# Question #e509f

Sep 25, 2016

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, ${\text{HAsO}}_{4}^{2 -}$, which is what you get when you place arsenic acid, ${\text{H"_3"AsO}}_{4}$, 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

$\text{ON"_ "H" + "ON"_ "As" + 4 xx "ON"_ "O} = 2 -$

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

$\left(+ 1\right) + \text{ON"_ "As} + 4 \times \left(- 2\right) = - 2$

Solve to find the oxidation state of arsenic

$\text{ON"_ "As} = - 2 - \left(+ 1\right) - 4 \times \left(- 2\right)$

$\text{ON"_ "As} = - 3 + 8 = + 5$

Therefore, arsenic has an oxidation state of $+ 5$ in the hydrogen arsenate anion, ${\text{HAsO}}_{4}^{2 -}$.