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Chapter 8: Problem 13

Calculate the molarity of \(\mathrm{Na}^{+}\) ions in each of the following solutions: a. \(0.29 M \mathrm{NaNO}_{3}\) b. \(0.33 \mathrm{g} \mathrm{NaCl}\) in \(25 \mathrm{mL}\) of solution c. \(0.88 M \mathrm{Na}_{2} \mathrm{SO}_{4}\) d. \(0.46 \mathrm{g} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in \(100 \mathrm{mL}\) of solution

Short Answer

Expert verified
Question: Determine the molarity of Na+ ions in each of the following solutions: a. 0.29 M NaNO3 b. 0.33 g NaCl in 25 mL of solution c. 0.88 M Na2SO4 d. 0.46 g Na3PO4 in 100 mL of solution Answer: a. 0.29 M Na+ ions b. 0.226 M Na+ ions c. 1.76 M Na+ ions d. 0.08412 M Na+ ions

Step by step solution

01

Each formula unit of NaNO3 dissociates into 1 Na+ ion and 1 NO3- ion when dissolved in water. Therefore, there is 1 mole of Na+ ions per mole of NaNO3. #Step 2: Calculate the molarity of Na+ ions#

Since there is 1 mole of Na+ ions per mole of NaNO3 and the original molarity of NaNO3 is 0.29 M, the molarity of Na+ ions is also 0.29 M. b. 0.33 g NaCl in 25 mL of solution #Step 1: Calculate the moles of NaCl#
02

To find the moles of NaCl, we will use the formula:

moles = mass / molar mass
03

The molar mass of NaCl is 58.44 g/mol, so moles of NaCl = 0.33 g / 58.44 g/mol = 0.00565 mol. #Step 2: Calculate the molarity of NaCl#

To find the molarity, we use the formula:
04

Molarity = moles / volume (in L)

Since there are 0.00565 moles of NaCl in 0.025 L (25 mL in liters), the molarity of NaCl is: 0.00565 mol / 0.025 L = 0.226 M. #Step 3: Determine the molarity of Na+ ions#
05

Since there is 1 mole of Na+ ions per mole of NaCl, the molarity of Na+ ions is also 0.226 M. c. 0.88 M Na2SO4 #Step 1: Identify the number of moles of Na+ ions per mole of Na2SO4#

Each formula unit of Na2SO4 dissociates into 2 Na+ ions and 1 SO42- ion when dissolved in water. Therefore, there are 2 moles of Na+ ions per mole of Na2SO4. #Step 2: Calculate the molarity of Na+ ions#
06

Since there are 2 moles of Na+ ions per mole of Na2SO4 and the original molarity of Na2SO4 is 0.88 M, the molarity of Na+ ions is 2 * 0.88 M = 1.76 M. d. 0.46 g Na3PO4 in 100 mL of solution #Step 1: Calculate the moles of Na3PO4#

To find the moles of Na3PO4, we will use the formula:
07

moles = mass / molar mass

The molar mass of Na3PO4 is 163.94 g/mol, so moles of Na3PO4 = 0.46 g / 163.94 g/mol = 0.002804 mol. #Step 2: Calculate the molarity of Na3PO4#
08

To find the molarity, we use the formula:

Molarity = moles / volume (in L)
09

Since there are 0.002804 moles of Na3PO4 in 0.1 L (100 mL in liters), the molarity of Na3PO4 is: 0.002804 mol / 0.1 L = 0.02804 M. #Step 3: Determine the molarity of Na+ ions#

Since there are 3 moles of Na+ ions per mole of Na3PO4, the molarity of Na+ ions is 3 * 0.02804 M = 0.08412 M.

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Key Concepts

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

Na+ ion concentration
Understanding the concentration of \(\mathrm{Na}^+\text{ ions}\) in different solutions is crucial in the study of chemistry, particularly in textbook exercises focused on molarity. The molarity of \(\mathrm{Na}^+\text{ ions}\) depends directly on the compound being dissolved and its dissociation in water. For example:
  • In \(\mathrm{NaNO}_3\), each molecule contributes 1 \(\mathrm{Na}^+\) ion.
  • In \(\mathrm{NaCl}\), each molecule also provides 1 \(\mathrm{Na}^+\) ion.
  • For \(\mathrm{Na}_2\mathrm{SO}_4\), each molecule gives 2 \(\mathrm{Na}^+\) ions.
  • In \(\mathrm{Na}_3\mathrm{PO}_4\), each molecule generates 3 \(\mathrm{Na}^+\) ions.
The different contributions of \(\mathrm{Na}^+\) ions from each chemical formula must be taken into account to accurately determine the molarity of the ions in solution. It is essential to identify how many \(\mathrm{Na}^+\) ions are released per formula unit when dissolved, in order to calculate the final molarity correctly.
Solution chemistry
Solution chemistry involves understanding the processes and principles that govern how substances dissolve and interact in solvents, primarily water. When a salt like \(\mathrm{NaCl}\) is dissolved, it separates into its constituent ions, \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\). This dissociation is key to explaining how substances like salts increase the concentration of ions in solutions.
Important concepts in solution chemistry:
  • Dissociation: Ionic compounds separate into ions when dissolved.
  • Saturation: The point at which no more solute can dissolve in a solvent at a given temperature.
  • Concentration: Measured in molarity, which is moles of solute per liter of solution.
These principles help chemists predict and calculate the outcomes of reactions and mixtures, allowing for precise and calculated experimentation and data interpretation.
Mole calculations
Mole calculations are essential for determining the quantitative aspects of chemical reactions, including finding the number of \(\mathrm{moles}\) of a given substance. This is calculated using the formula:\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\]Here's a breakdown:
  • First, find the molar mass of the compound (e.g., 58.44 g/mol for \(\mathrm{NaCl}\)).
  • Then, use the mass given in the problem to find the number of moles.
  • Convert the volume of the solution from mL to L for proper calculations.
Performing these steps accurately is crucial for solving problems related to molarity and concentration, ensuring students understand the relationship between mass, volume, and the number of molecules present.
Dissolution process
The dissolution process is a foundational concept in chemistry that describes how substances dissolve in solvents to form a solution. When an ionic compound like \(\mathrm{NaCl}\) is added to water, it undergoes a specific process:
  • The water molecules surround the \(\mathrm{NaCl}\) crystals.
  • Polar water molecules interact with the \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\) ions.
  • The ions are pulled into the solution, dissociating from each other.
This seamless incorporation into the solvent increases the concentration of ions, a critical factor in chemical reactions.
The efficiency of the dissolution process often depends on:
  • Temperature: Higher temperatures usually increase solubility.
  • Stirring: Helps disperse ions throughout the solution more evenly.
  • Nature of the solute and solvent: Polar or nonpolar compounds affect solubility.
By understanding dissolution, students can easily follow how ionic compounds behave in solutions, leading to a more comprehensive grasp of solubility and reaction mechanisms.

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Most popular questions from this chapter

A solution of table salt is a good conductor of electricity, but a solution containing an equal molar concentration of table sugar is not. Why?

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