Question

Part A Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.465 M solution of pyridine? (Assume that the temperature is 25 ∘C.) Express the pH numerically to two decimal places. Part B Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water: C6H5COOH⇌C6H5COO−+H+ The pKa of this reaction is 4.2. In a 0.80 M solution of benzoic acid, what percentage of the molecules are ionized? Express your answer to two significant figures and include the appropriate units.

Answer 1

part A)
use:
pKb = -log Kb
8.75-log Kb
Kb = 1.78E-9

C6H5NH2          -----> C6H5NH3+ +   OH-
0.4650                   0         0
0.4650-x                 x         x

Kb = [C6H5NH3+][OH-]/[C6H5NH2]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.78E-9)*0.4650) = 2.88E-5

pOH = -log [OH-] = -log (2.88E-5) = 4.54
PH = 14 - pOH = 14 - 4.54 = 9.46

part B)
use:
pKa = -log Ka
4.2-log Ka
Ka = 6.31E-5
for simplicity lets write weak acid as HA

C6H5COOH   <----->    C6H5COO− + H+
0.8000                 0         0
0.8000-x               x         x

Ka = [H+][C6H5COO-]/[C6H5COOH]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((6.31E-5)*0.8000) = 7.10E-3

% IONIZATION = 7.10E-3 * 100 / 0.80 =0.89 %