22. A-5 % acidity\" bottle of vinegar is approximately 0.375 M in acetic acid (K

Question

22. A-5 % acidity" bottle of vinegar is approximately 0.375 M in acetic acid (K.-1.8 × 10-5 ). a. Determine the approximate pH of this solution b. Some health experts recommend adding 1 tablespoon ( 1 ounce) of vinegar to 1 cup (8 total ounces) of water. Calculate the approximate pH of this solution What is the approximate percent ionization of acetic acid in the solution in part b? Based on the percent ionization was the approximation made to calculate the pH valid? (Answer both questions) c. Imagine that you are concerned about getting heartburn from the vinegar-water so you decide to add enough tablets of Alka-Seltzer© to neutralize the drink. Assume that the Alka-Seltzer@ acts like a strong base. What would be the resulting pH? a.

Explanation / Answer

a)

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.875 M; then

x^2 + (1.8*10^-5)x - 0.875 *(1.8*10^-5) = 0

solve for x

x =0.003959

substitute

[H+] = 0 + 0.003959= 0.003959 M

pH = -log(H+) = -log(0.003959) = 2.402

b)

Dilution law:

C1*V1 = C2*V2

0.875 * 1 = M2*8

M = 0.875 /8

M = 0.109375

from previous work

x^2 + Kax - M*Ka = 0

if M = 0.109375M; then

x^2 + (1.8*10^-5)x - 0.109375*(1.8*10^-5) = 0

x = 0.00139

pH = -log(0.00139) = 2.856

c)

%ionization = [H+]/M*!00% = (10^-pH)/(0.109375) * 100% =  (10^-2.856)/(0.109375) * 100% = 1.273%

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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