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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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Most popular questions from this chapter

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. a. ammonium sulfate and barium nitrate b. lead(II) nitrate and sodium chloride c. sodium phosphate and potassium nitrate d. sodium bromide and rubidium chloride e. copper(II) chloride and sodium hydroxide

The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times\) \(10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 113 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

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?

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?
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