Chapter 4: Problem 67
What is the total molar concentration of ions in each of the following solutions? (a) A \(1.250 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{OH}\) (b) A \(0.225 \mathrm{M}\) solution of $\mathrm{HClO}_{4}
Short Answer
Expert verified
(a) 0 M; (b) 0.450 M
Step by step solution
01
Identify the Compound in Part (a)
The compound given in part (a) is \(\mathrm{CH}_3\mathrm{OH}\), which is methanol. Methanol is a molecular compound and does not dissociate into ions in solution. It remains as neutral molecules.
02
Calculate Total Ion Concentration for Part (a)
Since \(\mathrm{CH}_3\mathrm{OH}\) does not dissociate into ions, the total molar concentration of ions in the solution is 0 M.
03
Identify the Compound in Part (b)
The compound given in part (b) is \(\mathrm{HClO}_4\), which is perchloric acid, a strong acid. Strong acids dissociate completely in water.
04
Write Dissociation Reaction for Part (b)
\[ \mathrm{HClO}_4 \rightarrow \mathrm{H}^+ + \mathrm{ClO}_4^- \] This indicates that one mole of \(\mathrm{HClO}_4\) produces one mole of \(\mathrm{H}^+\) ions and one mole of \(\mathrm{ClO}_4^-\) ions.
05
Calculate Total Ion Concentration for Part (b)
The initial concentration of \(\mathrm{HClO}_4\) is \(0.225 \mathrm{M}\). Since it dissociates completely, it produces \(0.225 \mathrm{M}\) of \(\mathrm{H}^+\) and \(0.225 \mathrm{M}\) of \(\mathrm{ClO}_4^-\). The total molar concentration of ions is therefore \(0.225 + 0.225 = 0.450 \mathrm{M}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Methanol
Methanol, also known as wood alcohol, is a simple alcohol with the chemical formula \(\mathrm{CH}_3\mathrm{OH}\). It is a molecular compound, which means it is made up of covalently bonded atoms. One of the key characteristics of methanol is that it does not dissociate into ions when dissolved in water.
This is because methanol's molecules do not break apart to form charged particles or ions. Instead, they stay together as neutral molecules. This means that solutions of methanol do not contribute any ions to the solution, and thus, the total molar concentration of ions in a methanol solution is zero.
For students learning chemistry, remember that molecular compounds like methanol lack ionic bonds, which results in no ion formation in solutions. This makes calculating the ionic concentration for substances like methanol straightforward because it remains 0 \(\mathrm{M}\).
This is because methanol's molecules do not break apart to form charged particles or ions. Instead, they stay together as neutral molecules. This means that solutions of methanol do not contribute any ions to the solution, and thus, the total molar concentration of ions in a methanol solution is zero.
For students learning chemistry, remember that molecular compounds like methanol lack ionic bonds, which results in no ion formation in solutions. This makes calculating the ionic concentration for substances like methanol straightforward because it remains 0 \(\mathrm{M}\).
Perchloric Acid
Perchloric acid, with the formula \(\mathrm{HClO}_4\), is known as a strong acid. Strong acids are substances that ionize completely in aqueous solutions. This means that when perchloric acid is dissolved in water, it breaks down into its constituent ions: hydrogen (\(\mathrm{H}^+\)) and perchlorate (\(\mathrm{ClO}_4^-\)).
One mole of \(\mathrm{HClO}_4\) results in the formation of one mole of \(\mathrm{H}^+\) ions and one mole of \(\mathrm{ClO}_4^-\) ions. This complete dissociation is characteristic of strong acids, explaining why they have such high conductivity in water.
One mole of \(\mathrm{HClO}_4\) results in the formation of one mole of \(\mathrm{H}^+\) ions and one mole of \(\mathrm{ClO}_4^-\) ions. This complete dissociation is characteristic of strong acids, explaining why they have such high conductivity in water.
- Strong acids fully dissociate in water.
- Their solutions contain a high concentration of ions.
Dissociation
Dissociation is the process by which compounds separate into smaller particles such as ions when dissolved in a solvent. This concept is crucial in understanding how certain solutions can conduct electricity and take part in reactions.
For example, when a strong acid like \(\mathrm{HClO}_4\) is dissolved in water, it completely dissociates into its ions \(\mathrm{H}^+\) and \(\mathrm{ClO}_4^-\). Unlike methanol which does not dissociate, dissociation in ionic compounds or strong acids facilitates the presence of charged particles that can move freely in a solution.
Understanding dissociation helps in calculating the total ion concentration in solutions. In cases where compounds dissociate completely, the molarity of the original compound often equals the total concentration of ions produced, considering stoichiometry is taken into account.
For example, when a strong acid like \(\mathrm{HClO}_4\) is dissolved in water, it completely dissociates into its ions \(\mathrm{H}^+\) and \(\mathrm{ClO}_4^-\). Unlike methanol which does not dissociate, dissociation in ionic compounds or strong acids facilitates the presence of charged particles that can move freely in a solution.
Understanding dissociation helps in calculating the total ion concentration in solutions. In cases where compounds dissociate completely, the molarity of the original compound often equals the total concentration of ions produced, considering stoichiometry is taken into account.
Molarity
Molarity is a measure of concentration, defined as the number of moles of solute per liter of solution. It is commonly denoted by \(M\) and usually gives us a straightforward way to understand solution concentration.
Understanding molarity is critical for students, as it lays the foundation for performing stoichiometric calculations in chemistry and aids in exploring the concentration relationships in chemical equations.
- A 1 M solution means there is one mole of a solute in one liter of solution.
- Molarity simplifies predicting the concentrations of ions in dissociated solutions.
Understanding molarity is critical for students, as it lays the foundation for performing stoichiometric calculations in chemistry and aids in exploring the concentration relationships in chemical equations.