Problem 14.78 Acrylic acid (C3H4O2) (Figure 1) is used in the manufacture of pai
ID: 1017629 • Letter: P
Question
Problem 14.78 Acrylic acid (C3H4O2) (Figure 1) is used in the manufacture of paints and plastics. The pKa of acrylic acid is 4.25. Part A Calculate the pH in 0.120 M acrylic acid. Express your answer using two decimal places.
Part B
Calculate the concentration of H3O+ in 0.120 M acrylic acid.
Express your answer to two significant figures and include the appropriate units.
Part C
Calculate the concentration C3H3O2 in 0.120 M acrylic acid.
Part D
Calculate the concentration of HC3H3O2 in 0.120 M acrylic acid.
Part E
Calculate the concentration of OH in 0.120 M acrylic acid.
Explanation / Answer
Problem 14.78 Acrylic acid (C3H4O2) (Figure 1) is used in the manufacture of paints and plastics. The pKa of acrylic acid is 4.25.
Part A Calculate the pH in 0.120 M acrylic acid. Express your answer using two decimal places.
The pKa of acrylic acid is 4.25. now calculate the value of Ka:
Pka=4.25
ka=10^-pka
=5.62*10^-5
Assume that the acrylic acid is HA. Now make a ICE table:
HA <---> H+ + A-
I 0..120.......0.............0
C -X..........+X...........+X
E 0.120-x.......x...........x
Ka = [H+][A-]/[HA]
5.62*10^-5=x^2/0.120-x
6.7*10^-6 -5.62*10^-5X = X^2
X^2 +5.62*10^-5X- 6.74*10^-6= 0
this a quadric function to solve this we will get:
x= 0.00256=[H+]
pH=-log[H+]=-log[0.00256]
=2.59
Part B
Calculate the concentration of H3O+ in 0.120 M acrylic acid.
x= 0.0026 M=[H+]
Express your answer to two significant figures and include the appropriate units.
Part C
Calculate the concentration C3H3O2 in 0.120 M acrylic acid.
C3H3O21= x= 0.0026 M
Part D
Calculate the concentration of HC3H3O2 in 0.120 M acrylic acid.
[HC3H3O2] =[0.12-0.0026]= [0.1174]M
Part E
Calculate the concentration of OH in 0.120 M acrylic acid.
Here pH = 2.59 then pOH = 14-2.59 = 11.41
pOH = - log [OH]-
[OH-]= 10^-11.41
= 3.89*10^-12 M