Suppose 25.00 mL of 0.0500 M CH 3 NH 2 (K b = 4.4 × 10 -4 ) is titrated with 0.0
ID: 895233 • Letter: S
Question
Suppose 25.00 mL of 0.0500 M CH3NH2 (Kb = 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?
Explanation / Answer
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
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
CH3NH3+ H2O CH3NH2 H3O+ I 0.0278 - 0 0 C -x - +x +x E 0.0278 - x - x xRelated Questions
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