art A Problem 14.78 Calculate the pH in 0.190 M acrylic acid Express your answer

Question

art A Problem 14.78 Calculate the pH in 0.190 M acrylic acid Express your answer using two decimal places. Acrylic acid (C3H402) (Figure 1) is used in the manufacture of paints and plastics. The pKa of acrylic acid is 4.25, pH= 2.49 Submit My Answers Give Up Answer Requested Part B Calculate the concentration of H3O+ in 0.190 M acrylic acid Express your answer to two significant figures and include the appropriate units. o Value Units [H30+1= Value Units Submit My Answers Give Up Figure 1 4 of 1 Part C Acidic hydrogen Calculate the concentration C3 H3 in 0.190 M acrylic acid C OH Express your answer to two significant figures and include the appropriate units. Acrylic acid C3H3Value Units

Explanation / Answer

first calculate the Ka value from Pka value

Ka = 10-pKa  = 10-4.25  = 5.62 x 10-5

acrylic acid is a week acid

lets construct the ICE table

C3H4O2 (aq) + H2O (l) ---> C3H4O2- H3O+

I 0.19 0 0

C -x +x +x

E 0.19-x +x +x

Ka = [H3O+] [C3H3O2-] / [C3H4O2]

5.62 x 10-5 = [x] [x] / [0.19-x]

x2 + x 5.62 * 10-5 - 1.07 x 10-5 = 0

solve the quadratic equation

x = 0.0103

from the ICE table [H3O+] = [C3H4O2-] = 0.0103

[C3H4O2] = 0.19 - x = 0.19 - 0.0103 = 0.1797 M

pH = -log[H3O+] = -log[0.0103] = 1.98 or 2

Part A = pH = 1.98 = 2

Part B [H3O+] = 0.0103 M

Part C = [C3H3O2-] = 0.0103 M

Part D = [C3H4O2] = 0.1797 M

Part E = [OH-] =

[H3O+][OH-] = 1.0 x 10-4

[OH-] = 1.0 x 10-4 / 0.0103

[OH-] = 0.0097 M

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