Question

You have 425 mL of an 0.15 M acetic acid solution. What volume (V) of 1.50 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.85? (The pKa of acetic acid is 4.76.)

Answer 1

the ph equation of a buffer:

pH = pKa + log(Acetat/Acid)

substitute known value

pH = pKa + log(Acetat/Acid)

4.85 = 4.76 + log(Acetate/acid)

Acetate/acid = 10^(4.75-4.76) = 0.9772

Acetate/acid = 0.9772

notice that

initial acid

MV = 0.15*425 = 63.75 mmol of acid

after addition of mmol base = Mbase*Vbase = 1.5*Vbase

then

acid = 63.75 - 1.5*Vbase

acetate = 0 + 1.5*Vbase

and we know that

Acetate/acid = 0.9772

(1.5*Vbase)/(63.75 - 1.5*Vbase) = 0.9772

(1.5*Vbase) = 0.9772*63.75 - 1.5*0.9772*Vbase

(1.5+1.5*0.9772)*Vbase = 0.9772*63.75

Vbase = (0.9772*63.75)/(1.5+1.5*0.9772) = 21.00ml of base needed