Question

If a polymer is prepared from 3.603 g of acrylic acid, and 2.25 mL of 8.0 M NaOH, what will the mass of the "dry" polymer be after all of the water is removed? Assume all reactions go to completion. Hint: You may want to construct a BCA table to determine the outcome of the neutralization step.

Answer 1

moles of acrylic acid = 3.603 g * (1mol / 72 g) = 0.05 moles

moles of NaOH = 0.00225 L * 8 M = 0.018 moles

Equilibrium of acrylic acid

Acrylic Acid + NaOH <-> Sodium Acrylate + H₂O

We make our BCA table:

	Acrylic Acid	+	NaOH	<->	Sodium Acrylate	+	H2O
B	0.05		0.018		0		0
C	-x		-x		+x		+x
A	0.05-x		0.018-x		x		x

Ka for acrylic acid is 5.6 x 10^-5

To get moles of polymer in equilibrium:

5.6 x 10^-5 = x² / (0.05-x) * (0.018-x)

x = 0.000222 moles

So as in equilibrium 0.000222 moles are produced, we get the mass:

0.000222 * 94 g/mol = 0.020 grams of sodium acrylate