(a) Which will have the highest concentration of potassium ion: \(0.20 M \mathrm{KCl}, 0.15 M \mathrm{K}_{2} \mathrm{CrO}_{4},\) or 0.080\(M \mathrm{K}_{3} \mathrm{PO}_{4} ?\) (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 \(\mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?\)

Short Answer

Expert verified
(a) The highest concentration of potassium ions is in 0.15 M K2CrO4 with 0.30 M K+ ions. (b) 30.0 mL of 0.15 M K2CrO4 contains the greater number of moles of potassium ions, with 0.009 moles K+ ions.

Step by step solution

01

Identify the number of potassium ions in each compound

In KCl, there is 1 potassium ion (K+); in K2CrO4, there are 2 potassium ions; and in K3PO4, there are 3 potassium ions.
02

Calculate the concentration of potassium ions in each solution

For KCl: The concentration of K+ ions will be equal to the concentration of KCl, which is 0.20 M. For K2CrO4: The concentration of K+ ions will be 2 times the concentration of K2CrO4. \(Concentration \, of \, K^{+} = 2 \times 0.15 M = 0.30 M\) For K3PO4: The concentration of K+ ions will be 3 times the concentration of K3PO4. \(Concentration \, of \, K^{+} = 3 \times 0.080 M = 0.24 M\)
03

Compare the concentrations of potassium ions

KCl: 0.20 M K2CrO4: 0.30 M K3PO4: 0.24 M The solution with the highest concentration of potassium ions is 0.15 M K2CrO4 with a concentration of 0.30 M K+ ions. #b) Find the solution with the greater number of moles of potassium ions#
04

Recall the relationship between moles, molarity, and volume

The formula to find moles is: \(moles = Molarity \times Volume\)
05

Calculate the number of moles of potassium ions in each solution

For 30.0 mL of 0.15 M K2CrO4: - Convert volume to liters: \(30.0 mL \times \frac{1 L}{1000 mL} = 0.030 L\) - Calculate the number of moles of K2CrO4: \(0.15 M \times 0.030 L = 0.0045 \, moles \, of \, K_{2}CrO_{4}\) - Calculate the number of moles of K+ ions: \(0.0045 \, moles \, of \, K_{2}CrO_{4} \times \frac{2 \, moles \, K^{+}}{1 \, mole \, K_{2}CrO_{4}} = 0.009 \, moles \, K^{+}\) For 25.0 mL of 0.080 M K3PO4: - Convert volume to liters: \(25.0 mL \times \frac{1 L}{1000 mL} = 0.025 L\) - Calculate the number of moles of K3PO4: \(0.080 M \times 0.025 L = 0.0020 \, moles \, of \, K_{3}PO_{4}\) - Calculate the number of moles of K+ ions: \(0.0020 \, moles \, of \, K_{3}PO_{4} \times \frac{3 \, moles \, K^{+}}{1 \, mole \, K_{3}PO_{4}} = 0.006 \, moles \, K^{+}\)
06

Compare the number of moles of potassium ions

30.0 mL of 0.15 M K2CrO4: 0.009 moles K+ 25.0 mL of 0.080 M K3PO4: 0.006 moles K+ The solution with the greater number of moles of potassium ions is 30.0 mL of 0.15 M K2CrO4, containing 0.009 moles of K+ ions.

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Key Concepts

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

Molarity and Concentration
Molarity is a measure of concentration, expressing the number of moles of a solute per liter of solution. It's signified by the letter 'M' and is calculated using the formula:
\[ Molarity = \frac{moles \text{ of solute}}{volume \text{ of solution in liters}} \]
In simpler terms, it tells us how 'strong' or 'concentrated' a solution is with respect to a particular solute. In the context of potassium ion concentration in various compounds like KCl, K2CrO4, and K3PO4, molarity directly affects the number of available potassium ions in the solution. When comparing solutions with different molarities, the ion concentration is not just about the stated molarity of compounds, but also about the stoichiometry, which is the ratio of ions released per formula unit of the compound. A higher molarity usually means a higher concentration of ions up to the point where the solution becomes saturated and no more solute can be dissolved.
Moles and Volume Relationship
Understanding the relationship between moles and volume is crucial for many chemistry problems, including calculating ion concentrations in solutions. A mole is a unit of measurement that represents a set number of particles, usually atoms or molecules. The volume is the space that the solution occupies, typically measured in liters or milliliters.
For dilute solutions, where the solute-solvent interactions don't significantly change the volume, the relationship between moles, molarity, and volume is linear and is given by the straightforward formula:
\[ moles = Molarity \times Volume \]
This relationship allows us to determine the number of moles of any component in a solution if we know the molarity and the volume. The accurate conversion between the units of volume (milliliters to liters) is critical, as molarity is defined per liter of solution. To avoid common mistakes, always check that the volume used in calculations is in liters.
Stoichiometry of Ionic Compounds
Stoichiometry deals with the quantitative relationships of the elements within compounds and the calculations based on these relationships. When it comes to ionic compounds, stoichiometry becomes particularly important in understanding how many ions are produced when the compound dissolves in water.
For each compound such as KCl, K2CrO4, and K3PO4, the proportion of potassium to the other elements is determined by the chemical formula. For example, KCl has 1 potassium ion for every chloride ion, while K2CrO4 and K3PO4 have 2 and 3 potassium ions, respectively, for each formula unit. When these compounds dissolve in water, each unit dissociates to release multiple potassium ions, impacting the concentration. The stoichiometric coefficients tell us the exact number of moles of ions released from one mole of ionic compound, playing a vital role in the calculations of ion concentration in solution.

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Most popular questions from this chapter

A solution of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{KOH}\) is mixed with a solution of 200.0 \(\mathrm{mL}\) of 0.150 \(\mathrm{M} \mathrm{MiSO}_{4}\) . (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Formic acid, HCOOH, is a weak electrolyte. What solutes are present in an aqueous solution of this compound? Write the chemical equation for the ionization of HCOOH.

(a) By titration, 15.0 \(\mathrm{mL}\) of 0.1008 \(\mathrm{M}\) sodium hydroxide is needed to neutralize a \(0.2053-\mathrm{g}\) sample of a weak acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C},\) and 23.5\(\% \mathrm{O}\) by mass. What is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C},\) and 23.5\(\% \mathrm{O}\) by mass. What is its molecular formula?

(a) A strontium hydroxide solution is prepared by dissolving 12.50 g of \(\operatorname{Sr}(\mathrm{OH})_{2}\) in water to make 50.00 \(\mathrm{mL}\) of solution. What is the molarity of this solution? (b) Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions. (c) If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 37.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl},\) (b) \(\mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) \mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{CdSO}_{4}\) .

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