How to balance this Oxidation Reduction Reaction (redox)?

Question

As2O3(s) + NO3-(aq) > H3AsO4 (aq) + N2O3 (aq)

By using the half reaction method, someone please walk me step by step through this. Also, does H3AsO4 ionize because the main equation say that it is aqueous? I know that it is a weak acid, but the (aq) makes me doubt myself.

Answer 1

As2O3    > 2 H3AsO4 & 4 e- lost .......(2 As+3 each --> 2 As+5 each)

2 (NO3)-1  & & 4 e- taken  >  N2O3 ... (2 N+5 each --> 2 N+3 each)

since they are balanced in electrons taken & lost, combine them:

1As2O3(s) + 2NO3-(aq) > 2H3AsO4 (aq) + N2O3 (aq)

the imbalances are:

zero H's <-> 6 H's

9 oxygens <-> 11 oxygens

-2 ion charge <-> no charge

I  will be able to later balance the need on the left for H's & O's with neutral H2O later, so I am going to balance the charges first, by adding 2 negative hydroxides to the right side:

As2O3(s) + 2NO3-(aq) > 2H3AsO4 (aq) + N2O3 (aq) & 2 OH-

with the charges now balanced, I notice that I now have an excess of 8 Hydrogens & 4 oxygens on the right , which can be balanced by adding 4 H2O's to the left:

1As2O3(s) + 2NO3-(aq) & 4 H2O > 2H3AsO4 (aq) + 1N2O3 (aq) & 2 OH-

that's your answer

===================================

as an edit... i forgot to answer your question:"does H3AsO4 ionize "

yes it will somewhat ionize

H3AsO4 --> 1 H+  & (H2AsO4)-1  ... Ka1 = 5.6 e-3

further ionization gets even less, as the

Ka2 = 1.0e-7    

Ka3 = 3.0e-12

however we won't need to bother with this , when we re-dox balance