Part A Pyridine is a weak base that is used in the manufacture of pesticides and
ID: 1048206 • Letter: P
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+H2OC5H5NH++OH
The pKb of pyridine is 8.75. What is the pH of a 0.220 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:
C6H5COOHC6H5COO+H+
The pKa of this reaction is 4.2. In a 0.62 M solution of benzoic acid, what percentage of the molecules are ionized?
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"
there are free OH- ions so, expect a basic pH
B + H2O <-> HB+ + OH-
The equilibirum Kb:
Kb = [HB+][OH-]/[B]
let "x" be OH- in solution
in equilibrium due to sotichiometry:
[HB+]= x= [OH-]
Acocunt for the dissolved base in solution vs. not in solution:
[B] = 0.05-x
Substitute in Kb
Kb = 10^-pKb = 10^-8.75 =1.77827*10^-9
Kb = [HB+][OH-]/[B]
1.77827*10^-9= x*x/(0.22-x)
Solve for x, using quadratic equation
x = OH- = 1.978*10^-5
pOH = -log(OH-) = -log( 1.978*10^-5) = 4.703
pH = 14-pOH = 14-4.703
pH = 9.30
B)
By definition, in equilibrium:
Ka = [H+][A-]/[HA]
Then
Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)
Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)
Substitute
Ka = (x*x)/(M-x)
Ka = 10^-pKa = 10^-4.2
6.309*10^-5 = x*x/(0.62-x)
solve for x (quadratic equation)
x = [H+] = 0.006223 M
[H+] = [A-] = x = 0.006223
The ionization:
% ionization = [H+] / M * 100% = 0.006223 /(0.62) * 100% = 1.003709 %
2 sig fig --< 1.0 %
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