What are the final concentrations of all the ions when following are mixed? \(50 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}, 100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{FeCl}_{3}\) and \(100 \mathrm{~mL}\) of \(0.26 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Short Answer

Expert verified
Final concentrations: [Fe3+] = 0.22 M, [NO3-] = 0.494 M, [Cl-] = 0.30 M, [Mg2+] = 0.104 M.

Step by step solution

01

Determine the Moles of Each Ion

First, calculate the moles of each ion before mixing. Use the concentration (in Molarity, M) and volume (in L) for each solution using the formula: Moles = Concentration * Volume. For Fe(NO3)3: Moles of Fe3+ = 0.12 M * 0.050 L and Moles of NO3- = 3 * Moles of Fe3+. For FeCl3: Moles of Fe3+ = 0.10 M * 0.100 L and Moles of Cl- = 3 * Moles of Fe3+. For Mg(NO3)2: Moles of Mg2+ = 0.26 M * 0.100 L and Moles of NO3- = 2 * Moles of Mg2+.
02

Sum of Moles After Mixing

The total volume for all solutions is 250 mL or 0.250 L. Add the moles of identical ions calculated in Step 1 to get the total moles of each ion after mixing. Total Moles of Fe3+ = Moles from Fe(NO3)3 + Moles from FeCl3. Total Moles of NO3- = Moles from Fe(NO3)3 + Moles from Mg(NO3)2. Total Moles of Cl- = Moles from FeCl3. Total Moles of Mg2+ = Moles from Mg(NO3)2.
03

Calculate Final Concentration

Final concentrations can be determined by dividing the total moles of each ion by the total volume. Concentration of Fe3+ = Total Moles of Fe3+ / 0.250 L. Concentration of NO3- = Total Moles of NO3- / 0.250 L. Concentration of Cl- = Total Moles of Cl- / 0.250 L. Concentration of Mg2+ = Total Moles of Mg2+ / 0.250 L.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is a term used in chemistry to describe the concentration of a solution. It's defined as the number of moles of solute (the substance to be dissolved) per liter of solution. The unit for molarity is moles per liter ((M/L) or molar ((M)). To calculate molarity, you can use the following formula:
\( \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \).
In the context of the exercise provided, molarity assists in understanding how much of each ion is present in a given volume of solution before and after mixing. For instance, a solution having a molarity of 0.12 M implies that there are 0.12 moles of the solute in every liter of the solution. Accurate molarity calculations are essential for the proper execution of stoichiometric calculations in chemistry.
Stoichiometry
Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the substances as they participate in chemical reactions. In essence, it's like a recipe for a reaction: telling you how much of each reactant you need and what amount of product you can expect.
When performing stoichiometric calculations, balance between the reactants and products in a chemical equation is crucial. This balance is based on the conservation of mass and the principle that atoms are neither created nor destroyed in a chemical reaction.
For the given exercise, stoichiometry is used to determine the moles of each ion in the solutions. By understanding the stoichiometric relationships in the compounds (for example, \( \mathrm{Fe(NO_{3})_{3}} \) has three \( \mathrm{NO_{3}^{-}} \) ions for every one \( \mathrm{Fe^{3+}} \) ion), we can accurately calculate the final concentrations of ions after mixing the solutions.
Solution Concentration
Solution concentration refers to the amount of solute present in a given quantity of solvent or solution. It's a measure of how 'concentrated' a solution is. There are several ways to express concentration, including molarity, molality, mass percent, and volume percent. Molarity, the measure used in the provided exercise, is one of the most common methods, especially in aqueous solutions, which are solutions where water is the solvent.
In practice, understanding solution concentration is paramount when preparing solutions for experiments, as well as when combining solutions, as seen in the exercise. Knowing the concentration of each solution helps to predict how the concentrations will change upon mixing. For instance, mixing equal volumes of solutions with different molarities will result in a final molarity that is the average of the two initial molarities.
The exercise precisely uses the concept of solution concentration to determine the final concentration of ions when the initial solutions are mixed. By calculating the final concentrations, we can predict the behavior of the ions in the new combined solution and use this knowledge for further chemical analysis or reactions.

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Most popular questions from this chapter

\(1.08 \mathrm{~g}\) of copper wire was allowed to react with nitric acid. The resulting solution was dried and ignited when \(1.35 \mathrm{~g}\) of copper oxide was obtained. In another experiment \(2.30 \mathrm{~g}\) of copper oxide was heated in presence of hydrogen yielding \(1.84 \mathrm{~g}\) of copper. Show that the above data are in accordance with law of constant proportion.

Calculate the volume of \(1.00 \mathrm{~mol} \mathrm{~L}^{-1}\) aqueous sodium hydroxide that is neutralized by \(200 \mathrm{~mL}\) of \(2.00 \mathrm{~mol} \mathrm{~L}^{-1}\) aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is; $$ \mathrm{NaOH} \text { (aq.) }+\mathrm{HCl} \text { (aq.) } \longrightarrow \mathrm{NaCl}(\mathrm{aq} .)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

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(a) When \(4.2 \mathrm{~g} \mathrm{NaHCO}_{3}\) is added to a solution \(\mathrm{of} \mathrm{CH}_{3} \mathrm{COOH}\) weighing \(10.0 \mathrm{~g}\), it is observed that \(2.2 \mathrm{~g} \mathrm{CO}_{2}\) is released into atmosphere. The residue is found to weigh \(12.0 \mathrm{~g}\). Show that these observations are in agreement with the low of conservation of weigh. (b) If \(6.3 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\) are added to \(15.0 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}\) solution. The residue is found to weigh \(18.0 \mathrm{~g}\). What is the mass of \(\mathrm{CO}_{2}\) released in this reaction.

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