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Chapter 12: Problem 196

A student dilutes \(75.0 \mathrm{~mL}\) of a \(2.00 \mathrm{M}\) solution of iron(III) nitrate with sufficient water to prepare \(2.00 \mathrm{~L}\) of solution. (a) What is the molar concentration of iron(III) nitrate in the diluted solution? Once in solution, the iron(III) nitrate exists not intact but rather as dissociated ions. What are the molar concentrations (b) of \(\mathrm{Fe}^{3+}(a q)\) in the diluted solution and (c) of \(\mathrm{NO}^{3-}(a q)\) in the diluted solution?

Short Answer

Expert verified
The molar concentrations in the diluted solution are: (a) Iron(III) nitrate: 0.075 M, (b) Fe³⁺: 0.075 M, and (c) NO₃⁻: 0.225 M.

Step by step solution

01

Use the dilution formula to find the molar concentration of the diluted solution.

The dilution formula is given by: \[ M_1V_1 = M_2V_2 \] Where: - \(M_1\) is the initial concentration of the solution. - \(V_1\) is the initial volume of the solution. - \(M_2\) is the final concentration of the solution after dilution. - \(V_2\) is the final volume of the solution after dilution. Given that the initial concentration is 2.00 M and the initial volume is 75.0 mL, we can plug these values into the dilution formula, keeping in mind that the final volume is 2.00 L. We want to find \(M_2\), the molar concentration of iron(III) nitrate after dilution, so let's rearrange the equation: \[ M_2 = \frac{M_1V_1}{V_2} \] Then plug in the known values and solve for \(M_2\).
02

Calculate the molar concentration of iron(III) nitrate after dilution.

Substituting the values: \[ M_2 = \frac{(2.00 \mathrm{~M})(75.0 \mathrm{~mL})}{2.00 \mathrm{~L}} \] Convert 75.0 mL to L: \[ 75.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.075 \mathrm{~L} \] Now, calculate \(M_2\): \[ M_2 = \frac{(2.00 \mathrm{~M})(0.075 \mathrm{~L})}{2.00 \mathrm{~L}} = 0.075 \mathrm{~M} \] The molar concentration of iron(III) nitrate in the diluted solution is 0.075 M.
03

Write the dissociation equation for iron(III) nitrate in water.

To find the molar concentrations of Fe³⁺ and NO₃⁻, we must first understand how iron(III) nitrate dissociates in water: \[ \mathrm{Fe(NO_3)_3} (aq) \rightarrow \mathrm{Fe^{3+}}(aq) + 3\mathrm{NO_3^-}(aq) \] From the stoichiometry of the equation, for every 1 mole of iron(III) nitrate that dissociates, 1 mole of Fe³⁺ and 3 moles of NO₃⁻ are produced.
04

Calculate the molar concentration of Fe³⁺ in the diluted solution.

Since the dissociation produces 1 mole of Fe³⁺ for every 1 mole of iron(III) nitrate, the molar concentration of Fe³⁺ in the diluted solution is equal to the molar concentration of iron(III) nitrate: \[ [\mathrm{Fe^{3+}}] = 0.075 \mathrm{~M} \]
05

Calculate the molar concentration of NO₃⁻ in the diluted solution.

Since the dissociation produces 3 moles of NO₃⁻ for every 1 mole of iron(III) nitrate, the molar concentration of NO₃⁻ in the diluted solution is three times the molar concentration of iron(III) nitrate: \[ [\mathrm{NO_3^-}] = 3 \times 0.075 \mathrm{~M} = 0.225 \mathrm{~M} \] So, the molar concentrations in the diluted solution are: a) Iron(III) nitrate: 0.075 M b) Fe³⁺: 0.075 M c) NO₃⁻: 0.225 M

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