What is the iron (iii) ion concentration in 0.4309 m iron (iii) sulfate solution? select one: a. 2.780 m b. 0.4309 m c. 0.8619 m d. 1.293 m

c. [Fe³⁺] = 0.8618 mol·L⁻¹.

              Fe₂(SO₄)₃(s) ⟶ 2Fe³⁺(aq)    +   3SO₄²⁻(aq)

I/mol:        0.4309                   0                    0

C/mol:     -0.4309          +2×0.4309       +3×0.4039

F/mol:       0                          0.8618              1.293

Each mole of iron(III) sulfate produces two moles of Fe³⁺.

since electrons are negative the answer is neagtaive