# 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?

Aug 3, 2018

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

#### Explanation:

We conduct a REDOX titration...

Oxalic acid is OXIDIZED to carbon dioxide...$\left(i\right)$

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

And for every oxidation there is a corresponding reduction...$\left(i i\right)$

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

And we take $5 \times \left(i\right) + 2 \times \left(i i\right)$

$5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} + 10 {e}^{-} \rightarrow 10 C {O}_{2} \left(g\right) \uparrow + 10 {e}^{-} + 2 M {n}^{2 +} + 8 {H}_{2} O$

After cancellation....

$5 {C}_{2} {O}_{4}^{2 -} + {\underbrace{2 M n {O}_{4}^{-}}}_{\text{purple"+16H^+ rarr underbrace(2Mn^(2+)+ 10CO_2(g)uarr ++8H_2O(l))_"colourless}}$

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