The concentration of oxalic acid is 'x' mol `"litre"^(-1)`. 40 mL of this solution reacts with 16 mL of 0.05 M acidified `KMnO_(4)`. What is the pH of 'x' M oxalic acid solution ?
(Assume that oxalic acid dissociates completely.)

Options

A. 1.3
B. 1.699
C. 1
D. 2

Answer

C

Explanation

`2KMnO_(4)+3H_(2)SO_(4)+5{:(COOH,),(|" ",),(COOH,):}toK_(2)SO_(4)+2MnSO_(4)+8H_(2)O+10CO_(2)`
`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2)){:((COOH,),(|" ",),(COOH,)):}`
`(0.05xx16)/(2)=(x xx40)/(5)`
x=0.05 M
`[H^(+)]=2xx0.05=0.1 M`
`pH= - " log" (H^(+))= - " log" (0.1)=1`