I really need help with the math of these two problems. I can set up an ICE char
ID: 899075 • Letter: I
Question
I really need help with the math of these two problems. I can set up an ICE chart but I can't figure it out. Thank you!
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 0.225 M solution of pyridine?
Express the pH numerically to two decimal places.
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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 0.50 M solution of benzoic acid, what percentage of the molecules are ionized?
Express your answer to two significant figures and include the appropriate units.
pH =Explanation / Answer
C5H5N(aq) +H2OC5H5NH+(aq) +OH(aq)
pKb = 8.75
If [C5H5N]= 0.225 M then find pH
First, set the equilibrium
Kb = [C5H5NH+][OH-]/[C5H5N]
[C5H5NH+]= [OH-] in the equilibrium (1:1 due to stoichiometry)
Let us asign "x" for each concnetration so
[C5H5NH+]= [OH-] = x
[C5H5N] = 0.225 - x (accounting for the dissociated base)
Note that we need kb
kb = 10^-pkb = 10^-8.75 = 1.778*10^-9
Substitute everythin in
Kb = [C5H5NH+][OH-]/[C5H5N]
1.778*10^-9 = x*x /(0.225 - x)
Solve for x (quadratic equatin)
x = 2*10^-5
which is [OH-] = x = 2*10^-5
we need pH so
pOH = -log(OH) = 4.69
pH = 14-pOH = 14-4.69 = = 9.30
pH = 9.30
PART B
Apply same logic here
C6H5COOHC6H5COO+H+
Ka = 10^-4.2 = 6.3*10^-5
[C6H5COOH] = 0.5 - x
[C6H5COO] = [H+] = x
Ka expression is given as
ka = [H+][C6H5COO-]/[C6H5COOH]
6.3*10^-5 = x*x / (0.5 - x)
substitute all
x = [H+] = 0.00558
pH = -log(H) = -log(0.00558) =2.253
pH = 2.25
But you need % ionization
% ionizatin = H+ / Acid before ionization * 100%
% ionizatin = 0.00558 / 0.5 * 100 = 1.116%
% ionization = 1.12%