A sample of \(12.7\) grams of sodium sulfate (Na2SO4) is dissolved in \(672 \mathrm{~mL}\) of distilled water. a. What is the molar concentration of sodium sulfate in the solution? b. What is the concentration of sodium ion in the solution?

Short Answer

Expert verified
The molar concentration of sodium sulfate is 0.0894 M and the concentration of sodium ion is 0.1788 M.

Step by step solution

01

Calculate moles of sodium sulfate

Use the molar mass of sodium sulfate (Na2SO4) to convert the mass of the sample to moles. The molar mass of Na2SO4 is 142.04 g/mol. Using the formula: moles = mass (g) / molar mass (g/mol), calculate the number of moles.
02

Convert volume to liters

Convert the volume of the solution from milliliters to liters by dividing by 1000, since 1000 ml is equivalent to 1 liter.
03

Calculate molar concentration of sodium sulfate

Use the formula: molarity (M) = moles of solute / volume of solution (L) to calculate the molar concentration of sodium sulfate.
04

Determine concentration of sodium ions

Since sodium sulfate dissociates into two sodium ions and one sulfate ion, each mole of sodium sulfate will produce two moles of sodium ions. Multiply the molar concentration of sodium sulfate by 2 to get the concentration of sodium ions.

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Key Concepts

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

Molarity
Molarity is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute divided by the total volume of the solution in liters. The unit for molarity is moles per liter (M). For example, if you dissolve a substance into water, the molarity tells you how concentrated that solution is. To calculate molarity, you use the following formula: \( M = \frac{\text{moles of solute}}{\text{volume of solution in L}} \).

Using molarity is useful because it allows chemists to predict how substances will react with one another based on how concentrated they are. Moreover, it's an easy way to measure and communicate the strength of a solution in a laboratory or industrial setting. A greater molarity indicates a higher concentration and vice versa.
Solution Concentration
Solution concentration is a general term referring to the amount of solute present in a given quantity of solvent or solution. Molarity is one method used to express solution concentration, but there are others such as molality, mass percent, and volume percent. The appropriate method to use often depends on the situation and laboratory requirements. For instance, molarity is temperature-dependent since it involves volume, which can change with temperature.

In the context of our exercise, when we refer to the concentration of sodium sulfate in water, we specifically mean the molarity of that solute within the solution. Higher solution concentrations imply a greater amount of solute dissolved in a fixed amount of solvent.
Moles Calculation
The concept of moles is fundamental in chemistry, as it allows for a standardized method of quantifying substances. One mole is equal to Avogadro's number of particles, which is approximately \(6.02 \times 10^{23}\) particles. These particles could be atoms, molecules, ions, or electrons.

To calculate the number of moles, you use the substance's molar mass, which is the mass of one mole of that substance. The formula is: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). This calculation is crucial because it helps you transition from the macroscopic world, which deals with grams and liters, to the microscopic world measured in moles, thereby enabling chemists to use the mole concept to balance reactions and find the stoichiometry among reactants and products.
Dissociation of Ionic Compounds
Ionic compounds, such as sodium sulfate (\(Na_2SO_4\)), dissociate in water to form ions. This process is crucial for calculating the concentration of ions in a solution. For every mole of sodium sulfate dissolved, two moles of sodium ions (\(Na^+\)) and one mole of sulfate ions (\(SO_4^{2-}\)) are formed.

To determine the concentration of individual ions, one must consider this dissociation. If the molar concentration of \(Na_2SO_4\) is, say, 1 M, then the concentration of \(Na^+\) ions is 2 M, because each unit of sodium sulfate provides two sodium ions. This same concept applies to other ionic compounds and is vital for understanding properties like electrical conductivity and osmotic pressure in solutions.

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Most popular questions from this chapter

Calculate the mass of sodium chloride required to make \(125.0 \mathrm{~mL}\) of a \(0.470 \mathrm{M} \mathrm{NaCl}\) solution. If you dissolve \(5.8 \mathrm{~g}\) of \(\mathrm{NaCl}\) in water and then dilute to a total of \(100.0 \mathrm{~mL}\), what will be the molar concentration of the resulting sodium chloride solution?

1\. A 1.50 mL aliquot of a \(0.177 \mathrm{M}\) solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) is diluted into \(10.0 \mathrm{~mL}\) of distilled water, to give solution A. A \(10.0 \mathrm{~mL}\) aliquot of \(\mathbf{A}\) is then diluted into \(50.0 \mathrm{~mL}\) of distilled water, to give solution \(\mathbf{B}\). Finally, \(10.0 \mathrm{~mL}\) of \(\mathbf{B}\) is diluted into \(900.0 \mathrm{~mL}\) of distilled water to give solution C. Additional distilled water is then added to \(\mathrm{C}\) to give a final volume of \(1.0000 \mathrm{~L}\). What is the final concentration of sulfuric acid in solution C? 2\. A solution was prepared by mixing \(250 \mathrm{~mL}\) of \(0.547 \mathrm{M} \mathrm{NaOH}\) with \(50.0 \mathrm{~mL}\) of \(1.62 \mathrm{M} \mathrm{NaOH}\) and then diluting to a final volume of \(1.50 \mathrm{~L}\). What is the molarity of \(\mathrm{Na}^{+}\) in this solution? To what final volume should \(75.00 \mathrm{~mL}\) of \(0.889\) M HCl(aq) be diluted to prepare \(0.800 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) ?

For each of the molecules of \(\mathrm{NH}_{3}\) and \(\mathrm{CO}_{2}\) indicate whether a molecular dipole exists. If a dipole does exist, use a dipole arrow to indicate the direction of the molecular dipole.

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