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Chapter 4: Problem 19

How many total moles of ions are released when each of the following dissolves in water? (a) \(0.805 \mathrm{~mol}\) of \(\mathrm{Rb}_{2} \mathrm{SO}_{4}\) (b) \(3.85 \times 10^{-3} \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(4.03 \times 10^{19}\) formula units of \(\mathrm{Sr}\left(\mathrm{HCO}_{3}\right)_{2}\)

Short Answer

Expert verified
(a) 2.415 mol of ions, (b) 7.05 \times 10^{-5} mol of ions, (c) 2.01 \times 10^{-4} mol of ions.

Step by step solution

01

Identify the Ionic Compounds

Determine the ions produced when each compound dissolves in water.
02

Calculate the Moles of Ions for Each Compound (a)

For \(\text{Rb}_2 \text{SO}_4\), it dissociates as follows: \[ \text{Rb}_2 \text{SO}_4 (s) \rightarrow 2 \text{Rb}^+ (aq) + \text{SO}_4^{2-} (aq) \] From one mole of \(\text{Rb}_2 \text{SO}_4\), you get 2 moles of \(\text{Rb}^+\) and 1 mole of \(\text{SO}_4^{2-}\), giving a total of 3 moles of ions per mole of \(\text{Rb}_2 \text{SO}_4\). Multiply this by the given amount (0.805 moles) of \(\text{Rb}_2 \text{SO}_4\): \[ 0.805 \text{ mol} \times 3 = 2.415 \text{ mol of ions} \]
03

Convert Mass to Moles of Compound (b)

For \(\text{Ca(NO}_3)_2\), it dissociates as follows: \[ \text{Ca(NO}_3)_2 (s) \rightarrow \text{Ca}^{2+} (aq) + 2 \text{NO}_3^- (aq) \] First, convert the mass (3.85 \times 10^{-3} \text{ g}) to moles by using the molar mass of \(\text{Ca(NO}_3)_2\) (molar mass = 164.1 \text{ g/mol}): \[ \frac{3.85 \times 10^{-3} \text{ g}}{164.1 \text{ g/mol}} = 2.35 \times 10^{-5} \text{ mol of } \text{Ca(NO}_3)_2 \] Each mole of \(\text{Ca(NO}_3)_2\) produces 3 moles of ions (1 mole of \(\text{Ca}^{2+}\) and 2 moles of \(\text{NO}_3^-\)). Therefore, the total moles of ions are: \[ 2.35 \times 10^{-5} \text{ mol} \times 3 = 7.05 \times 10^{-5} \text{ mol of ions} \]
04

Convert Formula Units to Moles of Compound (c)

For \(\text{Sr(HCO}_3)_2\), it dissociates as follows: \[ \text{Sr(HCO}_3)_2 (s) \rightarrow \text{Sr}^{2+} (aq) + 2 \text{HCO}_3^- (aq) \] Convert the number of formula units (4.03 \times 10^{19}) to moles using Avogadro鈥檚 number (6.022 \times 10^{23} \text{ units/mol}): \[ \frac{4.03 \times 10^{19} \text{ units}}{6.022 \times 10^{23} \text{ units/mol}} = 6.69 \times 10^{-5} \text{ mol of } \text{Sr(HCO}_3)_2 \] Each mole of \(\text{Sr(HCO}_3)_2\) produces 3 moles of ions (1 mole of \(\text{Sr}^{2+}\) and 2 moles of \(\text{HCO}_3^-\)). Therefore, the total moles of ions are: \[ 6.69 \times 10^{-5} \text{ mol} \times 3 = 2.01 \times 10^{-4} \text{ mol of ions} \]

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Key Concepts

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

ionic compounds
An ionic compound is formed when atoms transfer electrons to or from another atom, creating ions. These compounds consist of positively charged ions (cations) and negatively charged ions (anions). For example, the compound \(\text{Rb}_2 \text{SO}_4\) contains rubidium ions \(\text{Rb}^+\) and sulfate ions \(\text{SO}_4^{2-}\). Other examples include \(\text{Ca(NO}_3)_2\) and \(\text{Sr(HCO}_3)_2\). When these ionic compounds dissolve in water, they split into their respective ions, a process known as dissociation.
dissociation in water
Dissociation in water is the process by which an ionic compound separates into its ions when dissolved. For instance, \(\text{Rb}_2 \text{SO}_4\) dissociates as follows: \[ \text{Rb}_2 \text{SO}_4 (s) \rightarrow 2 \text{Rb}^+ (aq) + \text{SO}_4^{2-} (aq) \] Similarly, \(\text{Ca(NO}_3)_2\) and \(\text{Sr(HCO}_3)_2\) dissociate: \[ \text{Ca(NO}_3)_2 (s) \rightarrow \text{Ca}^{2+} (aq) + 2 \text{NO}_3^- (aq) \ \ \text{Sr(HCO}_3)_2 (s) \rightarrow \text{Sr}^{2+} (aq) + 2 \text{HCO}_3^- (aq) \] This phenomenon allows the ions to become uniformly dispersed in the water, making them free to move and carry electric current, which is why ionic solutions conduct electricity.
mole concept
The mole concept is a fundamental chemistry principle that relates the amount of a substance to its particles. One mole of any substance contains \(6.022 \times 10^{23}\) particles, atoms, molecules, or ions. This number is known as Avogadro's number. For example, one mole of \(\text{Rb}_2 \text{SO}_4\) contains \(6.022 \times 10^{23}\) formula units of \(\text{Rb}_2 \text{SO}_4\). When calculating the number of moles, we often use the relation: \[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Avogadro's number
Avogadro's number, \(6.022 \times 10^{23}\), is crucial for converting between atomic-scale measurements and macroscopic amounts of material. For example, to find the moles from the number of formula units, we use: \[ \text{Number of moles} = \frac{\text{Number of units}}{6.022 \times 10^{23}} \] If we have \(4.03 \times 10^{19}\) formula units of \(\text{Sr(HCO}_3)_2\), the calculation would be: \[ \frac{4.03 \times 10^{19}}{6.022 \times 10^{23}} = 6.69 \times 10^{-5} \text{ moles} \]
molar mass
Molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). It's determined by summing the atomic masses of all the atoms in the compound's formula. For \(\text{Ca(NO}_3)_2\), the atomic masses are: \[ \text{Ca} = 40.08 \text{ g/mol}, \text{N} = 14.01 \text{ g/mol} \text{ (2 atoms)}, \text{O} = 16.00 \text{ g/mol} \text{ (6 atoms)} \] So, the molar mass of \(\text{Ca(NO}_3)_2\) is: \[ 40.08 + 2 \times 14.01 + 6 \times 16.00 = 164.1 \text{ g/mol} \] To find moles from a given mass, use: \[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \]

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Most popular questions from this chapter

One of the first steps in the enrichment of uranium for use in nuclear power plants involves a displacement reaction between \(\mathrm{UO}_{2}\) and aqueous HF: $$ \mathrm{UO}_{2}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{UF}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l)[\text { unbalanced }] $$ How many liters of \(2.40 \mathrm{M}\) HF will react with \(2.15 \mathrm{~kg}\) of \(\mathrm{UO}_{2}\) ?

To study a marine organism, a biologist prepares a \(1.00-\mathrm{kg}\) sample to simulate the ion concentrations in seawater. She mixes \(26.5 \mathrm{~g}\) of \(\mathrm{NaCl}, 2.40 \mathrm{~g}\) of \(\mathrm{MgCl}_{2}, 3.35 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}, 1.20 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}, 1.05 \mathrm{~g}\) of \(\mathrm{KCl}, 0.315 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3},\) and \(0.098 \mathrm{~g}\) of \(\mathrm{NaBr}\) in distilled water. (a) If the density of the solution is \(1.025 \mathrm{~g} / \mathrm{cm}^{3}\). what is the molarity of each ion? (b) What is the total molarity of alkali metal ions? (c) What is the total molarity of alkaline earth metal ions? (d) What is the total molarity of anions?

When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write balanced molecular, total ionic, and net ionic equations: (a) Potassium carbonate + barium hydroxide (b) Aluminum nitrate \(+\) sodium phosphate

Calculate each of the following quantities: (a) Mass (g) of solute needed to make \(475 \mathrm{~mL}\) of \(5.62 \times 10^{-2} \mathrm{M}\) potassium sulfate (b) Molarity of a solution that contains \(7.25 \mathrm{mg}\) of calcium chloride in each milliliter (c) Number of \(\mathrm{Mg}^{2+}\) ions in each milliliter of \(0.184 \mathrm{M}\) magnesium bromide

In 1997 and \(2009,\) at United Nations conferences on climate change, many nations agreed to expand their research efforts to develop renewable sources of carbon-based fuels. For more than a quarter century, Brazil has been engaged in a program to replace gasoline with ethanol derived from the root crop manioc (cassava). (a) Write separate balanced equations for the complete combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and of gasoline (represented by the formula \(\mathrm{C}_{8} \mathrm{H}_{18}\) ). (b) What mass (g) of oxygen is required to burn completely \(1.00 \mathrm{~L}\) of a mixture that is \(90.0 \%\) gasoline \((d=0.742 \mathrm{~g} / \mathrm{mL})\) and \(10.0 \%\) ethanol \((d=0.789 \mathrm{~g} / \mathrm{mL})\) by volume? (c) If \(1.00 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) occupies \(22.4 \mathrm{~L}\), what volume of \(\mathrm{O}_{2}\) is needed to burn \(1.00 \mathrm{~L}\) of the mixture? (d) Air is \(20.9 \% \mathrm{O}_{2}\) by volume. What volume of air is needed to burn \(1.00 \mathrm{~L}\) of the mixture?
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