methylamine is a weak base with pkb=3.35 consider the titration of 30.00 mL of .
ID: 974025 • Letter: M
Question
methylamine is a weak base with pkb=3.35 consider the titration of 30.00 mL of .030 M methyl amine with .025 M HCL.
A. writethe appropriate equation for the reaction of methyl amine (CH3NH2) and HCL
b. Calculate K for the reaction in part (a) as you have written it.
c. Calculate the pH of the initial methyl amine solution (0 mL added HCL).
d. Calculate the pH of the solution following addition of 10 mL HCL.
e. Calculate the pH of the solution following addition of 20mL HCL
f Calculate the pH of the solution following addition of 35 mL HCL.
g Calculate the pH of the solution following addition of 36 mL HCL.
h Calculate the pH of the solution following addition 37 mL HCL
i Calculate the pH of the solution following addition 50 mL HCL.
Explanation / Answer
Titration
A. Equation : CH3NH2 + HCl ---> CH3NH3Cl
b. K = [CH3NH3Cl]/[CH3NH2][HCl]
c. Initial pH
CH3NH2 + H2O <==> CH3NH3+ + OH-
ley x amount has reacted
Kb = [CH3NH3+]/[CH3NH2][O-] = 4.46 x 10^-4 = x^2/0.03
x = [OH-] = 3.66 x 10^-3 M
pOH = -log[OH-] = 2.43
pH = 14 - pOH = 11.56
d. when 10 ml HCl added
moles of CH3NH2 = 0.03 M x 30 ml = 0.9 mmol
moles of HCl added = 0.025 M x 10 ml = 0.25 mmol
This is buffer
[CH3NH3+] = 0.25 mmol/40 ml = 0.00625 M
[CH3NH2] = 0.65 mmol/40 ml = 0.01625 M
pH = pKa + log(base/acid)
pKa = 14 - pKb = 14 - 3.35 = 10.65
pH = 10.65 + log(0.01625/0.00625) = 11.065
(e) when 20 ml HCl added
moles of CH3NH2 = 0.03 M x 30 ml = 0.9 mmol
moles of HCl added = 0.025 M x 20 ml = 0.5 mmol
This is buffer
[CH3NH3+] = 0.5 mmol/50 ml = 0.01 M
[CH3NH2] = 0.4 mmol/50 ml = 0.008 M
pH = pKa + log(base/acid)
pH = 10.65 + log(0.008/0.01) = 10.55
(f) after 35 ml HCl added
moles of CH3NH2 = 0.03 M x 30 ml = 0.9 mmol
moles of HCl added = 0.025 M x 35 ml = 0.875 mmol
This is buffer
[CH3NH3+] = 0.875 mmol/65 ml = 0.013 M
[CH3NH2] = 0.025 mmol/65 ml = 3.84 x 10^-4 M
pH = pKa + log(base/acid)
pH = 10.65 + log(3.84 x 10^-4/0.013) = 9.12
(g) after 36 ml HCl added
moles of CH3NH2 = 0.03 M x 30 ml = 0.9 mmol
moles of HCl added = 0.025 M x 36 ml = 0.9 mmol
This is equivalence point
[CH3NH3+] = 0.9 mmol/66 ml = 0.0136 M
CH3NH3+ + H2O <==> CH3NH2 + H3O+
Ka = 1 x 10^-14/4.467 x 10^-4 = x^2/0.0136
x = [H3O+] = 5.52 x 10^-7 M
pH = -log[H3O+] = 6.26
(h) after 37 ml HCl added
excess [HCl] = [H+] = 0.025 M x 1 ml/67 ml = 3.73 x 10^-4 M
pH = -log[H+] = 3.43
(i) after 50 ml HCl added
excess [HCl] = [H+] = 0.025 M x 20 ml/80 ml = 0.00625 M
pH = 2.20
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