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Chapter 4: Problem 110

Bone was dissolved in hydrochloric acid, giving \(50.0\) \(\mathrm{mL}\) of solution containing calcium chloride, \(\mathrm{CaCl}_{2}\). To precipitate the calcium ion from the resulting solution, an excess of potassium oxalate was added. The precipitate of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), weighed \(1.437 \mathrm{~g} .\) What was the molarity of \(\mathrm{CaCl}_{2}\) in the solution?

Short Answer

Expert verified
The molarity of \(\mathrm{CaCl_2}\) in the solution is 0.2244 M.

Step by step solution

01

Understanding the Reaction

The reaction involved is the precipitation of calcium ions as calcium oxalate. The chemical reaction is: \(\mathrm{Ca^{2+} + C_2O_4^{2-} \rightarrow CaC_2O_4}\). Each mole of \(\mathrm{CaCl_2}\) provides one mole of \(\mathrm{Ca^{2+}}\), which reacts with one mole of \(\mathrm{C_2O_4^{2-}}\) to form calcium oxalate.
02

Convert Mass of Precipitate to Moles

Calculate the molar mass of \(\mathrm{CaC_2O_4}\). It is composed of calcium (Ca) with a mass of 40.08 \(\mathrm{g/mol}\), carbon (C) with a mass of 12.01 \(\mathrm{g/mol}\), oxygen (O) with a mass of 16.00 \(\mathrm{g/mol}\): \[Molar \text{ Mass of } \mathrm{CaC_2O_4} = 40.08 + 2(12.01) + 4(16.00) = 128.10 \mathrm{g/mol}\]Now, use this molar mass to convert the 1.437 g of calcium oxalate to moles:\[Moles \text{ of } \mathrm{CaC_2O_4} = \frac{1.437 \text{ g}}{128.10 \mathrm{g/mol}} = 0.01122 \text{ moles}\]
03

Relate Moles of Precipitate to Moles of Calcium Chloride

Since each mole of \(\mathrm{Ca^{2+}}\) corresponds to one mole of \(\mathrm{CaC_2O_4}\), the moles of \(\mathrm{Ca^{2+}}\) are equal to the moles of \(\mathrm{CaC_2O_4}\). This means that the moles of \(\mathrm{CaCl_2}\) are also 0.01122 moles, as \(\mathrm{CaCl_2}\) dissociates completely in solution providing \(\mathrm{Ca^{2+}}\) ions.
04

Calculating Molarity

The molarity of a solution is defined as the number of moles of solute per liter of solution. The solution has a volume of 50.0 mL, which is 0.050 L. Calculate the molarity as:\[ \text{Molarity of } \mathrm{CaCl_2} = \frac{0.01122 \text{ moles}}{0.050 \text{ L}} = 0.2244 \text{ M} \]

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Key Concepts

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

Precipitation Reaction
A precipitation reaction occurs when two soluble substances in a solution combine to form an insoluble solid, known as the precipitate. In the context of the given exercise, the reaction between calcium ions from calcium chloride ( CaCl_2 ) and oxalate ions from potassium oxalate results in the formation of calcium oxalate ( CaC_2O_4 ) as a solid precipitate.
  • Precipitation reactions are a type of double displacement reaction.
  • The formed precipitate can be removed from the solution via filtration.
  • Indicators or changes in color can help identify when a reaction has reached completion.
This type of reaction is crucial in various chemical processes, including water treatment and mineral extraction. Understanding the stoichiometry of these reactions is essential in predicting the formation and amount of precipitate formed.
Calcium Oxalate
Calcium oxalate ( CaC_2O_4 ) is a compound formed by the combination of calcium and oxalate ions. It appears as a white crystalline solid, making it easily identifiable when it precipitates from a solution.
  • Calcium oxalate is commonly found in nature, including in plant tissues and as a component of kidney stones in humans.
  • In laboratory settings, it's often used to demonstrate precipitation reactions.
  • The solubility of calcium oxalate is quite low, causing it to form quickly as a precipitate when calcium and oxalate ions are present together in solution.
Understanding calcium oxalate's properties and behavior is important in a variety of fields, including medicine, where it helps in understanding kidney stone formation and treatment.
Molar Mass Calculation
The calculation of molar mass is a fundamental concept in chemistry that allows us to relate the mass of a substance to the number of moles of that substance. This relation is crucial for calculations involving chemical reactions and stoichiometry.
In the exercise, the molar mass of calcium oxalate (CaC_2O_4) is calculated by summing the atomic masses of its constituent elements:
  • Calcium (Ca): 40.08 \( ext{g/mol}\)
  • Carbon (C): 2 atoms x 12.01 \( ext{g/mol}\) = 24.02 \( ext{g/mol}\)
  • Oxygen (O): 4 atoms x 16.00 \( ext{g/mol}\) = 64.00 \( ext{g/mol}\)
Adding all these gives a total molar mass of 128.10 \( ext{g/mol}\) for calcium oxalate.
  • This value is used to convert between grams of a compound and moles, a necessary step in calculating reactant or product quantities in a chemical reaction.
  • Molar mass is determined using the atomic weights found on the periodic table, with each element contributing proportionally to the total.
Having a clear understanding of molar mass calculations is indispensable for any student attempting to solve chemical reaction problems.

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Most popular questions from this chapter

Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(81.15 \mathrm{~g}\) of mercury(II) nitrate and \(8.52 \mathrm{~g}\) of hydrogen sulfide are mixed in \(550.0 \mathrm{~g}\) of water to form \(58.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?

Barium carbonate is the source of barium compounds. It is produced in an aqueous precipitation reaction from barium sulfide and sodium carbonate. (Barium sulfide is a soluble compound obtained by heating the mineral barite, which is barium sulfate, with carbon.) What are the molecular equation and net ionic equation for the precipitation reaction? A solution containing \(33.9 \mathrm{~g}\) of barium sulfide requires \(21.2 \mathrm{~g}\) of sodium carbonate to react completely with it, and \(15.6 \mathrm{~g}\) of sodium sulfide is produced in addition to whatever barium carbonate is obtained. How many grams of barium sulfide are required to produce \(10.0\) tons of barium carbonate? (One ton equals 2000 pounds.)

An aqueous solution contains \(4.00 \% \mathrm{NH}_{3}\) (ammonia) by mass. The density of the aqueous ammonia is \(0.979 \mathrm{~g} / \mathrm{mL}\). What is the molarity of \(\mathrm{NH}_{3}\) in the solution?

Complete and balance each of the following molecular equations, including phase labels, if a reaction occurs. Then write the net ionic equation. If no reaction occurs, write \(N R\) after the arrow. a. \(\mathrm{Sr}(\mathrm{OH})_{2}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \longrightarrow\) b. \(\mathrm{NH}_{4} \mathrm{I}+\mathrm{CsCl} \longrightarrow\) c. \(\mathrm{NaNO}_{3}+\mathrm{CsCl} \longrightarrow\) d. \(\mathrm{NH}_{4} \mathrm{I}+\mathrm{AgNO}_{3} \longrightarrow\)

Balance the following oxidation-reduction reactions by the half-reaction method. a. \(\mathrm{FeI}_{3}(a q)+\mathrm{Mg}(s) \longrightarrow \mathrm{Fe}(s)+\mathrm{MgI}_{2}(a q)\) b. \(\mathrm{H}_{2}(g)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Ag}(s)+\mathrm{H}^{+}(a q)\)
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