Question

How do I find the First half equivalence point pH and the second half equivalence point pH of kmno4+h2c2o4+h2so4 titration ?

First half equivalence point pH=(pKa1+pKa2)/2 usually, But on the above redox reaction there are two acids H2SO4 and H2C204. So Is the pH=(pKa1+pKa2)/2 still valid for this reaction or is there anyother way to find the pH of the above reaction?

Answer 1

I think you might need to repropose this question...

Explanation:

We conduct a REDOX titration...

Oxalic acid is OXIDIZED to carbon dioxide...`(i)`

`stackrel(""^(+III))C_2O_4^(2-) rarr 2CO_2(g)uarr +2e^(-)`

And for every oxidation there is a corresponding reduction...`(ii)`

`stackrel(""^(+VII))MnO_4^(-)+8H^+ + 5e^(-) rarr Mn^(2+)+4H_2O`

And we take `5xx(i)+2xx(ii)`

`5C_2O_4^(2-) +2MnO_4^(-)+16H^+ + 10e^(-)rarr 10CO_2(g)uarr +10e^(-)+2Mn^(2+)+8H_2O`

After cancellation....

`5C_2O_4^(2-) +underbrace(2MnO_4^(-))_"purple"+16H^+ rarr underbrace(2Mn^(2+)+ 10CO_2(g)uarr ++8H_2O(l))_"colourless"`

And thus a redox titration rather than an acid-base titration is proposed. The redox reaction is performed in ACIDIC conditions...