Question

Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}\), (b) \(1.3 \times 10^{-2} \mathrm{MMgSO}_{4}\), (c) \(0.0150 \quad M \quad \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (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

For the given solutions, the concentrations are as follows: (a) \(0.25 M \mathrm{NaNO}\) - Sodium ion (\(\mathrm{Na}^+\)) concentration = \(0.25 M\) - Nitrate ion (\(\mathrm{NO}^-\)) concentration = \(0.25 M\) (b) \(1.3 \times 10^{-2} M \mathrm{MgSO}_{4}\) - Magnesium ion (\(\mathrm{Mg}^{2+}\)) concentration = \(1.3 \times 10^{-2} M\) - Sulfate ion (\(\mathrm{SO}_{4}^{2-}\)) concentration = \(1.3 \times 10^{-2} M\) (c) \(0.0150 M \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) - Glucose (\(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)) concentration = \(0.0150 M\) (d) Mixture of \(45.0 mL\) of \(0.272 M \mathrm{NaCl}\) and \(65.0 mL\) of \(0.0247 M\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\): - Sodium ion (\(\mathrm{Na}^+\)) concentration = \(0.111 \, mM\) - Chloride ion (\(\mathrm{Cl}^-\)) concentration = \(0.111 \, mM\) - Ammonium ion (\(\mathrm{NH}_{4}^+\)) concentration = \(0.0291 \, mM\) - Carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) concentration =\(0.0146 \, mM\)

Step-by-step solution

Solution - Part (a)

For the \(0.25 M\) \(\mathrm{NaNO}\): The compound \(\mathrm{NaNO}\) breaks down into Sodium ions (\(\mathrm{Na}^+\)) and Nitrate ions (\(\mathrm{NO}^-\)). The given concentration is molar, or moles/liter. Since there is one \(\mathrm{Na}^+\) ion and one \(\mathrm{NO}^-\) ion for each \(\mathrm{NaNO}\) compound, their concentration will be equal to the concentration of the \(\mathrm{NaNO}\) compound. Sodium ion (\(\mathrm{Na}^+\)) concentration = \(0.25 M\) Nitrate ion (\(\mathrm{NO}^-\)) concentration = \(0.25 M\)

Solution - Part (b)

For the \(1.3 \times 10^{-2} M \mathrm{MgSO}_{4}\): The compound \(\mathrm{MgSO}_{4}\) breaks down into Magnesium ions (\(\mathrm{Mg}^{2+}\)) and Sulfate ions (\(\mathrm{SO}_{4}^{2-}\)). Similar to part (a), there is one \(\mathrm{Mg}^{2+}\) ion and one \(\mathrm{SO}_{4}^{2-}\) ion for each \(\mathrm{MgSO}_{4}\) compound, so their concentrations will be equal to the concentration of the \(\mathrm{MgSO}_{4}\) compound. Magnesium ion (\(\mathrm{Mg}^{2+}\)) concentration = \(1.3 \times 10^{-2} M\) Sulfate ion (\(\mathrm{SO}_{4}^{2-}\)) concentration = \(1.3 \times 10^{-2} M\)

Solution - Part (c)

For the \(0.0150 M \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\): The compound \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) is a molecule that doesn't break down into ions; it is a sugar molecule (glucose). Thus, we can directly state the concentration: Glucose (\(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)) concentration = \(0.0150 M\)

Solution - Part (d)

For the mixture of \(45.0 mL\) of \(0.272 M \mathrm{NaCl}\) and \(65.0 mL\) of \(0.0247 M\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\): First, we need to find the moles of each ion in both solutions and then calculate their concentrations in the mixed solution. We can assume the volumes are additive: Total volume \(= 45.0 mL + 65.0 mL = 110.0 mL\) 1. For the \(0.272 M \mathrm{NaCl}\) solution: - Moles of \(\mathrm{Na}^+\) ions = \((45.0 \,mL) \times (0.272 \,mM) = 12.24 \, mmol\) - Moles of \(\mathrm{Cl}^-\) ions = \((45.0 \,mL) \times (0.272 \,mM) = 12.24 \, mmol\) 2. For the \(0.0247 M\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) solution: - Moles of \(\mathrm{NH}_{4}^+\) ions = \((65.0 \,mL) \times (2 \times 0.0247 \, mM) = 3.205 \, mmol\) - Moles of \(\mathrm{CO}_{3}^{2-}\) ions = \((65.0 \,mL) \times (0.0247 \,mM) = 1.6025 \, mmol\) Now we can find the concentrations in the mixed solution: - Sodium ion (\(\mathrm{Na}^+\)) concentration = \(\frac{12.24 \, mmol}{110.0 \, mL} =0.111 \, mM\) - Chloride ion (\(\mathrm{Cl}^-\)) concentration = \(\frac{12.24 \, mmol}{110.0 \, mL} = 0.111 \, mM\) - Ammonium ion (\(\mathrm{NH}_{4}^+\)) concentration = \(\frac{3.205 \, mmol}{110.0 \, mL} = 0.0291 \, mM\) - Carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) concentration =\(\frac{1.6025 \, mmol }{110.0 \, mL} = 0.0146 \, mM\) These are the concentrations of each ion in the mixed solution.

Key concepts

Ionic Compounds

Ionic compounds are made up of positive and negative ions held together by electrostatic forces. These compounds generally consist of metals and non-metals. In a solid state, they form crystal lattices where ions are arranged in a repeating pattern.
When ionic compounds are dissolved in water, they dissociate into their constituent ions. For example, when sodium nitrate (\(\mathrm{NaNO}_3\)) is dissolved, it separates into sodium ions (\(\mathrm{Na}^+\)) and nitrate ions (\(\mathrm{NO}_3^-\)).
Ionic compounds have certain properties such as high melting and boiling points, and they conduct electricity when molten or dissolved in water. This is because the free ions in solution can move around, allowing electric current to pass through. Understanding ionic compounds is fundamental, as they are involved in many chemical reactions, especially those in aqueous solutions.

• High melting and boiling points
• Conduct electricity in specific conditions
• Form crystals

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The symbol for molarity is \(M\).
To calculate the molarity, you divide the number of moles of solute by the volume of solution in liters. For example, a solution with 1 mole of solute in 1 liter has a molarity of 1 \(M\).
Knowing the molarity is essential for preparing solutions and performing stoichiometric calculations in chemistry. It allows chemists to predict how substances will interact when combined in a chemical reaction. Molarity is fundamental when discussing topics such as concentration and dilution.

• Expressed in moles per liter (\( \text{mol/L} \))
• Helps in solution preparation
• Crucial for stoichiometry

Chemical Dissociation

Chemical dissociation refers to the process where compounds break down into smaller constituents, often ions, when dissolved in a solvent like water.
This process is crucial for ionic compounds, as they dissociate into positive and negative ions. For example, \(\mathrm{MgSO}_4\) dissociates into magnesium ions (\(\mathrm{Mg}^{2+}\)) and sulfate ions (\(\mathrm{SO}_4^{2-}\)).
Dissociation is important because it determines the electrical conductivity and reactivity of solutions. For non-ionic compounds like sugar (\(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)), no dissociation occurs; they dissolve as intact molecules.

• Occurs with ionic compounds
• Produces ions in solution
• Affects solution conductivity

Solution Mixing

Solution mixing involves combining two or more solutions which may have different concentrations and compositions.
The key to understanding solution mixing is recognizing that the concentrations of ions will change when solutions are mixed. The resulting concentration depends on the initial concentrations and the volumes of each solution.
For example, when mixing sodium chloride solution with ammonium carbonate solution, volumes are additive, making it important to recalculate the concentration based on the total volume. This type of mixing calculation is fundamental in labs where precise concentrations are necessary for reactions.

• Relates to concentration changes
• Important for lab preparations
• Volume affects ion concentration