Consider a solution made by mixing 500.0 mL of 0.04182 M Na2HAsO4 with 500.0 mL of 0.03645 M NaOH. Complete the mass balance expressions below for Na and arsenate species in the final solution.
There are 4 arsenate species, but I only got three of them. I am unsure of what the last one could be.
The ones I have are HAsO4^2-, H3AsO4, AsO4^3-. I thought the last one would be H2AsO4^- but it wasn't correct.
Na2HAsO4
neutral compound
What about H2AsO4^-
To determine the last arsenate species, we can start by writing the balanced equation for the reaction occurring between Na2HAsO4 and NaOH:
Na2HAsO4 + NaOH ->
Based on the reactants, we can identify the products:
Na2HAsO4 + NaOH -> Na3AsO4 + H2O
So, the fourth arsenate species is Na3AsO4. It is formed as a product when Na2HAsO4 reacts with NaOH.
The complete mass balance expressions for Na and arsenate species in the final solution are:
Na:
Number of Na atoms in Na2HAsO4 = 2
Number of Na atoms in NaOH = 1
Number of Na atoms in Na3AsO4 = 3
Therefore, the mass balance expression for Na would be:
2 * [Na2HAsO4] + 1 * [NaOH] = 3 * [Na3AsO4]
Arsenate species:
Number of HAsO4^2- ions in Na2HAsO4 = 1
Number of H3AsO4 molecules = 1
Number of AsO4^3- ions in Na3AsO4 = 1
Therefore, the mass balance expression for arsenate species would be:
[Na2HAsO4] + [H3AsO4] = [Na3AsO4]
Make sure to use the given concentrations (0.04182 M and 0.03645 M) and volumes (500.0 mL and 500.0 mL) to calculate the actual values of the concentrations in the mass balance expressions.