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A saturated solution of \(\mathrm{CaF}_{2}\) contains \(.00168 \mathrm{~g}\) of \(\mathrm{CaF}_{2}\) per \(100 \mathrm{~g}\) of water. Determine the \(\mathrm{K}_{\mathrm{sp}}\).

Short Answer

Expert verified
The Ksp of 颁补贵鈧 is approximately 4.00 x 10^{-11}.

Step by step solution

01

Convert solubility into molarity

We need to convert the given solubility of 颁补贵鈧 (0.00168 g per 100 g of water) into molarity. First, we find the molar mass of 颁补贵鈧: Molar mass of 颁补贵鈧 = 40.08 g/mol (for Ca) + 2 * 19 g/mol (for F) = 78.08 g/mol Now, let's assume we have 100 g of water, which corresponds to approximately 100 mL in volume, since the density of water is close to 1 g/mL. Thus, we can calculate the molarity of the saturated 颁补贵鈧 solution: Moles of 颁补贵鈧 = (0.00168 g) / (78.08 g/mol) = 2.15 x 10^{-5} mol Molarity of 颁补贵鈧 = (2.15 x 10^{-5} mol) / (0.100 L) = 2.15 x 10^{-4} M
02

Write the balanced dissolution equation and Ksp expression

The balanced dissolution equation for 颁补贵鈧 is: 颁补贵鈧 (s) 鈬 Ca虏鈦 (aq) + 2F鈦 (aq) According to the stoichiometry of this equation, 1 mole of 颁补贵鈧 produces 1 mole of Ca虏鈦 ions and 2 moles of F鈦 ions when dissolved. Therefore, if the molarity of the dissolved 颁补贵鈧 is 2.15 x 10^{-4} M, the molarities of the ions will be: [Ca虏鈦篯 = 2.15 x 10^{-4} M [F鈦籡 = 2 (2.15 x 10^{-4}) M= 4.3 x 10^{-4} M Now, we can write the Ksp expression for 颁补贵鈧: Ksp = [Ca虏鈦篯 [F鈦籡虏
03

Substitute the concentrations and solve for Ksp

Substitute the molarities of the ions from step 2 into the Ksp expression: Ksp = (2.15 x 10^{-4})(4.3 x 10^{-4})虏 Now, calculate the Ksp value: Ksp 鈮 4.00 x 10^{-11} Therefore, the Ksp of 颁补贵鈧 is approximately 4.00 x 10^{-11}.

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Key Concepts

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

Molarity
Molarity is an important concept in chemistry that measures the concentration of a solute in a solution. In simple terms, it tells us how much of a particular substance is dissolved in a certain volume of liquid.
For instance, when we talk about a molar solution, we're referring to the number of moles of solute present in one liter of solution.
This unit is expressed in moles per liter (mol/L), commonly referred to as molarity (M). To calculate molarity, you use the following formula:
  • Molarity (M) = Moles of solute / Volume of solution in liters
Knowing the molarity of a solution allows chemists to predict how substances will react with each other in chemical reactions, as it directly relates to the number of particles available to interact.
In the context of our problem with 颁补贵鈧, we convert given solubility data to molarity to understand how much Ca虏鈦 and F鈦 ions are available in a saturated solution.
Dissolution Equation
The dissolution equation represents the process of a solid substance dissolving in a solvent to form a solution. This involves the separation of the solute into its individual ions or molecules, which then become surrounded by solvent particles.
In the case of calcium fluoride ( 颁补贵鈧), the dissolution equation is: 颁补贵鈧 (s) 鈬 Ca虏鈦 (aq) + 2F鈦 (aq) This balanced chemical equation tells us that each formula unit of 颁补贵鈧 that dissolves in water breaks down into one calcium ion (Ca虏鈦) and two fluoride ions (F鈦).
The double arrow signifies equilibrium between the dissolved ions and the undissolved solid.
The extent to which 颁补贵鈧 dissolves in water is governed by its solubility product constant (K_sp), which provides a quantitative measure of the solubility under specific conditions.
颁补贵鈧
Calcium fluoride ( 颁补贵鈧) is an ionic compound consisting of calcium and fluoride ions. It appears as a white crystalline solid with a highly stable ionic lattice structure.
This stability limits its solubility in water, making 颁补贵鈧 only slightly soluble. When 颁补贵鈧 is placed in water, a small portion of it dissolves to produce calcium ions (Ca虏鈦) and fluoride ions (F鈦). This dissolution process establishes an equilibrium between solid 颁补贵鈧 and the dissociated ions in solution.
The solubility product constant ( K_sp) is specific to 颁补贵鈧 and provides insight into how much of this compound will dissolve in water under given conditions. Understanding 颁补贵鈧's behavior in aqueous solutions is crucial for fields like material science and chemical engineering, where solubility affects both product formation and environmental impacts.
Ionic Concentrations
Ionic concentrations in a solution refer to the amount of each ion present as measured by their molarity.
These concentrations are critical in determining the solubility product constant ( K_sp) as well as predicting the outcome of chemical reactions in solutions. In the dissolution of 颁补贵鈧, we determined earlier that for each mole of 颁补贵鈧 that dissolves, one mole of Ca虏鈦 ions and two moles of F鈦 ions are produced:
  • Molar concentration of Ca虏鈦, [Ca虏鈦篯 = 2.15 x 10^{-4} M
  • Molar concentration of F鈦, [F鈦籡 = 2 x 2.15 x 10^{-4} M = 4.3 x 10^{-4} M
Calculating these ionic concentrations allows us to substitute them into the K_sp expression: K_sp = [Ca虏鈦篯 [F鈦籡虏 Knowing the K_sp value helps predict whether a precipitate will form in a solution under specific conditions, as it gives a measure of the equilibrium between the dissolved ions and the undissolved solid.

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Most popular questions from this chapter

\(0.01\) liter of \(0.3 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) is mixed with \(0.02\) liters of a solution that is initially \(0.1 \mathrm{M} \mathrm{ca}^{++}\) and \(0.1 \mathrm{M} \mathrm{Sr}^{++}\). Given that \(\mathrm{K}_{s p}=2.4 \times 10^{-5}\) for \(\mathrm{CaSO}_{4}\) and \(\mathrm{K}_{\mathrm{sp}}=7.6 \times 10^{-7}\) for \(\mathrm{SrSO}_{4}\) what is the final concentration of \(\mathrm{Ca}^{++}, \mathrm{Sr}^{++}\), and \(\mathrm{SO}^{--} 4\) in this solution?

The solubility of iodine in pure water is \(0.0279 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water. In a solution that is originally \(0.10 \mathrm{M} \mathrm{KI}\), it is possible to dissolve, at the maximum, \(1.14 \mathrm{~g}\) of iodine per \(100 \mathrm{~g}\) of solution. What is \(\mathrm{K}\) for the reaction \(\mathrm{I}_{2}+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}^{-}_{3} ?\)

Assuming the \(\mathrm{K}_{\mathrm{sp}}\) for radium sulfate is \(4 \times 10^{-11}\), what is its solubility in (a) pure water, and (b) \(1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) ?

How many grams of \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\) (calcium oxalate) will dissolve in water to form \(1.0\) liter of saturated solution? The \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\) is \(2.5 \times 10^{-9} \mathrm{~mole}^{2} / \mathrm{liter}^{2}\)

A chemist mixes equal volumes of \(.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) and \(.001 \mathrm{M} \mathrm{BaCl}_{2}\) together. Assuming the \(\mathrm{K}_{\mathrm{sp}}\) of \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) is \(1.2 \times 10^{-7}\), will a precipitate form?

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