With a 27.7 mL sample of a 0.400 M aqueous hypochlorous acid solution titrated with a 0.335 M aqueous potassium hydroxide solution, the pH value is mathematically given as
pH=12.90
Moles of HClO=0.329*21.5
Moles of HClO=7.0735
and
Moles of KOH=0.456*23.3
Moles of KOH=10.6248
Hence
The net mole of KOH=10.6248-7.0735
The net mole of KOH=3.5513
Volume of solution=21.5+23.3
Volume of solution=44.8ml
Generally, the equation for the Concentration of KOH is mathematically given as
CKOH=mole/volume
Therefore
CKOH=3.5513/44.8ml
CKOH=0.07927M
In conclusion,
pOH=-log{OH-}
pOH=-log{0.07927}
pOH=1.10
pH=14-110
pH=12.90
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