Part A Pyridine is a weak base that is used in the manufacture of pesticides and

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: The pKb of pyridine is 8.75. What is the pH of a 0.240 M solution of pyridine? Express the pH numerically to two decimal places. Hints pH- Submit My Answers Give Up 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 dissociates in water: The pKa of this reaction is 4.2. In a 0.72 M solution of benzoic acid, what percentage of the molecules are ionized? Express the percentage numerically using two significant figures. Hints percent ionized- Submit My Answers Give Up

Explanation / Answer

A)

we have below equation to be used:

pKb = -log Kb

8.75-log Kb

log Kb = -8.75

K = 10^(-8.75)

Kb = 1.778*10^-9

Lets write the dissociation equation of C5H5N

C5H5N +H2O -----> C5H5NH+ + OH-

0.24 0 0

0.24-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*0.24) = 2.066*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.066*10^-5 M

So, [OH-] = x = 2.066*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.066*10^-5)

= 4.68

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.68

= 9.32

Answer: 9.32

B)

we have below equation to be used:

pKa = -log Ka

4.2 = -log Ka

log Ka = -4.2

Ka = 10^(-4.2)

Ka = 6.31*10^-5

Lets write the dissociation equation of C6H5COOH

C6H5COOH -----> H+ + C6H5COO-

0.72 0 0

0.72-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*0.72) = 6.74*10^-3

since c is much greater than x, our assumption is correct

so, x = 6.74*10^-3 M

% dissociation = (x*100)/c

= 6.74*10^-3*100/0.72

= 0.936 %

Answer: 0.936 %

Feel free to comment below if you have any doubts or if this answer do not work

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.