Question

How much 5.50M NaOH must be added to 650.0 ml of a buffer that is 0.0215 M acetic acid and 0.0250 M sodium acetate to raise the pH to 5.75? In mL.

Answer 1

Vbase required if M = 5.5 M of NaOh is added to:

V = 650 mL buffer

initially

mmol of acetic acid = MV = 650*0.0215 = 13.975

mmol of sodium acetic= MV = 650*0.0250 = 16.25

then...

after addition of x mmol of NaOH = Mbase*Vbase = 5.5*Vbase

there is a reaction

mmol of acid react with storng base to form mmol of conjguate

mmol of acetic acid = 13.975 -  5.5*Vbase

mmol of conjguate acetate= 16.25 +  5.5*Vbase

substitute in the buffer equation

pH = pKa + log(Acetate/Acetic Acid)

pKa = 4.75 for acetic acid

pH = 5.75 required so

5.75 = 4.75 + log( (16.25 +  5.5*Vbase) / (13.975 -  5.5*Vbase))

10^(5.75-4.75) = (16.25 +  5.5*Vbase) / (13.975 -  5.5*Vbase)

10 *  (13.975 -  5.5*Vbase) = (16.25 +  5.5*Vbase)

139.75 - 55Vbase = 16.25 + 5.5Vbase

60.5Vbase = (139.75-16.25)

Vbase =  (139.75-16.25)/60.5 = 2.041322 mL

Volume of NaOh required = 2.041322 mL