Question #e509f

1 Answer
Sep 25, 2016

Answer:

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-)#.