Acrylic acid (C3H4O2) (Figure 1) is used in the manufacture of paints and plasti

Question

Acrylic acid (C3H4O2) (Figure 1) is used in the manufacture of paints and plastics. The pKa of acrylic acid is 4.25.

PA:

Calculate the pH in 0.110 M acrylic acid.
Express your answer using two decimal places.

B:

Calculate the concentration of H3O+ in 0.110 M acrylic acid.
Express your answer to two significant figures and include the appropriate units.

[H3O+] =   

PC:Calculate the concentration C3H3O2 in 0.110 M acrylic acid.
Express your answer to two significant figures and include the appropriate units.

PD:Calculate the concentration of HC3H3O2 in 0.110 M acrylic acid.
Express your answer to three significant figures and include the appropriate units.

PE:Calculate the concentration of OH in 0.110 M acrylic acid.
Express your answer to two significant figures and include the appropriate units.

PF:Calculate the percent dissociation in 0.0540 M acrylic acid.
Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

A. pKa of acrylic acid = 4.25

ka = 10^(-pka) = 10^-4.25 = 5.623*10^-5

acrylic acid is a weakacid

pH = 1/2(pka-logC)

c = concentration of acid = 0.11 M

    = 1/2(4.25-log0.11)

   = 2.6

B. [H3O+] = 10^-pH

          = 10^-2.6 = 0.002512 M

C. C3H4O2 + H2O <====> C3H3O2^- + H3O+

[H3O+]= [C3H3O2^-] = 0.002512 M

D. [C3H4O2] = 0.11 - 0.002512 = 0.1075 M

e. [OH-] = 10^-14/[H3O+]

         = 10^-14/(0.002512)

        =3.98*10^-12 M

f. ka = cx^2

   5.623*10^-5 = 0.11x^2

x = 0.0226 = degree of dissociation

percent dissociation = 0.0226*100 = 2.26%

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