Question

How many mL of 0.200 M NaOH must be added to 31.35 mL of a 0.0500 F solution of fumaric acid (trans-butenedioic acid) to make a buffer of pH 4.50?

Answer 1

mmol of base required = MV = 0.2*V

mmol of acid = MV = 31.35*0.50 = 15.675

then

pH = pKa + log(fumarate/fumaric acid)

4.50= 4.44 + log(F-2/ HF-)

this implies:

1st neutralizatino of H2F to HF-

mmol of of base for 1st neutralization = 15.675

now..

4.50= 4.44 + log(F-2/ HF-)

4.50= 4.44 + log( x / (15.675 -x))

solve for x

10^(4.50-4.44) = x / (15.675 -x)

1/1.148x = 15.675 -x

x = (15.675 )/(1/1.148+1) = 8.3775

then

total mmol = 15.675 +8.3775 = 24.0525 mmol of base

V = mmol/M = 24.0525/0.2 = 120.2625 mL