Question

Suppose 25.00 mL of 0.0500 M CH₃NH₂ (K_b = 4.4 × 10^-4) is titrated with 0.0625 M HCl.

How many millimoles of CH₃NH₂ are present initially?

How many millimoles of HCl are required to reach the equivalence point?

What is the equivalence point (in mL)?

What is the total volume (in mL) of analyte solution at the equivalence point?

At the equivalence point, which are the principal species present (excluding Cl^-)?

Select one:

a. H₂O and CH₃NH₂

b. H₂O and CH₃NH₃⁺

c. H₂O, H₃O⁺ and CH₃NH₃⁺

d. H₂O, H₃O⁺ and CH₃NH₂

e. H₂O, H₃O⁺, CH₃NH₂, and CH₃NH₃⁺

What is the pH at the equivalence point?

Answer 1

We are using 25 mL of 0.05 M CH3NH2

millimols of CH3NH2 = 25 mL * 1 L/1000 mL * 0.05 mol/L * 1000 millimol/ 1 mol = 1.25 millimol

millimols of CH3NH2 present initially are 1.25

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The chemical equation for the reaction between CH3NH2 and HCl is given as

CH3NH2 + HCl ------------> CH3NH3Cl

From the above equation we can say that the stoichiometry for CH3NH2 and HCl is 1:1

Therefore millimols of HCl required to reach the equivalence point is

1.25 millimol of CH3NH2 * 1 HCl/1 CH3NH2 = 1.25 millimol of HCl

millimols of HCl required to reach equivalence point are 1.25

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The volume of HCl at equivalence point is

1.25 millimol HCl * 1 mol/1000 millimols * 1 L/ 0.0625 mol * 1000 mL/ 1 L = 20 mL

Equivalence point is at 20 mL HCl

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We started with 25 mL of CH3NH2 solution and we added 20 mL of HCl to reach the equivalence point

Total volume therefore is 25 mL + 20 mL = 45 mL

Total volume of analyte solution at equivalence point is 45 mL

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The chemical equation is written as

CH3NH2 + HCl ---------> CH3NH3Cl

Net ionic equation is

CH3NH2 + H+ -------> CH3NH3^+

At equilibrium, all the reactants are consumed and we have only the product CH3NH3^+

CH3NH3^+ has a tendency to hydrolyze in presence of water as follows

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

So the principal species present at equilibrium are  H₂O, H₃O⁺, CH₃NH₂, and CH₃NH₃⁺

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The reaction that occurs at equivalence point is

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

Let us calculate Concentration of CH3NH3+ at equivalence point

mol of CH3NH3+ at Eq. point = 1.25 mmol = 1.25 x 10^-3 mol

volume of Eq. point = 45 mL = 0.045 L

Concentration of CH3NH3+ at Eq. point = 1.25 x 10^-3 mol/ 0.045 L = 0.0278 M

Let us draw ICE table for this reaction

	CH3NH3+	H2O	CH3NH2	H3O+
I	0.0278	-	0	0
C	-x	-	+x	+x
E	0.0278 - x	-	x	x

The above reaction is acid dissociation, therefore we need Ka

Ka = Kw/Kb

Ka = 1 x 10^-14 / 4.4 x 10^-4

Ka = 2.27 x 10^-11

Ka = [ H3O+] [ CH3NH2] / [CH3NH3+]

2.27 x 10^-11 = (x) (x) / 0.0278 - x ( Use approximation method and neglect x on the bottom)

2.27 x 10^-11 = x^2/0.278

x^2 = 6.32 x 10^-13

x = 7.95 x 10^-7

But x = [H3O+]

pH = -log [H3O+]

pH = - log ( 7.95 x 10^-7)

pH = 6.10

pH at equivalence point is 6.10