A solution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to a volumetric flask and brining in the volume to 1.0 L with water. The final pH is then adjusted and measured to be 5.25. What are the concentrations of acetate and acetic acid in solution? (given pKa of acetic acid is 4.75)

The concentration of

the Acetic Acid is 0.0325M

the Sodium Acetate is 0.1028M

Explanation:

Given

The dissociation of acetic acid is

CH3COOH + CH3COO- + H+ + HA -> A- + H+

We calculate the molarity of sodium acetate by using

Molarity = Moles/Volume

The formula weight of sodium acetate is 82.03 g/mol.

Moles = 11.1g/82.03g/mol

= 0.1353M

Since the volume is 1L, the molarity of sodium acetate will still be 0.1353M

The starting concentration of sodium acetate is 0.1353M

The final concentration of sodium acetate is the starting amount of sodium acetate, minus the amount that was protonated.

Assume that, x = concentration of acetic acid, which is the amount of sodium acetate that was protonated.

From the Henderson-Hasselbalch equation

PH = pKa + log[A-]/[HA]

[A-] = 0.1353 - x

[HA] = x

pKa = 4.75

PH = 5.25

Substitute in these values

5.25 = 4.75 + log(0.1353-x)/x

log(0.1353 - x)/x = 5.25 - 4.75

log(0.1353-x)/x = 0.5

log(0.1353 - x)/x = ½

(0.1353-x)/x = 10^½

(0.1353 - x)/x = 3.1623

0.1353 - x = 3.1623x

0.1353 = x + 3.1623x

0.1353 = 4.1623x

x = 0.1353/4.1623

x = 0.0325M

Therefore [HA] = x = 0.0325M

[A-] = 0.1353 - x = 0.1353 - 0.0325 =

[A-] = 0.1028M

d

Explanation: