# Which of the following is the buffer solution of strong acidic nature?

## a) $\text{HCOOH}$ $+$ ${\text{HCOO}}^{-}$ b) $\text{CH"_3"COOH}$ $+$ ${\text{CH"_3"COO}}^{-}$ c) ${\text{H"_2"C"_2"O}}_{4}$ $+$ ${\text{C"_2"O}}_{4}^{2 -}$ d) ${\text{H"_3"BO}}_{3}$ $+$ ${\text{BO}}_{3}^{2 -}$

Mar 9, 2018

C, oxalic acid, but there is a catch.

#### Explanation:

Oxalic has the lowest pKa at 1.46, so it could make a buffer in the range of pH=1.46 (+/- 1 pH unit). However, Oxalic acid has 2 pKas, so it can also buffer at pH=4.40

If you make a buffer with oxalic acid, it could have strong acidic nature, but if you use the 2nd pKa value, the buffer would be weaker in acidic nature than Formic Acid (A).

Also, you don't make a buffer in a diprotic acid (oxalic acid) with the fully acidic form and the fully deprotonated form. I'll show you what I mean (the question is a bit messed up and wrong)

${H}_{2} {C}_{2} {O}_{4}$ loses 1 Hydrogen at a time.
${H}_{2} {C}_{2} {O}_{4} = {H}^{+} + H {C}_{2} {O}_{4}^{-} 1$
the pKa of this process is 1.46, and so you can make a buffer with
${H}_{2} {C}_{2} {O}_{4} \mathmr{and} H {C}_{2} {O}_{4}^{-} 1$ at about pH = 1.46

THEN at pH = 4.40, you can make a different buffer:
$H {C}_{2} {O}_{4}^{-} 1 = {H}^{+} + {C}_{2} {O}_{4}^{-} 2$
This pKa is 4.40, and this will be the pH area for this buffer.

But you CAN'T make a buffer with ${H}_{2} {C}_{2} {O}_{4} \mathmr{and} {C}_{2} {O}_{4}^{-} 2$, as the problem suggests (it actually is incorrect in that sense for C and D