Question

Consider a solution made by mixing 500.0 mL of 0.03158 M Na2HAsO4 with 500.0 mL of 0.04202 M NaOH. Complete the mass balance expressions below for Na and arsenate species in the final solution.

[Na+] = ____ M

[HAsO42-] + [__] + [__] + [__] = ____ M

Answer 1

moles of NaOH = 0.04202 * 0.5 = 0.02101 mol

moles of Na2HAsO4 = 0.03158 * 0.5 = 0.01579 mol

moles of Na+ in the Na2HAsO4 = 0.01579 * 2 = 0.03158 mol

so total moles of Na+ = 0.02101 mol + 0.03158 mol = 0.05259 mol

[Na+] = 0.05259 mol/0.5 L = 0.10518 M

the other species will react with sodium hydroxide to form AsSO4^3-
so to find the moles, we subtract the moles of OH- from the moles of Na2HAsO4
0.03158 - 0.02101 = 0.01057 mol
Other possible species are [H3AsO4], [H2AsO4^- ]

[AsSO4^3-] = 0.01057 mol /0.5 L = 0.02114 M

[HAsO32-] + [AsSO4^3-] + [H3AsO4]+[H2AsO4^-] = total moles of Na2HAsO4 / volume in L
                                           = 0.03158 mol/ 0.5 L
                                           = 0.06316 M