Question

Ignoring protolysis reactions, indicate the concentration of each ion or molecule present in the following solutions: \(\left(\right.\) a) \(0.35 M \mathrm{~K}_{3} \mathrm{PO}_{4},(\mathbf{b}) 5 \times 10^{-4} \mathrm{MCuCl}_{2},(\mathbf{c}) 0.0184\) \(\mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) a mixture of \(35.0 \mathrm{~mL}\) of \(0.010 \mathrm{MNa}_{2} \mathrm{CO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{MK}_{2} \mathrm{SO}_{4}\). Assume the volumes are additive.

Short answer

(a) \([\mathrm{K}^+] = 1.05 \text{ M}, [\mathrm{PO}_4^{3-}] = 0.35 \text{ M}\); (b) \([\mathrm{Cu}^{2+}] = 5 \times 10^{-4} \text{ M}, [\mathrm{Cl}^-] = 1 \times 10^{-3} \text{ M}\); (c) \([\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}] = 0.0184 \text{ M}\); (d) \([\mathrm{Na}^+] = 0.00824 \text{ M}, [\mathrm{CO}_3^{2-}] = 0.00412 \text{ M}, [\mathrm{K}^+] = 0.235 \text{ M}, [\mathrm{SO}_4^{2-}] = 0.118 \text{ M}\).

Step-by-step solution

Determine Ionization for Solution (a)

A solution of potassium phosphate, \(\mathrm{K}_3\mathrm{PO}_4\), dissociates completely in water into its ions: 3 potassium ions \(\mathrm{K}^+\) and 1 phosphate ion \(\mathrm{PO}_4^{3-}\). Given the concentration of \(\mathrm{K}_3\mathrm{PO}_4\) is 0.35 M, we calculate the concentration of each ion: - \([\mathrm{K}^+] = 3 \times 0.35\, \text{M} = 1.05\, \text{M}\) - \([\mathrm{PO}_4^{3-}] = 0.35\, \text{M}\)

Determine Ionization for Solution (b)

Copper(II) chloride, \(\mathrm{CuCl}_2\), dissociates into one copper ion \(\mathrm{Cu}^{2+}\) and two chloride ions \(\mathrm{Cl}^-\) in solution. Given a concentration of \(5 \times 10^{-4}\, \text{M}\), the concentration of each ion is computed as:- \([\mathrm{Cu}^{2+}] = 5 \times 10^{-4}\, \text{M}\)- \([\mathrm{Cl}^-] = 2 \times 5 \times 10^{-4}\, \text{M} = 1 \times 10^{-3}\, \text{M}\)

Determine Concentration for Solution (c)

Ethanol does not ionize or dissociate in solution, so the concentration of ethanol \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\) in the solution remains as given, which is 0.0184 M.

Determine Ionization for Mixed Solution (d)

First, calculate the total volume of the mixed solution: \(35.0\, \text{mL} + 50.0\, \text{mL} = 85.0\, \text{mL}\).For \(\mathrm{Na}_2\mathrm{CO}_3\):- Initial moles = \(35.0\, \text{mL} \times 0.010\, \text{M} = 0.350\, \text{mmol}\)For \(\mathrm{K}_2\mathrm{SO}_4\):- Initial moles = \(50.0\, \text{mL} \times 0.200\, \text{M} = 10.0\, \text{mmol}\)New concentrations after mixing:- \([\mathrm{Na}^+]= \frac{2 \times 0.350\, \text{mmol}}{85.0\, \text{mL}} = 0.00824\, \text{M}\)- \([\mathrm{CO}_3^{2-}] = \frac{0.350\, \text{mmol}}{85.0\, \text{mL}} = 0.00412\, \text{M}\)- \([\mathrm{K}^+]= \frac{2 \times 10.0\, \text{mmol}}{85.0\, \text{mL}} = 0.235\, \text{M}\)- \([\mathrm{SO}_4^{2-}] = \frac{10.0\, \text{mmol}}{85.0\, \text{mL}} = 0.118\, \text{M}\)

Key concepts

Solution Chemistry

Solution chemistry deals with the study of solute particles that dissolve in a solvent to form a solution. A solution is a homogeneous mixture where a solute is dissolved in a solvent, which is typically a liquid such as water. In these scenarios, knowing how substances interact in a solution provides insight into their chemical behavior.

Key aspects to understand include:

• Solubility: Not all substances dissolve evenly; solubility varies from one compound to another.
• Concentration: The amount of solute present per unit of solution is vital. It is usually indicated in moles per liter, also termed as molarity (M).
• Dissolution: This is the process where solute particles separate and disperse uniformly in the solvent.

Understanding solution chemistry is fundamental for predicting how different substances will behave when mixed with solvents.

Dissociation Reactions

In solution chemistry, dissociation reactions play a crucial role. Dissociation is when a compound breaks into its individual ions in a solvent, usually water. For ionic compounds, this is a common process that helps in understanding the behavior of ions in a solution.

For example, potassium phosphate ( K_3PO_4) dissociates into 3 potassium ions ( K^+) and 1 phosphate ion ( PO_4^{3-}).

• Cation and Anion Formation: Dissociation results in the creation of positively charged cations and negatively charged anions.
• Complete vs. Partial Dissociation: Some compounds dissociate completely, while others only partially. This affects the concentration of ions in the solution.

This concept is essential in predicting ion concentration and understanding the resultant solution properties.

Molarity Calculations

Molarity, denoted by 'M', is a measure of the concentration of a solute in a solution. It defines the number of moles of solute per liter of solution. Calculating molarity is critical in preparing solutions with desired properties.

Here's a quick guide on calculating molarity:

• Formula: Molarity (M) = moles of solute/liters of solution.
• Units: Molarity is expressed in moles per liter (mol/L).
• Application: Used to calculate concentrations of ions following dissociation.

Mastery of molarity calculations allows for precise manipulation of solution concentrations, such as in lab settings.

Chemical Equations

Chemical equations represent a chemical reaction, showing reactants transforming into products. They use symbols and form a concise way to convey the details of chemical reactions.

Elements to note in chemical equations include:

• Reactants and Products: The substances on the left of the arrow are reactants, while products are on the right.
• Balancing Equations: Ensures the same number of each type of atom on both sides.
• States of Matter: Indicated by subscripts, such as (s), (l), (g), and (aq).

Understanding chemical equations is pivotal for grasping stoichiometry and quantitative aspects of reactions. They are the backbone of documenting chemical processes systematically.