Question

A scientist is performing a titration of HClO in bleach samples (10.00 mL aliquot) with 0.0500 M NaOH.  She does the initial titration with a pH meter, and it is found to take 32.1 mL of base to reach the endpoint, which shows a pH of 10.07.  She now wants to prepare a buffered reference solution with the same pH and same ionic strength as the titration solution at the equivalence point (so the titration can be done routinely without the need of a pH meter).  If the scientist decides to make the buffer using 1.00 M NH₃ and 1.00 M HCl, calculate how many mL of each solution should be used to make 100.0 mL of the buffer

Answer 1

mmol of NaOH = MV = 0.05*32.1 = 1.605 mmol of OH-

pH = 10.07 pOH = 14-10.07 = 3.93

then

pOH = pKb + log(NH4+/NH3)

3.93 = 4.75 + log(NH4+/NH3)

find V of each for V = 100 mL

so

mmol of NH3 = MV = 1*Vbase

mmol of NH4+ = 1*Vbase - 1*Vacid

Vbase+ Vacid = 100

3.93 = 4.75 + log((1*Vbase - 1*Vacid / 1*Vbase)

now solve for Vbase

10^(3.93-4.75) = (1*Vbase - 1*Vacid / 1*Vbase)

0.15135*Vbase = Vbase - Vacid

(0.15135-1)*Vbase = -Vacid

Vbase = 1.17834*Vacid

now

Vbase+ Vacid = 100

1.17834*Vacid+ Vacid = 100

2.17834Vacid = 100

Vacid = 100/2.17834 = 45.906 mL of acid

Vbase = 100-45.906 = 54.094mL of NH3