pH Calculations of Weak Bases

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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	K_b
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 K_b 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 = pK_w - pOH,	$\mathrm pH \approx 14.00 - 4.7 = 9.3$

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