Question

Acrylic acid is the conjugate acid of the weak base sodium acrylate, NaC₃H₃O₂. The K_a for acrylic acid is 5.5 X 10^-5. Using this information, complete the following: (3 points)

A.) Write a balanced net ionic equation for the reaction of the acrylate anion with water, which illustrates why this reaction makes the solution basic.

B.) Calculate K_b for this reaction

C.) Find the pH of a solution prepared by dissolving 1.61 g of sodium acrylate in enough water to make 835.0 mL of solution.

Answer 1

C3H3O2^-   + H2O -----------------> HC3H3O2 + OH^-

HC3H3O2 is weak acid so dissociate partially.

OH^-  is strongly basic so resulting solution act as basic.

B) Ka x Kb = Kw

Kb = Kw/Ka = 1.0 x 10^-14 / 5.5 x 10^-5

Kb = 1.82 x 10^-10

C) first calculate the molarity of solution

Molarity = (W/MW) (1000 / volume of solution in mL)

Molarity = (1.61/94.04) (1000 / 835)

Molarity = 0.020 M

pH = 1/2[pKw + pKa + log C]

pKa = - log Ka = - log [5.5 x 10^-5] = 4.26

pH = 1/2 [14 + 4.26 + log 0.020]

pH = 8.28