m 2 Part 2 Extra Cred sExam%202A%20p 2%20F 1-1 7%20xtra%20( 1 ).pdf 22. A \"5 %
ID: 580560 • Letter: M
Question
m 2 Part 2 Extra Cred sExam%202A%20p 2%20F 1-1 7%20xtra%20( 1 ).pdf 22. A "5 % acidity" bottle of vinegar is approximately 0.875 M in acetic acid ( K, = 1.8x 10" a. Determine the approximate pH of this solution )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. d. Imagine that you are concerned about getting heartburn from the vinegar-water so you decide to add enough tablets of Alka-SeitzerO to neutralize the drink, Assume that the Alka-Seltzer base. What would be the resulting pH? acts like a strongExplanation / Answer
Q22
a)
pH approx:
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 ) = 2.40
b)
dilution --> 1/8*0.875 = 0.109375 M
x^2 + Kax - M*Ka = 0
if M =0.109375 M; then
x^2 + (1.8*10^-5)x - 0.109375 *(1.8*10^-5) = 0
solve for x
x =0.001393
substitute
[H+] = 0 + 0.001393= 0.001393M
pH = -log(H+) = -log(0.001393) = 2.86
c)
ionization of b:
%ion = [H+]/HA * 100% = 0.001393/0.109375 *100 = 1.2736 %
d)
if neutralzied.
[Acetate ion] = 0.8 M approx
then
pH = will be basic, since
A- + H2O <-> HA + OH-
form hydrolysis
pH wil be approx = 8-9
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