Question

Question #e509f

Answer 1

Here's what I got.

Explanation:

It's very important to realize that you're dealing with an ion here, not with a neutral compound.

More specifically, this is the hydrogen arsenate anion, `"HAsO"_4^(2-)`, which is what you get when you place arsenic acid, `"H"_3"AsO"_4`, in weakly basic conditions.

That said, the first thing to note here is that because you're dealing with a polyatomic ion, you must make sure that the sum of the oxidation number of each atom is equal to the overall charge of the ion.

In this case, you have

`"ON"_ "H" + "ON"_ "As" + 4 xx "ON"_ "O" = 2-`

So, start with what you know. Oxygen has an oxidation state of `-2` in almost all compounds it forms, and this case is no exception.

Similarly, hydrogen has an oxidation state of `+1` in almost all compounds it forms, and this case is no exception.

This means that you can write

`(+1) + "ON"_ "As" + 4 xx (-2) = -2`

Solve to find the oxidation state of arsenic

`"ON"_ "As" = -2 - (+1) - 4 xx (-2)`

`"ON"_ "As" = -3 + 8 = +5`

Therefore, arsenic has an oxidation state of `+5` in the hydrogen arsenate anion, `"HAsO"_4^(2-)`.