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Chapter 15: Problem 18

Determine the molarity of each of the ions present in the following aqueous salt solutions: (assume \(100 \%\) ionization) (a) \(2.25 \mathrm{M} \mathrm{} \mathrm{FeCl}_{3}\) (b) \(1.20 \mathrm{M} \mathrm{MgSO}_{4}\) (c) \(0.75 \mathrm{M} \mathrm{NaH}_{2} \mathrm{PO}_{4}\) (d) \(0.35 \mathrm{M} \mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}\)

Short Answer

Expert verified
FeCl3: Fe^{3+} = 2.25 M, Cl^{-} = 6.75 M; MgSO4: Mg^{2+} = 1.20 M, SO4^{2-} = 1.20 M; NaH2PO4: Na^{+} = 0.75 M, H2PO4^{-} = 0.75 M; Ca(ClO3)2: Ca^{2+} = 0.35 M, ClO3^{-} = 0.70 M.

Step by step solution

01

Determine the dissociation equation

Write the dissociation equation for each salt in water.
02

Calculate the ion ratios

Using the dissociation equation, determine the ratio in which the salt dissociates into ions.
03

Calculate the molarity of each ion

Multiply the initial molarity of the salt by the number of ions produced from one formula unit of the salt for each ion present.
04

Solution for (a): FeCl3

Dissociation equation: \[ \mathrm{FeCl}_{3} \rightarrow \mathrm{Fe}^{3+} + 3\mathrm{Cl}^{-} \] Starting molarity: 2.25 M Ion molarities: - For Fe^{3+}: \[ 2.25 \mathrm{M} \] - For Cl^{-}: \[ 3 \times 2.25 \mathrm{M} = 6.75 \mathrm{M} \]
05

Solution for (b): MgSO4

Dissociation equation: \[ \mathrm{MgSO}_{4} \rightarrow \mathrm{Mg}^{2+} + \mathrm{SO}_{4}^{2-} \] Starting molarity: 1.20 M Ion molarities: - For Mg^{2+}: \[ 1.20 \mathrm{M} \] - For SO_{4}^{2-}: \[ 1.20 \mathrm{M} \]
06

Solution for (c): NaH2PO4

Dissociation equation: \[ \mathrm{NaH}_{2} \mathrm{PO}_{4} \rightarrow \mathrm{Na}^{+} + \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \] Starting molarity: 0.75 M Ion molarities: - For Na^{+}: \[ 0.75 \mathrm{M} \] - For H_{2} \mathrm{PO}_{4}^{-}: \[ 0.75 \mathrm{M} \]
07

Solution for (d): Ca(ClO3)2

Dissociation equation: \[ \mathrm{Ca(ClO}_{3})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{ClO}_{3}^{-} \] Starting molarity: 0.35 M Ion molarities: - For Ca^{2+}: \[ 0.35 \mathrm{M} \] - For ClO_{3}^{-}: \[ 2 \times 0.35 \mathrm{M} = 0.70 \mathrm{M} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissociation Equation
The dissociation equation describes how a salt splits into its ions when it dissolves in water. This is crucial in determining ion concentration.
If we take the salt \(\text{FeCl}_3\) as an example, its dissociation equation is \[ \text{FeCl}_3 \rightarrow \text{Fe}^{3+} + 3\text{Cl}^{-} \].
This equation shows that one unit of \(\text{FeCl}_3\) produces one iron ion (Fe^{3+}) and three chloride ions (Cl^{-}). Understanding this helps us figure out the ion concentrations.
Different salts dissociate differently, so writing the correct dissociation equation for each compound is very important.
Ion Concentration
Ion concentration tells us how many ions are present in a solution. To calculate it, we use the initial molarity of the salt and the dissociation equation.
For example, with \(\text{FeCl}_3\) at 2.25 M, its dissociation produces one \(\text{Fe}^{3+}\) and three \(\text{Cl}^{-}\) ions. Since one unit of \(\text{FeCl}_3\) yields three chloride ions, their concentration will be \[ 3 \times 2.25 \text{ M} = 6.75 \text{ M} \].
This means we have 2.25 M \(\text{Fe}^{3+}\) and 6.75 M \(\text{Cl}^{-}\) in the solution. Repeat these steps for other salts to find their ion concentrations.
Aqueous Solutions
Aqueous solutions are those where water is the solvent. When a salt dissolves in water, it forms an aqueous solution with free ions.
This is why knowing how salts dissociate in water is key. For instance, in water, \(\text{NaH}_2\text{PO}_4\) dissociates to \[ \text{Na}^{+} + \text{H}_2\text{PO}^{4-} \].
These free ions are responsible for the solution’s properties. Understanding how to calculate the molarity in aqueous solutions helps in various applications, such as chemistry labs and real-world scenarios like water treatment.
Chemical Reactions
Chemical reactions involve the transformation of substances through breaking and forming of bonds. In the context of aqueous solutions, dissociation is a chemical reaction.
Taking \(\text{MgSO}_4\) for example, the dissociation reaction is \[ \text{MgSO}_4 \rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \].
It shows how magnesium sulfate breaks into its ions in water. Understanding these reactions is crucial for predicting the behavior of salts in solutions, which is important in fields like pharmacology and environmental science.
Stoichiometry
Stoichiometry explains the quantitative relationships between reactants and products in a chemical reaction. When calculating ion concentrations, stoichiometry guides how we use molar ratios.
For instance, the dissociation of \(\text{Ca(ClO}_3)_2\) gives us \[ \text{Ca(ClO}_3)_2 \rightarrow \text{Ca}^{2+} + 2\text{ClO}_3^{-} \].
If we start with 0.35 M of \(\text{Ca(ClO}_3)_2\), we get 0.70 M of \(\text{ClO}_3^{-}\) due to the 1:2 ratio. Stoichiometry ensures accurate concentration calculations from the dissociation equations.

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Most popular questions from this chapter

In an acid-base titration, \(25.22 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) were used to neutralize \(35.22 \mathrm{~mL}\) of \(0.313 \mathrm{M} \mathrm{NaOH}\). What was the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution? $$ \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O} $$

Determine whether each of the following is a strong acid, weak acid, strong base, or weak base. Then write an equation describing the process that occurs when the substance is dissolved in water. (a) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}\) (c) \(\mathrm{HClO}_{4}\) (d) \(\mathrm{HBr}\)

Write an equation showing how sodium sulfate dissociates in water.

Calculate the \(\mathrm{pH}\) of (a) black coffee, \(5.0 \times 10^{-5} \mathrm{MH}^{+}\) (b) limewater, \(3.4 \times 10^{-11} \mathrm{M} \mathrm{H}^{+}\) (c) fruit punch, \(2.1 \times 10^{-4} \mathrm{MH}^{+}\) (d) cranberry apple drink, \(1.3 \times 10^{-3} \mathrm{M} \mathrm{H}^{+}\)

Sulfuric acid reacts with NaOH: (a) Write a balanced equation for the reaction producing \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). (b) How many milliliters of \(0.10 \mathrm{M} \mathrm{NaOH}\) are needed to react with \(0.0050 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) ? (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) will also form?
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