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pH Calculation for Weak  Bases

 

WEAK BASES

Basic Information

1) Weak bases are less than 100% ionized in solution.

2) Ammonia (formula = NH3) is the most common weak base example used by instructors.

3) The acquire H+ in aqueous solutions.

4) They have a small value for Kb

5) weak bases acquire hydrogen ions from water leaving hydroxide

ex. \mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_6N^+ (aq) + OH^- (aq)}

Name
Formula
Kb
Ammonia
NH3
1.8 x 10-5
Methylamine
CH3NH2
4.38 x 10-4
Ethylamine
C2H5NH2
5.6 x 10-4
Diethylamine
(C2H5)2NH
1.3 x 10-3
Aniline
C6H5NH2
3.8 x 10-10
Pyridine
C5H5N
1.7 x 10-9

 

A typical pH problem

Calculate the pH of a 0.20 M aqueous solution of pyridine, C5H5N.

The Kb for C5H5N is 1.8 x 10-9

First, write the proton transfer equilibrium:

\mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_6N^+ (aq) + OH^- (aq)}

K_b=\mathrm{[C_5H_6N^+][OH^-]\over [C_5H_5N]}

The equilibrium table, with all concentrations in moles per liter, is

  C5H5N C5H6N+ OH-
initial normality .20 0 0
change in normality -x +x +x
equilibrium normality .20 -x x x


Substitute the equilibrium molarities into the basicity constant K_b=\mathrm {1.8 \times 10^{-9}} = {x \times x \over .20-x}
Assume that x << .20.

 

\mathrm {1.8 \times 10^{-9}} \approx {x^2 \over .20}
Solve for x. \mathrm x \approx \sqrt{.20 \times (1.8 \times 10^{-9})} = 1.9 \times 10^{-5}
Check the assumption that x << .20

(or Kb x [C5H5N]<< .001 (disregard x in the denominator))

\mathrm 1.9 \times 10^{-5} \ll .20; so the approximation is valid
Find pOH from pOH = -log [OH-] with [OH-]=x \mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7
From pH = pKw - pOH, \mathrm pH \approx 14.00 - 4.7 = 9.3

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