The concentration of oxalic acid is 'x' mol `L^(-1)` 40 mL of this solution reacts with 16 mL of 0.05M acidified KMNO4. 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

A

Explanation

Volume of oxalic acid,`V_(1) = 40 mL`
Molarity of `KMnO_(4),M_(2) = 0.05M`
Volume of `KmnO_(4),V_(2) = 16 mL`
Let molarity of oxalic acid, `M_(1)V_(1) = M_(2)V_(2)`
`x xx 40 = 0.05 xx 16`
`:. x= 1/50`
Normality N = Molarity `xx` valency
`N= M xx 2` (`:'` oxaic acid is a dibasic acid, its valency is 2)
` = 1/50 xx 2/4 = 1/25`
The normality of oxalic acid represents the hydorgen ion concentration
`pH = -"log"[H^(+)] = -"log"(1/25) = 1.39`