A 112.8 mL sample of 0.100 M methylamine (CH3NH2, Kb=3.7x10^-4) is titrated with

Question

A 112.8 mL sample of 0.100 M methylamine (CH3NH2, Kb=3.7x10^-4) is titrated with 0.270 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.... a). 0.0 mL b). 20.9 mL c). 41.8 mL d). 62.7 mL A 112.8 mL sample of 0.100 M methylamine (CH3NH2, Kb=3.7x10^-4) is titrated with 0.270 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.... a). 0.0 mL b). 20.9 mL c). 41.8 mL d). 62.7 mL A 112.8 mL sample of 0.100 M methylamine (CH3NH2, Kb=3.7x10^-4) is titrated with 0.270 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.... a). 0.0 mL b). 20.9 mL c). 41.8 mL d). 62.7 mL

Explanation / Answer

Let us calcualte the pH at all the given volumes of HNO3

a) 0 ml HNO3 ,

The methyl amine hydorlyze as

CH3NH2 + H2O <-------> CH3NH3+ + OH-,

Kb = [CH3NH3+][OH-] / [CH3NH2] = 3.7 X 10-4

3.7 X 10-4 X [CH3NH2] = [CH3NH3+][OH-]

Let the concentration of base hydrolyzed is x

3.7 X 10-4 X [0.1 - x] = x^2

x = 6.082 X 10-3 = [OH-]

pOH = 2.22

pH = 14 - pOH = 14- 2.22 = 11.78

b) 20.9 mLHNO3 added,

Moles of HNO3 added = Molarity X volume = 0.270 X 20.9 mL = 5.643 mmoles

Moles of CH3NH2 present = Molarity X volume = 0.1 X 112.8 = 11.28 millimoles

Millimoles of CH3NH3+ formed = 5.643 mmoles

Millimoles of CH3NH2 left = 11.28 - 5.643 = 5.637 millimoles

The solution will act as buffer, the pH of buffer is calcualted using Hendersen Hassalbalch equation

pOH = pkb + log[CH3NH3+]/[CH3NH2]

pkb = -log(3.7 x10-4) = 3.43,

pOH = 3.43 + log [5.643 / 5.637] = 3.43 (approx)

pH = 14 - 3.43 = 10.57

c)41.8 mL of HNO3 added

Moles of HNO3 added = Molarity X volume = 0.270 X 41.8 mL = 11.28 mmoles

Moles of CH3NH2 present = Molarity X volume = 0.1 X 112.8 = 11.28 millimoles

Moles of CH3NH3+ formed = 11.28 millimoles

[CH3NH3+]= Millimoles / total volume in mL= 11.28 / 154.6 = 0.073 M

Moles of CH3NH2 left = 0

The salt will hydrolyze as

CH3NH3+ + H2O ,<-----> CH3NH2 + H3O+ ,

Ka = Kw/Kb = 10-14/(3.7 x10-4) = 2.7 x10-11,

Let x hydrolyzed

Ka = 2.7 x10-11 =x^2 / (0.073 -x)

x<<1 so can be ignored in denominator

x2 = 1.971 X 10^-12

x = 1.404 X 10^-6 = [H3O+]

pH = -log[H3O+] = 5.85

d) 62.7 ml HNO3 is added

Moles of HNO3 added = Molarity X volume = 0.270 X 62.7 mL = 16.929 mmoles

Moles of CH3NH2 present = Molarity X volume = 0.1 X 112.8 = 11.28 millimoles

Excess of HNO3 = 5.65 millmoles

[HNO3] = 5.65 millimoles / 112.8 + 62.7 mL = 0.0322 M

pH = -log(0.0322) = 1.49

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.