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Chapter 4: Problem 69

(a) Which will have the highest concentration of potassium ion: \(0.20 \mathrm{M} \mathrm{KCl}, 0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4},\) or \(0.080 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of potassium ion: \(30.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) or \(25.0 \mathrm{~mL}\) of \(0.080 \mathrm{MK}_{3} \mathrm{PO}_{4} ?\)

Short Answer

Expert verified
(a) \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\) has the highest concentration of potassium ions with a concentration of \(0.30\, \mathrm{M}\) of \(\mathrm{K}^+\) ions. (b) \(30.0\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\) contains a greater number of moles of potassium ions (0.0090 moles) than \(25.0\, \mathrm{mL}\) of \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\) (0.0060 moles).

Step by step solution

01

Calculate the molar concentration of potassium ions in each solution

To calculate the molar concentration of potassium ions in each solution, we will multiply their molarity by the number of moles of potassium ions in each compound. For \(0.20\, \mathrm{M} \,\mathrm{KCl}\), there is one mole of \(\mathrm{K}^+\) in each mole of \(\mathrm{KCl}\), so the molar concentration of \(\mathrm{K}^+\) ions is: \(0.20\, \mathrm{M} \, \mathrm{KCl} \cdot \frac{1}{1} = 0.20\, \mathrm{M}\). For \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\), there are two moles of \(\mathrm{K}^+\) in each \(\mathrm{K}_{2}\, \mathrm{CrO}_{4}\), so the molar concentration of \(\mathrm{K}^+\) ions is: \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4} \cdot \frac{2}{1} = 0.30\, \mathrm{M}\). For \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\), there are three moles of \(\mathrm{K}^+\) in each \(\mathrm{K}_{3}\, \mathrm{PO}_{4}\), so the molar concentration of \(\mathrm{K}^+\) ions is: \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4} \cdot \frac{3}{1} = 0.24\, \mathrm{M}\).
02

Determine the solution with the highest concentration of potassium ions

Comparing the molar concentrations of potassium ions in each solution, we have: \(0.20\,\mathrm{M}\, \mathrm{KCl}\): \(0.20\, \mathrm{M}\) \(\mathrm{K}^+\) ions \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\,\mathrm{CrO}_{4}\): \(0.30\, \mathrm{M}\) \(\mathrm{K}^+\) ions \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\): \(0.24\, \mathrm{M}\) \(\mathrm{K}^+\) ions Thus, the highest concentration of potassium ions in the solutions is in \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\), with a concentration of \(0.30\, \mathrm{M}\) of \(\mathrm{K}^+\) ions.
03

Calculate the number of moles of potassium ions in each given volume

For part (b), we will calculate the number of moles of potassium ions in the given volumes of each solution. For \(30.0\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\): \(\mathrm{Moles\, of \, K^+} =\frac{30.0\, \mathrm{mL} \, \mathrm{K}_{2}\, \mathrm{CrO}_{4}}{1000} \cdot 0.30\, \mathrm{M} = 0.0090\, \mathrm{moles}\, \mathrm{K}^+\) For \(25.0\, \mathrm{mL}\) of \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\): \(\mathrm{Moles\, of \, K^+} =\frac{25.0\, \mathrm{mL} \, \mathrm{K}_{3}\, \mathrm{PO}_{4}}{1000} \cdot 0.24\, \mathrm{M} = 0.0060\, \mathrm{moles}\, \mathrm{K}^+\)
04

Determine which solution contains a greater number of moles of potassium ions

Comparing the number of moles of potassium ions in each solution, we have: \(30.0\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\): \(0.0090\, \mathrm{moles}\) \(\mathrm{K}^+\) ions \(25.0\, \mathrm{mL}\) of \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\): \(0.0060\, \mathrm{moles}\) \(\mathrm{K}^+\) ions Thus, \(30.0\, \mathrm{mL}\) of \(0.15\, \mathrm{M}\, \mathrm{K}_{2}\, \mathrm{CrO}_{4}\) contains a greater number of moles of potassium ions (0.0090 moles) than \(25.0\, \mathrm{mL}\) of \(0.080\, \mathrm{M}\, \mathrm{K}_{3}\, \mathrm{PO}_{4}\) (0.0060 moles).

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Key Concepts

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

Molar Concentration
Understanding the concept of molar concentration is essential when studying chemistry. It refers to the amount of a substance, measured in moles, present in one liter of solution. Usually denoted as molarity (\( M \), units of moles per liter, mol/L), it helps chemists to quantify the concentration of an element or compound in a given solution. For example, if a solution is labeled as 0.20 M KCl, this means that there are 0.20 moles of potassium chloride dissolved in every liter of the solution.

Calculating molar concentration involves using the formula:
\( \text{Molarity} (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \).
When dealing with ionic compounds, remember that they dissociate in water to produce ions. Each compound releases a certain number of ions depending on its chemical formula. Therefore, to find the concentration of a specific ion, you must multiply the molarity of the compound by the number of ions it releases upon dissociation.
Mole Calculations
A cornerstone of chemistry is moles calculations, which allow scientists to work with the submicroscopic particles that make up materials. Since atoms and molecules are too small to count individually, chemists use the mole (approximately 6.022 x 1023 entities) as a convenient unit to express quantities. The number of moles can be used to relate the mass of a substance to the number of atoms or molecules it contains.

To convert between moles and particles, Avogadro’s number is used. Also, the relationship between moles, mass, and molecular weight is crucial in these calculations. For instance, the molecular weight (g/mol) of a substance can serve as a conversion factor to determine how many moles are in a given mass of the substance, using the formula:
\( \text{Number of moles} = \frac{\text{Mass}}{\text{Molecular weight}} \).
Stoichiometry
Stoichiometry is an area of chemistry that involves using balanced chemical equations to calculate the relative amounts of reactants and products involved in a chemical reaction. The coefficients in a balanced equation indicate the proportions of reactants and products and are central to stoichiometric calculations. These proportions can be used in mole-to-mole conversions, helping to predict how much product will form from a given amount of reactant and vice versa.

To use stoichiometry to solve problems, you must first write and balance the chemical equation for the reaction. Then, use the mole ratio derived from the coefficients of the balanced equation to convert between the amounts of different substances. This method is vital in ensuring that a reaction is as efficient as possible, with minimal waste of materials.
Chemical Solutions
In the study of chemical solutions, it’s important to understand not only the concentration but also the behavior of solutes and solvents. A solution is a homogeneous mixture consisting of a solute dissolved in a solvent. The solute is the substance that is being dissolved, whereas the solvent is the substance that does the dissolving, typically present in a greater amount.

The process of dissolving can involve a physical change (e.g., sugar in water) or a chemical change (e.g., salt in water, which dissociates into ions). When calculating the concentration of ions in a solution, as in the exercise example, we must consider how each compound dissociates in water because it affects the total number of dissolved particles and, consequently, the properties of the solution.

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Most popular questions from this chapter

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide in the following fashion: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow$$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of \(0.115 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00-\) qt sample of this vinegar?

Using modern analytical techniques, it is possible to detect sodium ions in concentrations as low as \(50 \mathrm{pg} / \mathrm{mL}\). What is this detection limit expressed in (a) molarity of \(\mathrm{Na}^{+}\) (b) \(\mathrm{Na}^{+}\) ions per cubic centimeter?

Because the oxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the following reaction: \(\mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) Based on the equation in part (a), write the net ionic equation for the reaction that occurs between \(\mathrm{NiO}(s)\) and an aqueous solution of nitric acid.

Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},\) (b) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},\) (c) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

The metal cadmium tends to form Cd \(^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?
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