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Chapter 4: Problem 27

Try and answer the following questions without using a calculator. a. A solution is made by mixing \(1.0 \mathrm{~L}\) of \(0.5 \mathrm{M} \mathrm{NaCl}\) and \(0.5 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{CaCl}_{2}\). Which ion is at the highest concentration in the solution? b. Another solution is made by mixing \(0.50 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{KBr}\) and \(0.50 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4}\). What is the concentration of each ion in the solution?

Short Answer

Expert verified
a. Ion Cl鈦� has the highest concentration in the first solution. b. Ion concentrations are 2.0 M (K鈦�), 0.5 M (Br鈦�), and 0.5 M (PO鈧劼斥伝).

Step by step solution

01

Determine Moles of Ions from NaCl and CaCl2

For NaCl: 1.0 L of 0.5 M NaCl results in 0.5 moles of NaCl, yielding 0.5 moles of Na鈦� ions. For CaCl鈧�: 0.5 L of 1.0 M CaCl鈧� results in 0.5 moles of CaCl鈧�. Since each molecule of CaCl鈧� provides 2 moles of Cl鈦� ions, this means 1.0 mole of Cl鈦� ions, and 0.5 moles of Ca虏鈦� ions.
02

Total Volume and Concentration in NaCl and CaCl2 Solution

The total volume of the mixed solution is 1.0 L + 0.5 L = 1.5 L.\[\text{Concentration of Na}^+ = \frac{0.5 \text{ moles}}{1.5 \text{ L}} = 0.33 \text{ M}\]\[\text{Concentration of Ca}^{2+} = \frac{0.5 \text{ moles}}{1.5 \text{ L}} = 0.33 \text{ M}\]\[\text{Concentration of Cl}^- = \frac{1.0 \text{ mole}}{1.5 \text{ L}} = 0.67 \text{ M}\]
03

Determine Moles of Ions from KBr and K3PO4

For KBr: 0.5 L of 1.0 M KBr results in 0.5 moles of KBr, yielding 0.5 moles of K鈦� and Br鈦� ions each. For K鈧働O鈧�: 0.5 L of 1.0 M K鈧働O鈧� results in 0.5 moles of K鈧働O鈧�, yielding 1.5 moles of K鈦� ions and 0.5 moles of PO鈧劼斥伝 ions.
04

Total Volume and Concentration in KBr and K3PO4 Solution

The total volume of the mixed solution is 0.5 L + 0.5 L = 1.0 L.\[\text{Concentration of K}^+ = \frac{0.5 + 1.5 \text{ moles}}{1.0 \text{ L}} = 2.0 \text{ M}\]\[\text{Concentration of Br}^- = \frac{0.5 \text{ moles}}{1.0 \text{ L}} = 0.5 \text{ M}\]\[\text{Concentration of PO}_4^{3-} = \frac{0.5 \text{ moles}}{1.0 \text{ L}} = 0.5 \text{ M}\]
05

Identify Ions with Highest Concentration

From the NaCl and CaCl鈧� solution, the ion with the highest concentration is Cl鈦� at 0.67 M. From the KBr and K鈧働O鈧� solution, the ion with the highest concentration is K鈦� at 2.0 M.

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Key Concepts

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

Molarity
When working with solutions in chemistry, understanding molarity is key. Molarity, often denoted as M, is a way of expressing concentration. It is defined as the number of moles of solute per liter of solution. For instance, if you have a 0.5 M NaCl solution, this means that there are 0.5 moles of sodium chloride dissolved in every liter of that solution.

Calculating molarity involves using the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]This formula helps to determine how concentrated a given solution is.

When solutions of different concentrations are mixed, the total volume and molarity will change. You need to calculate the number of moles present initially and then divide that by the new total volume to find the new molarity.
Ion Concentration
Ion concentration is crucial when dealing with ionic solutions. It tells you how many ions are present in a specific volume of solution. Understanding ion concentration involves breaking down the solute into its constituent ions.

For example, dissolving NaCl in water entails separating it into Na\(^+\) and Cl\(^-\) ions. The concentration of each ion depends on the original concentration of the solute.

To find the concentration of each ion, consider the stoichiometry of the dissociation reaction. In the problem involving NaCl and CaCl鈧�, the concentration of ions like Na\(^+\), Ca\(^{2+}\), and Cl\(^-\) are determined by the initial molarity and the nature of the ionic species in the compounds.

If a reaction produces more than one ion of a particular type per formula unit of solute, adjust the calculations accordingly, such as with CaCl鈧�, which produces two Cl\(^-\) ions per molecule.
Chemical Mixing
Mixing different chemical solutions results in a new solution with potentially different concentrations. The total volume of the mixtures plays a significant role in determining the final concentrations of ions in the solution.

When combining solutions, it's important to first calculate the total volume. This new volume will be the denominator in determining the concentration of each ion present.

For example, mixing 1.0 L of a NaCl solution with 0.5 L of a CaCl鈧� solution results in a total volume of 1.5 L.

You must also account for the different molarities and volumes of the solutions being mixed. The total moles of each type of ion from each solution are summed, and the concentration is then calculated by dividing the total moles by the total volume.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in chemical reactions. It allows you to predict the amounts of products and reactants involved in a chemical process.

In the context of ion concentration, stoichiometry helps to calculate how many moles of ions are produced when an ionic compound dissolves.

For instance, in a solution of CaCl鈧�, stoichiometry reveals that each molecule dissociates into one Ca\(^{2+}\) ion and two Cl\(^-\) ions. This stoichiometric ratio is critical when determining ion concentrations from a dissolved compound.

Always align stoichiometric coefficients with the actual chemical equation of dissociation to ensure accurate ion concentration calculations.

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Most popular questions from this chapter

Obtain the oxidation number for the element noted in each of the following. a. \(\mathrm{N}\) in \(\mathrm{NO}_{2}^{-}\) b. \(\mathrm{Cr}\) in \(\mathrm{Cr} \mathrm{O}_{4}^{2-}\) c. \(\mathrm{Zn}\) in \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}\) d. As in \(\mathrm{H}_{2} \mathrm{As} \mathrm{O}_{3}^{-}\)

Identify each of the following reactions as being a neutralization, precipitation, or oxidation-reduction reaction. a. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) b. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) $$ \mathrm{HgSO}_{4}(s)+2 \mathrm{NaNO}_{3}(a q) $$ c. \(\mathrm{CsOH}(a q)+\mathrm{HClO}_{4}(a q) \longrightarrow\) $$ \mathrm{Cs}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{ClO}_{4}^{-}(a q) $$ d. \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(g)+\mathrm{Na}_{2} \mathrm{~S}(a q) \longrightarrow \mathrm{MgS}(s)+2 \mathrm{NaNO}_{3}(a q)\)

Calculate the concentrations of each ion present in a solution that results from mixing \(50.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{NaClO}_{3}(a q)\) solution with \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\). Assume that the volumes are additive.

A \(3.33-\mathrm{g}\) sample of iron ore is transformed to a solution of iron(II) sulfate, \(\mathrm{FeSO}_{4}\), and this solution is titrated with \(0.150 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (potassium dichromate). If it requires \(43.7 \mathrm{~mL}\) of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is $$ \begin{aligned} &6 \mathrm{FeSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ &3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+7 \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{aligned} $$

A barium mineral was dissolved in hydrochloric acid to give a solution of barium ion. An excess of potassium sulfate was added to \(50.0 \mathrm{~mL}\) of the solution, and \(1.128 \mathrm{~g}\) of barium sulfate precipitate formed. Assume that the original solution was barium chloride. What was the molarity of \(\mathrm{BaCl}_{2}\) in this solution?
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