Question

A buffer solution is PARTIALLY neutralizing hypochlorous acid with sodium hydroxide; the K_a of HClO is 3.5 x 10^-8 . Use the Henderson-Hasselbalch equation.

a) If 50.0 mL of 1.00M HClO is used to make the buffer, what volume of 0.500M NaOH is needed for the final pH of the buffer to be equal to the pKa of the acid?

b) Assuming no density change upon mixing, what are the concentrations of the wek acid (HClO) and its conjugate base (ClO^-) of this buffer

c) If 20 mL of the buffer solution made in (a) is diluted to a new volume of 200 mL, what will be the new pH? explain why

d) comparing the two buffer solutions made in (a) and (c), which has the higher buffing capacity if equal volumes of each are used? explain why

e) Calculate the pH of a buffer made using 30.0 mL of the acid and 10.0 mL of the base

Answer 1

a) no of mole of HClO = 50*1 = 50 mmol

   pH = pka + log(salt(or)base/acid)

to be pH = pka , concentration of salt = acid

no of mole of NaOH must take = 50 mmol

volume of NaOH required = n/M = 50/0.5 = 100 ml

b) on mixing

concentration of weakacid = 50/(50+100) = 0.33 M

as pH = pka , salt concentration = acid concentration

   concentration of ClO- = 0.33 M

c) in a (ClO-/HClO) = 1 , NO effect of dilution on pH. because on dilution the ratio of (ClO-/HClO) do not change.

d) if pH = pka , buffering capacity is more. and as the concentration of a is more the capacity may be high.