Question

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

Answer 1

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⁺.