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A \(1.00-\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) contains \(0.0061 \mathrm{~g}\) of \(\mathrm{CaC}_{2} \mathrm{O}_{4} .\) Calculate the solubilityproduct constant for this salt at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility product constant (Ksp) for calcium oxalate (CaC2O4) in a 1.00 L solution containing 0.0061 g of the salt at 25°C can be determined by first calculating the molar mass of CaC2O4, which is \(40.08 + (2 \times 12.01) + (4 \times 16.00)\,\text{g/mol}\). Then, find the number of moles of CaC2O4 by dividing its mass by its molar mass, followed by finding the concentration of Ca²⁺ and C₂O₄²⁻ ions in the solution, which will be equal to the concentration of CaC2O4. Finally, calculate the Ksp using the formula Ksp = [Ca²⁺][C₂O₄²⁻], which is equal to the square of the concentration of ions.

Step by step solution

01

Find the molar mass of calcium oxalate

To find the concentration of the ions in the solution, we need the molar mass of calcium oxalate (CaC2O4). The molar mass of CaC2O4 can be found by adding up the molar masses of its constituent elements: Calcium (Ca): 40.08 g/mol Carbon (C): 12.01 g/mol (there are two Carbon atoms, so multiply by 2) Oxygen (O): 16.00 g/mol (there are four Oxygen atoms, so multiply by 4) Molar mass of CaC2O4 = \(40.08 + (2 \times 12.01) + (4 \times 16.00)\)
02

Calculate the number of moles of calcium oxalate

Now that we have the molar mass of calcium oxalate, we can find the number of moles of the compound in the 1.00 L solution. Moles of CaC2O4 = mass / molar mass = \(0.0061\,\text{g} / \left( 40.08 + (2 \times 12.01) + (4 \times 16.00) \right)\,\text{g/mol}\)
03

Find the concentration of Ca²⁺ and C₂O₄²⁻ ions in the solution

Calcium oxalate (CaC2O4) dissociates into one Ca²⁺ ion and one C₂O₄²⁻ ion. Thus, the concentration of each ion will be equal to the concentration of CaC2O4 in the solution. Concentration of ions = moles / volume = moles / 1.00 L
04

Calculate the solubility product constant (Ksp)

Now we have found the concentration of each ion, we can calculate the solubility product constant (Ksp) using the following equation: Ksp = [Ca²⁺][C₂O₄²⁻] Ksp = (Concentration of ions)² Use the formula and the concentration of ions calculated in step 3 to find the Ksp.

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Key Concepts

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

Calcium Oxalate Dissociation
Calcium oxalate, described by the chemical formula \(\text{CaC}_2\text{O}_4\), is a slightly soluble compound. When it is dissolved in water, it undergoes dissociation. This process involves splitting the compound into its constituent ions.
The dissociation equation is as follows: \[\text{CaC}_2\text{O}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{C}_2\text{O}_4^{2-} (aq)\]In this equation, the solid (\(s\)) form of calcium oxalate separates into one calcium ion (\(\text{Ca}^{2+}\)) and one oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)) in aqueous solution (\(aq\)).
This is key to understanding its solubility behavior. Such equilibrium processes determine the concentration of ions in a solution. It's important to note that a saturated solution contains the maximum amount of solute that can dissolve at a given temperature. Once the saturation point is reached, the compound will either stay as ions or begin to precipitate back into the solid form. This balance is crucial in calculating the solubility product constant (Ksp).
The Ksp will help quantify how much of this compound can be dissolved before forming a precipitate.
Molar Mass Calculation
Molar mass is fundamental when calculating the number of moles of a compound from a given mass. To find the molar mass of calcium oxalate \((\text{CaC}_2\text{O}_4)\), we add the molar masses of its constituent elements:
  • Calcium (Ca) has a molar mass of \(40.08\ \text{g/mol}\).
  • Carbon (C), with two atoms in the compound, contributes \(2 \times 12.01\ \text{g/mol}\).
  • Oxygen (O), with four atoms, adds \(4 \times 16.00\ \text{g/mol}\).
The formula to sum these values is:\[\text{Molar mass of } \text{CaC}_2\text{O}_4 = 40.08 + (2 \times 12.01) + (4 \times 16.00)\]After performing the calculations, the molar mass of \(\text{CaC}_2\text{O}_4\) is determined. Understanding this value allows us to convert from grams to moles, which is essential in determining the solute concentration in the solution. This step plays a critical role in further calculations of ion concentrations and ultimately the solubility product constant.
Ion Concentration in Solutions
Ion concentration refers to the number of ions of a given species in a solution. For calcium oxalate, the solubility in water results in the formation of calcium ions \((\text{Ca}^{2+})\) and oxalate ions \((\text{C}_2\text{O}_4^{2-})\).
The concentration of ions in a solution directly relates to the number of moles of the dissolved compound. We first calculate the number of moles by dividing the mass of calcium oxalate in the solution by its molar mass:\[\text{Moles of } \text{CaC}_2\text{O}_4 = \frac{0.0061\ \text{g}}{\text{Molar mass of } \text{CaC}_2\text{O}_4}\]In a 1.00 L solution, this number of moles corresponds to the molar concentration of the ions.
Since calcium oxalate dissociates into one calcium ion and one oxalate ion per formula unit, the concentration of \(\text{Ca}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\) will be equal to each other, and equal to the concentration of \(\text{CaC}_2\text{O}_4\) itself. Finally, the solubility product constant \(\text{K}_{sp}\) is calculated by the formula:\[\text{K}_{sp} = [\text{Ca}^{2+}][C_2O_4^{2-}]\]This equation helps us understand how soluble a compound is in solution, with higher \(\text{K}_{sp}\) values indicating greater solubility.

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Most popular questions from this chapter

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .(\mathrm{a})\) Write the equilibrium equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and \(B\) ? (c) What is the value of \(K_{b}\) for B?

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 \mathrm{M}\) HCl. Calculate the pH after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},(\mathbf{c}) 59.0 \mathrm{~mL},(\mathrm{~d}) 60.0 \mathrm{~mL},\) (e) \(61.0 \mathrm{~mL}\) (f) \(65.0 \mathrm{~mL}\).

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20} .\) The \(\mathrm{AsO}_{4}{ }^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{p} K_{a 1}=2.22\right.\); \(\left.\mathrm{p} K_{a 2}=6.98 ; \mathrm{p} K_{a 3}=11.50\right) .\) When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{s p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}^{3-}\right]\) Why was this a mistake?

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\) (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+},\) (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+}\). Suggest how each mixture might be separated.

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