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Ch.16 - Acids and Bases
Chapter 16, Problem 107

Calculate the concentration of all species in a 0.15 M KF solution.

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Identify the dissociation reaction of KF in water: \( \text{KF} \rightarrow \text{K}^+ + \text{F}^- \).
Recognize that KF is a strong electrolyte, meaning it dissociates completely in solution.
Determine the initial concentration of KF, which is given as 0.15 M.
Since KF dissociates completely, the concentration of \( \text{K}^+ \) ions will be equal to the initial concentration of KF, which is 0.15 M.
Similarly, the concentration of \( \text{F}^- \) ions will also be 0.15 M, as each formula unit of KF produces one \( \text{F}^- \) ion.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Dissociation of Ionic Compounds

Ionic compounds, such as KF, dissociate into their constituent ions when dissolved in water. In the case of KF, it separates into potassium ions (K⁺) and fluoride ions (F⁻). Understanding this dissociation is crucial for calculating the concentration of each species in the solution.
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Concentration Units

Concentration is a measure of the amount of solute in a given volume of solution, commonly expressed in molarity (M), which is moles of solute per liter of solution. For a 0.15 M KF solution, this means there are 0.15 moles of KF per liter of solution, which directly influences the concentrations of the resulting ions.
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Stoichiometry of Dissociation

The stoichiometry of the dissociation reaction provides the ratio of the ions produced from the solute. For KF, one mole of KF produces one mole of K⁺ and one mole of F⁻. Therefore, in a 0.15 M KF solution, the concentration of both K⁺ and F⁻ ions will also be 0.15 M, allowing for straightforward calculations of ion concentrations.
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