How many grams of sodium acetate were added ?

Question

To make a solution with a pH = 4.00 a student used the following procedure :
a certain amount of sodium acetate along with 0.320 moles of acetic acid is added to enough water to make a solution of 1.00 L

Ka = 1.80 x 10-5

Answer 1

according to the Henderson's eqn,

pH = pKa + log [salt]/[acid]
or,4 = - log (1.80 x 10^-5) + log [salt]/0.32
or, - 0.7447 = log [salt]/0.32
or, [salt]/0.32 = 10^(- 0.7447) = 0.18
or, [salt] = 0.32*0.18 mole/lt = 0.0576 mole/lt

since vol of the soln = 1 lt
no of moles of Na-acetate to be added = 0.0576 mole
= 82 * 0.0576 gm Na- acetate (molecular wt of Na-acetate=82)

=4.7232 gm (ans)

Answer 2

pH = pKa + log [Salt]/[Acid]

or [Salt] = 0.320x10^.18 = .484 mol