A)Pyridine is a weak base that is used in the manufacture of pesticides and plas
ID: 1026174 • Letter: A
Question
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+H2OC5H5NH++OH
The pKb of pyridine is 8.75. What is the pH of a 0.330 M solution of pyridine?
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 dissociates in water:
C6H5COOHC6H5COO+H+
The pKa of this reaction is 4.2. In a 0.71 M solution of benzoic acid, what percentage of the molecules are ionized?
Explanation / Answer
A)
use:
pKb = -log Kb
8.75= -log Kb
Kb = 1.778*10^-9
C5H5N dissociates as
C5H5N + H2O -----> C5H5NH+ + OH-
0.33 0 0
0.33-x x x
Kb = [C5H5NH+][OH-]/[C5H5N]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.778*10^-9)*0.33) = 2.422*10^-5
since c is much greater than x, our assumption is correct
so, x = 2.422*10^-5 M
use:
pOH = -log [OH-]
= -log (2.422*10^-5)
= 4.6157
use:
PH = 14 - pOH
= 14 - 4.6157
= 9.3843
Answer: 9.38
B)
use:
pKa = -log Ka
4.2 = -log Ka
Ka = 6.31*10^-5
C6H5COOH dissociates as:
C6H5COOH -----> H+ + C6H5COO-
0.71 0 0
0.71-x x x
Ka = [H+][C6H5COO-]/[C6H5COOH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((6.31*10^-5)*0.71) = 6.693*10^-3
since c is much greater than x, our assumption is correct
so, x = 6.693*10^-3 M
% dissociation = (x*100)/c
= 6.693*10^-3*100/0.71
= 0.9427 %
Answer: 0.943 %
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