Question

Calculate how many milliliters of 0.100 M HCl should be added to how many grams of sodium acetate dihydrate (NaOAc 2 H₂O, FM 118.06) to prepare 250.0 mL of 0.100 M buffer, pH 5.00.

Answer 1

we know that

moles = moalrity x volume (L)

so

total moles of acetate = 0.1 x 250 x 10-3

total moles of acetate = 25 x 10-3

now

all the moles of acetate comes from the added sodium acetate dihydrate

so

total moles of acetate = moles of Na0Ac . 2H20 added

so

moles of Na0Ac . 2H20 added = 25 x 10-3

now

we know that

moles = mass / molar mass

so

25 x 10-3 = mass / 118.06

mass = 2.9515

so

2.9515 grams of Na0Ac . 2H20 should be taken

now

let y ml of HCl be added

so

moles of HCl added = 0.1 x y x 10-3 = y x 10-4

now

the reaction is

CH3COONa + HCl ---> CH3COOH + NaCl

moles of CH3COONa reacted = moles of HCL added = y x 10-4

so

moles of CH3COONa left = 25 x 10-3 - y x 10-4

moles of CH3COOH formed = moles of HCl added = y x 10-4

now

we know that

for a buffer

pH = pKa + log [ conjugate base / acid ]

in this case

pH = pKa + log [ CH3COONa / CH3COOH]

also

we know that

pKa for acetic acid is 4.76

so

5 = 4.76 + log ( [ 25 x 10-3 - y x 10-4 ]/ y x 10-4 )

solving we get

y = 91.31

so

91.31 ml of HCl should be added