Question

Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}_{3}\), (b) \(1.3 \times 10^{-2} \mathrm{M} \mathrm{MgSO}_{4},(\mathrm{c}) 0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (d) a mixture of \(45.0 \mathrm{~mL}\) of \(0.272 \mathrm{M} \mathrm{NaCl}\) and \(65.0 \mathrm{~mL}\) of \(0.0247 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\). Assume that the volumes are additive.

Short answer

The concentrations of ions and molecules in the solutions are: (a) 0.250 M NaNO₃ Na⁺: 0.25 M NO₃⁻: 0.25 M (b) 1.3 x 10^{-2} M MgSO₄ Mg²⁺: 1.3 x 10^{-2} M SO₄²⁻: 1.3 x 10^{-2} M (c) 0.0150 M C₆H₁₂O₆ C₆H₁₂O₆: 0.0150 M (d) A mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH₄)₂CO₃ Na⁺: 0.1113 M Cl⁻: 0.1113 M NH₄⁺: 0.02912 M CO₃²⁻: 0.01456 M

Step-by-step solution

Identify ions in the solution

In NaNO₃, we have Na⁺ ions and NO₃⁻ ions.

Calculate the concentration of ions

Since it is a 1:1 ratio, the concentration of Na⁺ and NO₃⁻ ions would also be 0.25 M. Na⁺: 0.25 M NO₃⁻: 0.25 M #b) Finding the concentration of ions in 1.3 x 10^(-2) M MgSO₄ solution. #

Identify ions in the solution

In MgSO₄, we have Mg²⁺ ions and SO₄²⁻ ions.

Calculate the concentration of ions

Since it is a 1:1 ratio, the concentration of Mg²⁺ and SO₄²⁻ ions would also be 1.3 x 10^(-2) M. Mg²⁺: 1.3 x 10^(-2) M SO₄²⁻: 1.3 x 10^(-2) M #c) Finding the concentration of molecules in 0.0150 M C₆H₁₂O₆ solution. #

Identify molecules in the solution

In C₆H₁₂O₆, there are no ions, it's a covalent molecular compound (glucose).

Calculate the concentration of molecules

Glucose does not dissociate into ions, so it remains as a molecule at a concentration of 0.0150 M. C₆H₁₂O₆: 0.0150 M #d) Finding the concentration of ions in the mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH₄)₂CO₃ solution. #

Identify ions in the solutions

In NaCl, we have Na⁺ ions and Cl⁻ ions. In (NH₄)₂CO₃, we have 2 NH₄⁺ ions and 1 CO₃²⁻ ion.

Calculate the moles of ions

For NaCl: moles of Na⁺ and Cl⁻ ions = 0.272 M × 0.045 L = 0.01224 mol For (NH₄)₂CO₃: moles of NH₄⁺ ions = 2 × (0.0247 M × 0.065 L) = 0.003203 mol moles of CO₃²⁻ ions = 0.0247 M × 0.065 L = 0.001602 mol

Calculate the total volume of the mixture

Total volume = 45.0 mL + 65.0 mL = 110.0 mL

Calculate the concentration of ions in the mixture

Concentration of Na⁺ ions = 0.01224 mol / 0.110 L = 0.1113 M Concentration of Cl⁻ ions = 0.01224 mol / 0.110 L = 0.1113 M Concentration of NH₄⁺ ions = 0.003203 mol / 0.110 L = 0.02912 M Concentration of CO₃²⁻ ions = 0.001602 mol / 0.110 L = 0.01456 M The concentrations of ions and molecules in the solutions are: (a) 0.250 M NaNO₃ Na⁺: 0.25 M NO₃⁻: 0.25 M (b) 1.3 x 10^(-2) M MgSO₄ Mg²⁺: 1.3 x 10^(-2) M SO₄²⁻: 1.3 x 10^(-2) M (c) 0.0150 M C₆H₁₂O₆ C₆H₁₂O₆: 0.0150 M (d) A mixture of 45.0 mL 0.272 M NaCl and 65.0 mL 0.0247 M (NH₄)₂CO₃ Na⁺: 0.1113 M Cl⁻: 0.1113 M NH₄⁺: 0.02912 M CO₃²⁻: 0.01456 M

Key concepts

Molar Concentration

Understanding molar concentration is crucial when studying chemistry. It indicates how much of a substance is present in a given volume of solution. It’s often expressed in moles per liter (M), which can be seen in the exercise with various substances like NaNO3 and MgSO4. To calculate the molar concentration, you divide the number of moles of solute by the volume of the solution in liters. For instance, if you have 0.25 moles of NaNO3 in 1 liter of solution, the molar concentration is 0.25 M. This calculation is a backbone of many chemical calculations and provides a standard way to discuss the concentration of solutions.

When tackling exercises involving molar concentration, remember that accuracy in measuring volume and quantity of the solute is key. Also, consider the solute's molar mass to convert between grams and moles if needed.

Ion Dissociation

Ion dissociation is a process where ionic compounds separate into individual ions when dissolved in water. This phenomenon is observed in salts like NaCl, which, when in water, dissociate into Na⁺ and Cl⁻ ions. The ability to predict the degree of dissociation is essential, as it allows students to accurately determine the concentration of individual ions in a solution. For example, NaNO3 dissociates entirely into Na⁺ and NO3⁻ ions, each with a concentration of 0.25 M in a 0.25 M NaNO3 solution.

It's important to note that the dissociation process follows the stoichiometry of the compound. This means for every one formula unit of NaNO3 that dissociates, you will have one Na⁺ ion and one NO3⁻ ion in solution. This 1:1 relationship makes calculating the resulting ion concentrations straightforward.

Molecular Compounds

Molecular compounds, such as C₆H₁₂O₆ in our exercise, consist of molecules held together by covalent bonds, rather than ions. Unlike ionic compounds, they do not dissociate into ions in solution. Hence, when such compounds dissolve in water, the molar concentration of the solution refers to the intact molecules. For instance, a 0.0150 M solution of C₆H₁₂O₆ means that there are 0.0150 moles of glucose molecules in every liter of solution. These compounds are typically non-electrolytes and do not conduct electricity in solution because they lack free-moving ions.

Concentration of Mixtures

Mixtures are a combination of two or more substances. When dealing with the concentration of mixtures, as in the fourth part of the exercise, we need to calculate the molar concentration of each component in the final combined solution. This involves first determining the moles of each solute from their individual solutions and then dividing by the total volume of the mixed solution.

Example of Calculating the Concentration in Mixtures

In the mixture of NaCl and (NH₄)₂CO₃, the moles of each ion are calculated before they were mixed. Then, by adding the volumes of both solutions and assuming they are additive, the total volume is attained. The concentration of each ion is found by dividing its moles by the new total volume. Remember, the molar concentration changes when substances are mixed due to volume changes, even though the moles of the substance remain constant.