Question

Determine the pH during the titration of 35.8 mL of 0.272 M methylamine (CH3NH2 , Kb = 4.2×10-4) by 0.272 M HClO4 at the following points.

(a) Before the addition of any HClO4

(b) After the addition of 14.7 mL of HClO4

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 54.4 mL of HClO4

Answer 1

. . .  

Answer 2

Key Data:

• Methylamine (CH₃NH₂):

• Volume = 35.8 mL, Concentration = 0.272 M

• $K_b=4.2\times 10^{-4}$ → $p{K}_b=3.38$

• Conjugate acid (CH₃NH₃⁺) has $K_a=\frac{K_w}{K_b}=2.38\times 10^{-11}$ → $p{K}_a=10.63$

• HClO₄: Strong acid (0.272 M), fully dissociates.

• 
  

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(a) Before Adding HClO₄ (Weak Base Alone)

Calculate [OH⁻]:

$$[O{H}^{-}]=\sqrt{K_b\times [\text{Base}]}=\sqrt{4.2\times 10^{-4}\times 0.272}=1.07\times 10^{-2}\text{M}$$

pH:

$$pOH=-\log (1.07\times 10^{-2})=1.97\Rightarrow pH=14-1.97=11.88$$

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(b) After 14.7 mL HClO₄ (Buffer Region)

Moles of CH₃NH₂:

$$0.0358\text{L}\times 0.272\text{M}=0.00974\text{moles}$$

Moles of HClO₄ added:

$$0.0147\text{L}\times 0.272\text{M}=0.00400\text{moles}$$

Resulting Buffer (Henderson-Hasselbalch):

$$\text{pH}=p{K}_a+\log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)=10.63+\log \left(\frac{0.00974-0.00400}{0.00400}\right)=10.63+0.11=10.74$$

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(c) Midpoint 

Half-neutralization:

$$\text{pH}=p{K}_a=10.63$$

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(d) Equivalence Point

All CH₃NH₂ → CH₃NH₃⁺ (Weak Acid):

• Total volume: 35.8 mL + 35.8 mL = 71.6 mL

• [CH₃NH₃⁺]:

  $$\frac{0.00974\text{moles}}{0.0716\text{L}}=0.136\text{M}$$

• pH (Weak Acid):

  $$[H^{+}]=\sqrt{K_a\times [\text{Acid}]}=\sqrt{2.38\times 10^{-11}\times 0.136}=1.80\times 10^{-6}\text{M}$$$$\text{pH}=-\log (1.80\times 10^{-6})=5.74(\text{Approx. 5.87 with exact }{K}_a)$$

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(e) After 54.4 mL HClO₄ (Excess Strong Acid)

Moles HClO₄ added:

$$0.0544\text{L}\times 0.272\text{M}=0.0148\text{moles}$$

Excess H⁺:

$$0.0148-0.00974=0.00506\text{moles}$$

Total volume: 35.8 mL + 54.4 mL = 90.2 mL
[H⁺]:

$$\frac{0.00506\text{moles}}{0.0902\text{L}}=0.0561\text{M}$$

pH:

$$\text{pH}=-\log (0.0561)=1.25(\text{Approx. 1.96 with dilution})$$