Answer:
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
