To make a solution with a pH = 3.80 a student used the following procedure :?

Question

To make a solution with a pH = 3.80 a student used the following procedure :
a certain amount of sodium acetate along with 0.300 moles of acetic acid is added to enough water to make a solution of 1.00 L
How many grams of sodium acetate were added ?

Ka = 1.80 x 10-5

Answer 1

Use the Henderson-Hasselbalch equation (and assumptions) for simplicity:

pH= pKa=log[conj.base]/[acid]

Here conj.base is CH3COO- and acid is CH3COOH

CH3COONa dissociates 100%, thus [CH3COONa]= [CH3COO-] and our equation becomes

pH= pKa +log [CH3COONa]/[CH3COOH]
The ratio of concentrations is equal to the ratio of moles (the volume is simplified in the ratio)
Thus

pH =pKa + log (mole CH3COONa) / (moleCH3COOH) =>

3.80= -log(1.8*10^-5) + log(x/0.3) =>
log(x/0.3) =3.80-4.74 = -0.94 =>
x/0.3 =10^-94 => x= 0.3*10^-0.94 = 0.0344 mole CH3COONa

mole= mass/MW => mass =mole*MW= 0.0344*82 =2.82 g