Question

56.0 mL of 2.50 M Fe(NO₃)₂ is combined with 25.0 mL of 0.0525 M Na₂CO₃.

K_sp of FeCO₃ is 2.1 × 10^-11.

What mass of FeCO₃(s) will be produced? in grams

What is the [Fe²⁺] in the solution final solution? in M

What is the [CO₃^2-] in the solution final solution? in M

Answer 1

Given that 56 mL of 2.5 M Fe(NO₃)₂

                = 56 mL x 2.5 mole / 1000 mL

                = 0.14 moles

Given that 25 mL of 0.0525 M Na₂CO₃

                = 25 mL x 0.0525 mole / 1000 mL

                = 0.0013125 moles

Consider the equation

                                                  Fe(NO₃)₂         +         Na₂CO₃       FeCO₃      +      2 NaNO₃

Initial moles                                   0.14            0.0013125    0         0

after reaction                  (0.14 - 0.0013125)    0              0.0013125    0.002625

                                           = 0.1387

mass of FeCO₃ produced = 0.0013125 x molecular mass of FeCO₃

                                     = 0.0013125 x 115.85 = 0.152 g

Total volume after reaction = 56 + 25 = 81 mL = 0.081 L

Therefore concentration of Fe(NO₃)₂ after reaction = 0.1387 mole / 0.081 L

                                                                         = 1.712 M

Therefore concentration of Fe²⁺ in the resultant solution = 1.712 M

This is due to the excess of Fe(NO₃)₂

Given that the solubility product of FeCO₃ is 2.1 x 10^-11

                                      [Fe] [CO₃] = 2.1 x 10^-11

Consider the equation FeCO₃    Fe²⁺ + CO₃^2-

                                                      [x]                   [ x]

Where [x] is the concentration of Fe₂₊ CO₃^2- and in the solution

But already we have the concentration of Fe2+ due to the excess of Fe(NO₃)₂ = 1.712 M

Therefore the total conc of Fe²⁺ = [Fe²⁺] = [1.712 + x]

                          and         [CO₃^2-] = [x]

K_sp = 2.1 x 10^-11 = [1.712 + x] * [x]

                        = 1.712x        [since x is small compared to 1.712 it is neglected in [1.712 + x]

therefore x = 2.1 x 10^-11 / 1.712 = 1.23 x 10^-11 M

Thus

The mass of FeCO₃ produced = 0.152 g

[Fe²⁺] in the final solution = 1.712 M

[CO₃^2-] in the final solution = 1.23 x 10^-11 M