Chapter 6: Problem 34
Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\), \(50.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2},\) or \(75.0 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4} ?\)
Short Answer
Expert verified
The 75.0 mL of 0.150 M Na3PO4 solution contains the largest number of ions, with a total of 0.045 moles of ions.
Step by step solution
01
Calculate number of moles for each solution
First, we need to determine the number of moles of the electrolyte in each of the given solutions by multiplying the volume (in Liters) by the concentration (in moles/Liter).
For 100.0 mL of 0.100 M NaOH:
\(n_{NaOH} = 0.1 \mathrm{L} \times 0.100 \frac{\mathrm{moles}}{\mathrm{L}} = 0.01 \mathrm{moles}\)
For 50.0 mL of 0.200 M BaCl2:
\(n_{BaCl_2} = 0.05 \mathrm{L} \times 0.200 \frac{\mathrm{moles}}{\mathrm{L}} = 0.01 \mathrm{moles}\)
For 75.0 mL of 0.150 M Na3PO4:
\(n_{Na3PO_4} = 0.075 \mathrm{L} \times 0.150 \frac{\mathrm{moles}}{\mathrm{L}} = 0.01125 \mathrm{moles}\)
02
Calculate the number of ions produced by each electrolyte
Now, we need to determine the number of ions produced by each of the electrolyte when it dissociates in water.
NaOH dissociates into 2 ions: 1 Na+ ion and 1 OH- ion.
BaCl2 dissociates into 3 ions: 1 Ba2+ ion and 2 Cl- ions.
Na3PO4 dissociates into 4 ions: 3 Na+ ions and 1 PO4(3-) ion.
03
Calculate the total number of ions in each solution
Next, multiply the number of moles of electrolyte by the number of ions that electrolyte produces when it dissociates:
For NaOH:
\(Total\; ions\; = 0.01\; moles \times 2\; ions/mole\; = 0.02\;moles\; of\; ions\)
For BaCl2:
\(Total\; ions\; = 0.01\; moles \times 3\; ions/mole\; = 0.03\; moles\; of\; ions\)
For Na3PO4:
\(Total\; ions\; = 0.01125\; moles \times 4\; ions/mole\; = 0.045\; moles\; of\; ions\)
04
Compare the total ion concentrations
Now, we just need to compare the total ion concentrations of each solution to determine which solution has the largest number of ions.
0.02 moles of ions in NaOH solution.
0.03 moles of ions in BaCl2 solution.
0.045 moles of ions in Na3PO4 solution.
Since 0.045 moles of ions in the Na3PO4 solution is greater than the number of ions in the other solutions, the 75.0 mL of 0.150 M Na3PO4 solution contains the largest number of ions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity and Volume Calculations
Understanding how to calculate the molarity and volume of a solution is foundational to many concepts in chemistry, especially when analyzing the properties of an electrolyte solution. Molarity is defined as the number of moles of solute per liter of solution, giving us a concentration measurement. This is expressed mathematically as:
\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]
To apply this in a practical context, let's say we have 100.0 mL of 0.100 M NaOH. The first step involves converting the volume from milliliters to liters because molarity uses liters in its formula. Hence, 100.0 mL is 0.100 liters. Next, we multiply the volume in liters by the molarity to find the number of moles of NaOH:
\[ n_{\text{NaOH}} = 0.100 \text{ L} \times 0.100 \text{ M} = 0.01 \text{ moles} \]
A lucid comparison of molarity and volume among different solutions can reveal the solution with the higher quantity of solute or help in preparing solutions of desired concentrations for experiments.
\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]
To apply this in a practical context, let's say we have 100.0 mL of 0.100 M NaOH. The first step involves converting the volume from milliliters to liters because molarity uses liters in its formula. Hence, 100.0 mL is 0.100 liters. Next, we multiply the volume in liters by the molarity to find the number of moles of NaOH:
\[ n_{\text{NaOH}} = 0.100 \text{ L} \times 0.100 \text{ M} = 0.01 \text{ moles} \]
A lucid comparison of molarity and volume among different solutions can reveal the solution with the higher quantity of solute or help in preparing solutions of desired concentrations for experiments.
Dissociation of Strong Electrolytes
When strong electrolytes dissolve in water, they fully dissociate into their respective ions. This dissociation is critical in determining the total ion concentration in a solution. For instance, sodium hydroxide (NaOH) dissociates into one sodium ion (Na+) and one hydroxide ion (OH-). So, for every mole of NaOH dissolved, we get 2 moles of ions. Barium chloride (BaCl2), on the other hand, dissociates into 3 ions: one barium ion (Ba2+) and two chloride ions (Cl-).
Impact of Dissociation on Ion Concentration
The amount of ions generated from the dissolution and dissociation process directly impacts the ion concentration and the solution's physical and chemical properties. This is why solutions of strong electrolytes, like those used in the example problem, can significantly vary in their overall ionic strength even if they have similar molarities.Comparing Ion Concentrations
Comparing ion concentrations between multiple solutions involves understanding both molarity and the nature of dissociation. By calculating the total number of moles of ions in each solution, as shown in the problem, you gain insight into which solution contains the largest number of ions—a higher total ion concentration means a higher number of ions.
For instance, let’s look at Na3PO4, which dissociates into four ions: three sodium ions (Na+) and one phosphate ion (PO4(3-)). With 0.01125 moles of the compound yielding 0.045 moles of ions after dissociation, we can see that the quantity of ions is greater than in solutions of NaOH or BaCl2 of similar or even greater molarities.
For instance, let’s look at Na3PO4, which dissociates into four ions: three sodium ions (Na+) and one phosphate ion (PO4(3-)). With 0.01125 moles of the compound yielding 0.045 moles of ions after dissociation, we can see that the quantity of ions is greater than in solutions of NaOH or BaCl2 of similar or even greater molarities.