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Chapter 4: Problem 32

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 \(0.150 \mathrm{M} \mathrm{Na}_{4} \mathrm{PO}_{4}\) ?

Short Answer

Expert verified
The solution containing \(75.0\mathrm{~mL}\) of \(0.150\mathrm{M}\ \mathrm{Na}_{4}\mathrm{PO}_{4}\) has the largest number of ions at \(56.25\mathrm{~mmol}\).

Step by step solution

01

Write down the given information

The given information is as follows: - \(100.0\mathrm{~mL}\) of \(0.100\mathrm{M}\ \mathrm{NaOH}\) - \(50.0\mathrm{~mL}\) of \(0.200\mathrm{M}\ \mathrm{BaCl}_{2}\) - \(75.0\mathrm{~mL}\) of \(0.150\mathrm{M}\ \mathrm{Na}_{4}\mathrm{PO}_{4}\)
02

Write down the number of ions released per formula unit for each compound when dissolved in water

For each of the three solutions: - Sodium hydroxide (\(\mathrm{NaOH}\)) dissociates into 1 sodium ion (\(\mathrm{Na^{+}}\)) and 1 hydroxide ion (\(\mathrm{OH^{-}}\)), resulting in a total of 2 ions. - Barium chloride (\(\mathrm{BaCl}_{2}\)) dissociates into 1 barium ion (\(\mathrm{Ba^{2+}}\)) and 2 chloride ions (\(\mathrm{Cl^{-}}\)), resulting in a total of 3 ions. - Sodium phosphate (\(\mathrm{Na}_{4}\mathrm{PO}_{4}\)) dissociates into 4 sodium ions (\(\mathrm{Na^{+}}\)) and 1 phosphate ion (\(\mathrm{PO_{4}^{3-}}\)), resulting in a total of 5 ions.
03

Calculate the number of ions in each solution

Now multiply the volume by the concentration and then the number of ions: 1. For NaOH: \[\text{Ions}_{\text{NaOH}} = 100.0\mathrm{~mL} \times 0.100\mathrm{M} \times 2\] \[\text{Ions}_{\text{NaOH}} = 20.0\mathrm{~mmol}\] 2. For BaCl2: \[\text{Ions}_{\text{BaCl}_2} = 50.0\mathrm{~mL} \times 0.200\mathrm{M} \times 3\] \[\text{Ions}_{\text{BaCl}_2} = 30.0\mathrm{~mmol}\] 3. For Na4PO4: \[\text{Ions}_{\text{Na}_4\mathrm{PO}_4} = 75.0\mathrm{~mL} \times 0.150\mathrm{M} \times 5\] \[\text{Ions}_{\text{Na}_4\mathrm{PO}_4} = 56.25\mathrm{~mmol}\]
04

Compare the results and find the solution with the largest number of ions

Now we compare the number of ions in each solution: - \(\text{Ions}_{\text{NaOH}} = 20.0\mathrm{~mmol}\) - \(\text{Ions}_{\text{BaCl}_2} = 30.0\mathrm{~mmol}\) - \(\text{Ions}_{\text{Na}_4\mathrm{PO}_4} = 56.25\mathrm{~mmol}\) Based on the calculations, the solution containing \(75.0\mathrm{~mL}\) of \(0.150\mathrm{M}\ \mathrm{Na}_{4}\mathrm{PO}_{4}\) has the largest number of ions (\(56.25\mathrm{~mmol}\)).

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Key Concepts

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

Ionic Dissociation
When a compound dissolves in water and separates into its constituent ions, it undergoes a process called ionic dissociation. This is crucial for understanding electrolyte solutions, as it determines the number of ions present in the solution.
For example, when sodium hydroxide ( aOH) dissolves in water, it dissociates into two ions: a single sodium ion ( a^+ ) and a hydroxide ion ( OH^- ). Similarly, barium chloride ( BaCl_2 ) separates into one barium ion ( Ba^{2+} ) and two chloride ions ( Cl^- ), resulting in three ions per formula unit.
This knowledge is essential for predicting how many ions a strong electrolyte will produce in a solution, influencing properties like conductivity and reactivity.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, denoted as mol/L or M. This unit helps quantify how much of a substance is contained in a given volume.
For example, a solution with a molarity of 0.100 M means there are 0.100 moles of solute in every liter of solution.
In practical exercises like the one mentioned, molarity allows us to calculate the total ions present in a solution by multiplying the molarity by the volume and then by the number of ions produced by dissociation.
Ion Concentration
Ion concentration refers to the amount of ions in a solution, usually given in terms of moles per liter. Understanding ion concentration is critical in chemical reactions, where reaction rates may depend on the concentration of reacting ions.
In the given exercise, calculating ion concentration involves several steps:
  • Identify the dissociation pattern for each compound to know how many ions are produced.
  • Multiply the molarity of the solution by the volume of the solution.
  • Multiply the result by the number of ions released from one formula unit.
The result gives you the total moles of ions present, which is critical for comparing solutions like aOH, BaCl_2, and a_4PO_4.
Strong Electrolytes
Strong electrolytes are substances that dissociate completely into ions in solution. This means when a strong electrolyte is dissolved in water, virtually all of its formula units separate into their respective ions, enhancing the solution鈥檚 conductivity.
Compounds like aOH, BaCl_2, and a_4PO_4 are considered strong electrolytes. They provide multiple sets of ions per formula unit, vital for calculations involving ion concentration.
By understanding the behavior of strong electrolytes, one can predict the number of available ions in solution, which is important for determining property changes in chemical reactions or physical processes.

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Most popular questions from this chapter

You are given a \(1.50-\mathrm{g}\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{~mL}\) of water and then add an excess of \(0.500 \mathrm{M}\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of \(0.641 \mathrm{~g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100 M\) solution of barium chloride is mixed with \(100.0\) \(\mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of iron(III) sulfate?

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{l} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 M\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl}\) ?

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-} \longrightarrow \mathrm{X}^{2-}+\mathrm{H}_{2} \mathrm{O} $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

What volume of \(0.0521 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly \(14.20 \mathrm{~mL}\) of \(0.141 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) ? Phosphoric acid contains three acidic hydrogens.
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