Question

A buffer solution is made by mixing 0.100 liter each of 0.400 M acetic acid and 0.200 M sodium acetate. For acetic acid, Ka=1.8 x 10^-5

a. What is the pH of the buffer?

b. Assuming no change in volume, what is the pH of the solution after the addition of 0.0100 moles of KOH molecules are added? (KOH is a strong base in water solution)

Answer 1

a)

the equation of acidic buffer, via henderson hasselbach

pH = pKa + log(A-/HA)

so..

pKa = -log(Ka) = -log(1.8*10^-5) =4.74472

so

pH = 4.74472 + log(0.2/0.4) =4.44369

b)

if we add

mol of KOH = 0.01

then

find

mol of acid = MV = 0.1*0.4 = 0.04 mol of acid

mol of acetate = MV = 0.2*0.1 = 0.02 mol of acetate

after addition of 0.01 mol of base..

acid reacts = 0.04-0.01 = 0.03

the conjguate increases = 0.02 + 0.1 = 0.03

so

pH = pKa + log(A-/HA)

pH = 4.75 + log(1)

pH = 4.75