Question

Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following.

1) the pH at one-half of the equivalence point

2) the pH at the equivalence point

3) the pH after adding 4.0 mL of acid beyond the equivalence point

Answer 1

1) At first we determine the K_a value ,

K_b for CH₃NH₂ = 4.4 x 10^-4 , K_w = 1.0 x 10^-14

K_w = K_a x K_b

=> K_a = (1.0 x 10^-14) / (4.4 x 10^-4) = 2.27 x 10^-11

So, Pk_a = - log Ka = - log ( 2.27 x 10^-11) = - ( 0.36-11) = 10.64

At half equivalence point , the mol ratio of base : acid will be equal.

Therefore it is 1:1. pH = pK_a

pH = pKa + log ([base]/[acid])

=> pH = 10.64 + log (1/1)

=> pH = 10.64

2) Initial mol CH₃NH₂
0.175 mol CH₃NH₂ / L * 0.026 L = 0.00455 mol CH₃NH₂

Volume of acid added to reach the equivalence point is--]

M₁V₁ = M₂V₂

=> 0.175 M * 26 mL = 0.155 M * V₂

=> V₂ = 29.4 mL

At the equivalence point, we have reacted all of the 0.00455 moles that we had initially to form 0.00455 moles of CH₃NH₃⁺.

Total volume = 26.0 mL + 29.4 mL = 55.4 mL = 0.0554 L

[CH₃NH₃⁺] = moles CH₃NH₃⁺ / L = 0.00455 mol / 0.0554 L = 0.0821 M

The CH₃NH₃⁺ solution will be slightly acidic . Hence, it hydrolyzes in water as shown below---

Molarity . . . . . .CH₃NH₃⁺ + H₂O <==> H₃O⁺ + CH₃NH₂
Initial . . . . . . . . . .0.0821 . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
Equilibrium . .0.0821 - x . . . . . . . . . . . . x . . . . . .x

K_a CH₃NH₃⁺ = [H₃O⁺][CH₃NH₂] = x² / (0.0821 - x) = 2.3 x 10^-11

Because K_a is so small, the -x term after 0.0821 can be ignored, so we will have---

x² / 0.0821= 2.3 x 10^-11

=> x² = 0.18 x 10^-11

=> x² = 1.8 x 10^-12

=> x = 1.34x 10^-6 = [H₃O⁺]

pH = -log [H₃O⁺] = -log (1.34 x 10^-6) =-(0.13 -6) = 5.87 (slightly acidic)

3) pH after 4.0 mL of acid is added beyond the equivalence point
After the endpoint, the pH is controlled by the amount of excess strong acid (HBr) added.

mL HBr after endpoint = 4.0
moles excess HBr = M HBr x L HBr = (0.155)(0.0040) = 6.2 x 10^-4 moles HBr

Since HBr is a strong acid,
[HBr] = [H₃O+] = 6.2 x 10^-4 moles

The volume is now 55.4 mL + 4.0 mL = 59.4 mL = 0.0594 L

[H₃O⁺] = moles H₃O⁺ / L
= 6.2 x *10^-4 moles / 0.0594 L
= 1.044 x 10^-2 M

pH = -log [H₃O+]

= -log (1.044 x 10^-2 M)

= - ( 0.0187 - 2) = 1.98