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Chapter 4: Problem 37

A solution is prepared by dissolving \(10.8 \mathrm{~g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{~mL}\) of stock solution. \(\overline{\mathrm{A}}\) \(10.00-\mathrm{mL}\) sample of this stock solution is added to \(50.00 \mathrm{~mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Short Answer

Expert verified
The final concentration of ammonium ions (\(NH_4^+\)) in the solution is \(0.622~M\), while the final concentration of sulfate ions (\(SO_4^{2-}\)) in the solution is \(0.311~M\).

Step by step solution

01

Find the initial concentration of the stock solution

First, we need to find the concentration of the stock solution (ammonium sulfate) in moles per liter (M). To do this, we will convert the mass of ammonium sulfate (10.8 g) to moles and then divide by the volume of the solution (100.0 mL) converted to liters. The molar mass of ammonium sulfate: \((NH_4)_2SO_4 = (2 \times (14.01 + 4 \times 1.01)) + (32.07 + 4 \times 16.00)\) \((NH_4)_2SO_4 = (2 \times (14.01 + 4.04)) + (32.07 + 64.00)\) Now, we can find the moles of ammonium sulfate: moles \(= \frac{mass}{molar~mass}\)
02

Find the concentration of the stock solution

Next, we need to find the concentration of the stock solution in moles per liter (M). To do this, divide the number of moles found in step 1 by the volume of the solution (converted to liters). Concentration \(= \frac{moles}{volume(in~liters)}\)
03

Calculate the concentration of ions in the final solution

For every one molecule of ammonium sulfate, there are two ammonium ions (\(NH_4^+\)) and one sulfate ion (\(SO_4^{2-}\)). Therefore, the concentrations of ammonium ions and sulfate ions in the stock solution will be twice and equal to the concentration of the ammonium sulfate, respectively. Now, to find the concentrations of both ions in the final solution, we will have to use the dilution formula: \(C_1V_1 = C_2V_2\) Where: \(C_1\) = initial concentration of the ion \(V_1\) = initial volume of the solution \(C_2\) = final concentration of the ion \(V_2\) = final volume of the solution
04

Calculate the final concentrations of ammonium ions and sulfate ions

Using the dilution formula from Step 3, we can find the final concentrations of the ammonium and sulfate ions in the final solution by substituting their initial concentrations and volumes. For ammonium ions: \(C_1 = 2 \times \text{concentration of ammonium sulfate}\) \(V_1 = 10.00~mL\) \(V_2 = 10.00 + 50.00~mL\) For sulfate ions: \(C_1 = \text{concentration of ammonium sulfate}\) \(V_1 = 10.00~mL\) \(V_2 = 10.00 + 50.00~mL\) The final concentrations can be determined by substituting these values into the dilution formula.

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Most popular questions from this chapter

What mass of silver chloride can be prepared by the reaction of \(100.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) silver nitrate with \(100.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

A \(50.00-\mathrm{mL}\) sample of aqueous \(\mathrm{Ca}(\mathrm{OH})_{2}\) requires \(34.66 \mathrm{~mL}\) of a \(0.944-M\) nitric acid for neutralization. Calculate the concentration (molarity) of the original solution of calcium hydroxide.

What volume of \(0.100 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from \(150.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) ?

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\). A \(0.205-g\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is \(0.298 \mathrm{~g}\). What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}{ }^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?
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