Question

An aqueous solution contains 0.481 M hypochlorous acid. How many mL of 0.247 M potassium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 6.990?

Answer 1

According to Henderson-Hasselbalch equation

Ph = Pka + log(salt/acid)

Ph = 6.990

Pka of hypochlorous acid = 7.53

no.of moles acid = molarity*volume in lt = 0.247*0.125 = 0.031 moles

now

6.990 = 7.53 + log(salt/0.031)

0.288= (salt/0.031)

no.of moles of salt(NaOCl) = 0.009 moles

HOCl + NaOH ............> NaOCl + H2O

1 mole of NaOCl formed from 1 mole of NaOH

so 0.009 moles of NaOCl formed from ................?

   = 0.009*1/1 = 0.009 moles

conc. of NaOH = 0.247

Volume of NaOH = no.of moles/molarity = 0.009/0.247 = 0.036 lt

Volume of NaOH required is 36 ml