 2.028

pH = 14 - pOH

pH = 11.972

Part 2: after addition of 35.0 ml of 0.100 M HCl:

mols HCl added = 35.0 ml x 1 L / 1000 ml x 0.100 mol / L = 0.00350 moles

CH₃NH₂ + HCl ==> CH₃NH₃⁺

0.02.........0.0035..........0...........Initial

-0.0035...-0.0035.......+0.0035...Change

0.0165.......0................0.0035....Equilibrium

Henderson Hasselbaclh equation: pOH = pKb + log (conj.acid/base) and pKb = -log Kb = 3.36

pOH = 3.36 + log (0.0035/0.0165) = 3.36 - 0.67 = 2.69

pH = 11.31

Part 3: after addition of 60 ml 0.1 M HCl:

mols HCl added = 60 ml x 1 L / 1000 ml x 0.1 mol / L = 0.0060 mol

CH₃NH₂ + HCl ==> CH₃NH₃⁺

0.02.........0.0060..........0.........Initial

-0.0060....-0.0060......+0.0060..Change

0.0140..........0............+0.0060..Equilibrium

pOH = 3.36 + log (0.0060/0.0140) = 3.36 - 0.37 = 2.99

pH = 11.01

Part 4: at the equivalence point:

At the equivalence point, mols HCl = mols CH₃NH₂ so all 0.02 mols initially present have been converted to 0.02 mols CH₃NH₃⁺ in 0.100 L or 0.200 M CH₃NH₃⁺. The reaction is thus...

CH₃NH₃⁺ + H₂O ==> H₃O⁺ + CH₃NH₂ and Ka = 1x10^-14/4.4x10^-4 = 2.27x10^-11

2.27x10^-11 = (x)(x) / 0.2

x = 2.13x10^-6 M = [H₃O⁺]

pH = -log [H₃O⁺]

pH = 5.67

Part 5: after addition of 300 ml of 0.100 M HCl:

At this point, the solution is essentially HCl. 300 ml 0.100 M HCl added to 100 ml = 400 ml

[HCl] = (300 ml)(0.1 M) = (400 ml)(x M) and x = 0.075 M

pH = 1.12

Part 6: volume HCl to achieve pH = 10.64

pH = 10.64 is same as pOH = 3.36 and this is the same as the pKb

When pH = pKb, this is at 1/2 the equivalence point when [CH₃NH₃⁺] = [CH₃NH₂]

This occurs after the addition of 0.01 mols HCl (initial CH₃NH₂ = 0.02 mols)

Volume HCl = 0.01 mols HCl x 1 L / 0.1 mol = 0.1 L = 100 mls HCl