Question

Describe (with calculations) how you would prepare 50.0 mL of a 0.100 M pH 6.0 sodium citrate buffer using trisodium citrate and a 1.00 M aqueous solution of HCl.

Answer 1

V = 50 mL of 0.1 M

pH = 6

Na3Citrate

nearest pKa value is 5.40; for 6

so this is the third ionization

A-3 + H+ <-> HA-2

then

pH = pKA3 + log(A-3/HA-2)

6.0 = 5.40 + log(A-3/HA-2)

(A-3/HA-2) = 10^(6-5.4) = 3.981

(A-3/HA-2) = 3.981

and we knot, total concentration

mmol = MV = 50*0.1 = 5 mmol

A-3 + HA-2 = 5

HA-2 = 5- (A-3)

(A-3/HA-2) = 3.981

(A-3) = 3.981*(5- (A-3))

(3.981+1)(A-3) = 3.981*5

A-3 = (3.981*5) / (3.981+1) = 3.993

HA-2 = 5- (A-3) = 5-3.993 = 1.007

then...

total A-3 + HA2- = 5 mmol

mass of Na3Citrate = mmol*MW = 5*258.06 = 1290.3 mg = 1.2903 g

add:

1.2903 g of Na3Citrate ( trisodium citrate )

then,

add

mmol of acid = 1.007

Vacid = mmol/M = (1.007)/(1) = 1.007 mL of HCl

then, mix V up to V = 50 mL

stirr

this will be pH = 6