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Chapter 4: Problem 34

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?

Short Answer

Expert verified
The molarity of each ion is calculated in step 4. The total alkali metal ion molarity is 0.4825 M, total alkaline earth metal ion molarity is 0.0657 M, and total anion molarity is 0.6203 M.

Step by step solution

01

- Calculate the Volume of the Solution

Given the density (\rho) of the solution is \(1.025 \mathrm{~g/cm}^{3}\). The mass of the solution is \(1.00 \mathrm{~kg} = 1000 \mathrm{~g}\). Using the formula \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\), we find the volume: \[ \text{Volume} = \frac{1000\mathrm{~g}}{1.025\mathrm{~g/cm}^{3}} = 975.61 \mathrm{~cm}^{3} \approx 0.9756 \mathrm{~L} \].
02

- Convert Mass of Each Compound to Moles

Use the molar mass of each compound: \(\mathrm{NaCl}: 26.5 \mathrm{~g} / 58.44 \mathrm{~g/mol}\), \(\mathrm{MgCl}_{2}: 2.4 \mathrm{~g} / 95.21 \mathrm{~g/mol}\), \(\mathrm{MgSO}_{4}: 3.35 \mathrm{~g} / 120.37 \mathrm{~g/mol}\), \(\mathrm{CaCl}_{2}: 1.2 \mathrm{~g} / 110.99 \mathrm{~g/mol}\), \(\mathrm{KCl}: 1.05 \mathrm{~g} / 74.55 \mathrm{~g/mol}\), \(\mathrm{NaHCO}_{3}: 0.315 \mathrm{~g} / 84.01 \mathrm{~g/mol}\), \(\mathrm{NaBr}: 0.098 \mathrm{~g} / 102.89 \mathrm{~g/mol}\). Find the moles of each compound: \[ \text{Moles of NaCl} = \frac{26.5}{58.44} = 0.4539 \mathrm{~mol} \] \[ \text{Moles of MgCl}_2 = \frac{2.4}{95.21} = 0.0252 \mathrm{~mol} \] \[ \text{Moles of MgSO}_4 = \frac{3.35}{120.37} = 0.0278 \mathrm{~mol} \] \[ \text{Moles of CaCl}_2 = \frac{1.2}{110.99} = 0.0108 \mathrm{~mol} \] \[ \text{Moles of KCl} = \frac{1.05}{74.55} = 0.0141 \mathrm{~mol} \] \[ \text{Moles of NaHCO}_3 = \frac{0.315}{84.01} = 0.00375 \mathrm{~mol} \] \[ \text{Moles of NaBr} = \frac{0.098}{102.89} = 0.00095 \mathrm{~mol} \].
03

- Determine Ion Contributions

For each compound, determine how many moles of each ion are produced: \(\mathrm{NaCl}\) produces \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\), \(\mathrm{MgCl}_{2}\) produces \(\mathrm{Mg}^{2+}\) and \(2\mathrm{Cl}^-\), \(\mathrm{MgSO}_{4}\) produces \(\mathrm{Mg}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{CaCl}_{2}\) produces \(\mathrm{Ca}^{2+}\) and \(2\mathrm{Cl}^-\), \(\mathrm{KCl}\) produces \(\mathrm{K}^+\) and \(\mathrm{Cl}^-\), \(\mathrm{NaHCO}_{3}\) produces \(\mathrm{Na}^+\) and \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{NaBr}\) produces \(\mathrm{Na}^+\) and \(\mathrm{Br}^-\).
04

- Calculate Molarity of Each Ion

Using the volume found in Step 1 and the moles of each ion from Step 3, calculate the molarity for each ion: \[ \mathrm{[Na}^+] = \frac{0.4539 + 0.00375 + 0.00095}{0.9756} = 0.468 \mathrm{~M} \] \[ \mathrm{[Mg}^{2+}] = \frac{0.0252 + 0.0278}{0.9756} = 0.0546 \mathrm{~M} \] \[ \mathrm{[Ca}^{2+}] = \frac{0.0108}{0.9756} = 0.0111 \mathrm{~M} \] \[ \mathrm{[K}^+] = \frac{0.0141}{0.9756} = 0.0145 \mathrm{~M} \] \[ \mathrm{[Cl}^-] = \frac{0.4539 + 2 \times 0.0252 + 0.0278 + 2 \times 0.0108 + 0.0141}{0.9756} = 0.587 \mathrm{~M} \] \[ \mathrm{[SO}_{4}^{2-}] = \frac{0.0278}{0.9756} = 0.0285 \mathrm{~M} \] \[ \mathrm{[HCO}_{3}^{-}] = \frac{0.00375}{0.9756} = 0.00384 \mathrm{~M} \] \[ \mathrm{[Br}^{-}] = \frac{0.00095}{0.9756} = 0.000974 \mathrm{~M} \].
05

- Calculate Total Molarity of Alkali Metal Ions

Sum the molarities of \(\mathrm{Na}^+\) and \(\mathrm{K}^+\) to find the total molarity: \[ \mathrm{[Total~Alkali~Metal~Ions]} = 0.468 + 0.0145 = 0.4825 \mathrm{~M} \].
06

- Calculate Total Molarity of Alkaline Earth Metal Ions

Sum the molarities of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\): \[ \mathrm{[Total~Alkaline~Earth~Metal~Ions]} = 0.0546 + 0.0111 = 0.0657 \mathrm{~M} \].
07

- Calculate Total Molarity of Anions

Sum the molarities of \(\mathrm{Cl}^-\), \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{Br}^-\): \[ \mathrm{[Total~Anions]} = 0.587 + 0.0285 + 0.00384 + 0.000974 = 0.6203 \mathrm{~M} \].

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Key Concepts

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

Density
Density is a measure of how much mass is contained in a given volume. It's calculated using the formula: \[Density = \frac{Mass}{Volume}\] In this exercise, the density of the solution is given as \ 1.025 \mathrm{~g/cm^3}, with a mass of \ 1.00 \mathrm{~kg} \ or \ 1000 \mathrm{~g}. To find the volume of the solution, we divide the mass by the density: \[Volume = \frac{1000 \mathrm{~g}}{1.025 \mathrm{~g/cm^3}} = 975.61 \mathrm{~cm^3} \approx 0.9756 \mathrm{~L}\]. This volume is crucial for subsequent calculations, including determining ion molarity.
Molar Mass
Molar mass refers to the mass of one mole of a given substance, expressed in grams per mole (g/mol). It is a critical component in converting the mass of a compound to moles. For instance, in the exercise, the molar mass of \ NaCl is \frac{58.44 \text{g/mol}} \ and the mass is \ 26.5 \mathrm{~g} \. To find moles of \ NaCl, we use: \[ \text{Moles of NaCl} = \frac{26.5 \mathrm{~g}}{58.44 \mathrm{~g/mol}} = 0.4539 \mathrm{~mol} \]. This conversion is performed for each compound in the solution, setting up critical data for determining ion concentrations.
Ion Concentration
Ion concentration, or molarity, measures the number of moles of an ion in a solution per liter of solution. It's determined using the moles of each ion and the solution's volume. For example, the moles of \ \text{Na}^+ \ from \ NaCl, NaHCO_{3} \ and \ NaBr \ are summed and divided by the volume: \[ \text{[Na}^{+}] = \frac{0.4539 + 0.00375 + 0.00095}{0.9756} = 0.468 \mathrm{~M} \]. This concept is applied to calculate the molarity of each ion in the solution, including \ Cl^{-}, Mg^{2+}, Ca^{2+}, K^{+}, SO_{4}^{2-}, HCO_{3}^{-}, and Br^{-}.
Alkali Metal Ions
Alkali metals are elements in group 1 of the periodic table, which include \ Li, Na, K, Rb, Cs, \ and \ Fr \. In this exercise, \ Na \ and \ K \ are the alkali metal ions present. Their total molarity is the sum of their individual molarities: \[ \text{[Total~Alkali~Metal~Ions]} = 0.468 + 0.0145 = 0.4825 \mathrm{~M} \]. This tells us how concentrated these alkali metal ions are in the solution.
Alkaline Earth Metal Ions
Alkaline earth metals are found in group 2 of the periodic table, and include \ Be, Mg, Ca, Sr, Ba, \ and \ Ra \. Here, \ \text{Mg}^{2+} \ and \ \text{Ca}^{2+} \ ions are present. Their total molarity is calculated by summing their individual molarities: \[ \text{[Total~Alkaline~Earth~Metal~Ions]} = 0.0546 + 0.0111 = 0.0657 \mathrm{~M} \]. Understanding this helps in determining the concentration of these metal ions in the solution.
Anions
Anions are negatively charged ions. In this context, the key anions are \ Cl^{-}, SO_{4}^{2-}, HCO_{3}^{-}, \ and \ Br^{-} \. Each anion's concentration is calculated by finding the moles from each source compound and dividing by the solution volume. The total molarity of anions is the sum of their concentrations: \[ \text{[Total~Anions]} = 0.587 + 0.0285 + 0.00384 + 0.000974 = 0.6203 \mathrm{~M} \]. This figure represents the overall concentration of anions in the solution.
Solution Volume
The solution volume is an essential factor for calculating molarity. It's derived from the mass and density of the solution as: \[ \text{Volume} = \frac{1000 \mathrm{~g}}{1.025 \mathrm{~g/cm^3}} = 0.9756 \mathrm{~L} \]. This volume is used in all subsequent molarity calculations, forming the denominator for converting moles of each ion into molarity. Ensuring accurate volume measurements is crucial for precise chemical calculations.

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How many grams of chlorine gas can be produced from the electrolytic decomposition of \(874 \mathrm{~g}\) of calcium chloride? Name and calculate the mass (in \(\mathrm{g}\) ) of the other product.

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