Question

What mass, in grams , of sodium acetate (Na=CH3CO2-) must be added to 500 ml of 0.50M acetic acid (CH3CO2H, Ka= 1.8*10^-5)?

Answer 1

Assuming we have to find the amount of sodium actetate required to be added to the acetic acid to say, form a pH of 5,we proceed in the following way,

volume of acetic acid, CH3COOH = 500.0 mL = 0.5000 L
[CH3COOH] = 0.500 M

[CH3COO^-] = ? to have a buffer solution with pH = 5.00

We can use the Henderson-Hasselbalch equation to solve this problem.
pH = pKa + log{[CH3COO^-]/[CH3COOH]}
pKa [CH3COOH] = 4.75

5.00 = 4.75 + log[CH3COO^-] - log[CH3COOH]
log[CH3COOH] = log(0.500) = -0.301
0.25 = log[CH3COO^-] - (-0.301)
log[CH3COO^-] = 0.25 - 0.301 = -0.051
[CH3COO^-] = 10^(-0.45) = 0.889 M

Now, mass needed.
moles NaCH3COO = (0.889 mol/L)(0.5000 L) = 1.778 moles
molar mass NaCH3COO = 82.03 g/mol
mass NaCH3COO = (1.778 mole)(82.03 g)/(1 mole) = 145.87 g sodium acetate must be added.