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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, $\mathrm pH \approx 14.00 - 4.7 = 9.3$