# If HSO3- and HC2O4- are both put into water what will the equilibrium equation be?

## We are told to ignore reactions between the ions and water. The only thing I am confused by is that both the substances are amphoteric and both predominately act as acids in water so how do I write the products?

Feb 10, 2018

$H S {O}_{3}^{-} + H {C}_{2} {O}_{4}^{-} r i g h t \le f t h a r p \infty n s S {O}_{3}^{2 -} + {C}_{2} {O}_{4} {H}_{2} \left(a q\right)$

#### Explanation:

...else...

$H S {O}_{3}^{-} + H {C}_{2} {O}_{4}^{-} r i g h t \le f t h a r p \infty n s {H}_{2} S {O}_{3} \left(a q\right) + {C}_{2} {O}_{4}^{2 -}$

Where will the equilibrium lie? I dunno. And unless you quote data for the dissociation constants of oxalic and sulphurous acid you don't know either.

Feb 10, 2018

Here's what I get.

#### Explanation:

Both substances are acids, and both are amphoteric.

When both of them are together, the stronger acid will protonate the weaker acid, which will act as a Brønsted base.

Which is the stronger acid?

We have the following equilibria:

$\text{HSO"_3^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "SO"_3^"2-"; color(white)(ml)K_text(a₂) = 6.2 ×10^"-8}$

$\text{HC"_2"O"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "C"_2"O"_4^"2-"; K_text(a₂) = 6.5 ×10^"-5}$

The hydrogen oxalate ion is the stronger acid. It will protonate the hydrogen sulfite ion.

Write the equation for the equilibrium

overbrace(underbrace("HC"_2"O"_4^"-")_color(red)("stronger acid"))^(color(brown)("Brønsted acid")) + overbrace(underbrace("HSO"_3^"-")_color(red)("weaker acid"))^(color(brown)("Brønsted base")) ⇌ "C"_2"O"_4^"2-" + "H"_2"SO"_3^"-"