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Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.

Short Answer

Expert verified
The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} M\).

Step by step solution

01

Write Balanced Chemical Equation

For the dissolution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in water, we can write the balanced chemical equation as follows: \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca^{2+}}(aq) + 2\mathrm{PO^{3-}_{4}}(aq)\]
02

Write the Solubility Product Expression

Given the balanced chemical equation, we can now write the solubility product expression \(K_{sp}\) for the reaction. \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO^{3-}_{4}}]^2\]
03

Set Up the ICE Table

To find the equilibrium concentrations, we can set up an ICE (Initial, Change, Equilibrium) table for the reaction: \[\begin{array}{c|ccc} & \mathrm{Ca}^{2+} & \mathrm{PO}_{4}^{3-} \\ \hline \text{Initial} & 0 & 0.20 \\ \text{Change} & +3s & +2s \\ \text{Equilibrium} & 3s & 0.20 + 2s \end{array}\] Where s is solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).
04

Substituting the Equilibrium Concentrations into \(K_{sp}\) Value

Substituting the equilibrium concentrations into the solubility product expression, and using the given \(K_{sp} = 1.3 \times 10^{-32}\): \[1.3 \times 10^{-32} = (3s)^3(0.20 + 2s)^2\]
05

Solve for Solubility

Since \(K_{sp}\) is a very small number, we can approximate that \(0.20 + 2s \approx 0.20\). Therefore, the equation becomes: \[1.3 \times 10^{-32} = (3s)^3(0.20)^2\] Solve for s: \(s \approx 4.95 \times 10^{-11} \mathrm{M}\) The solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is approximately \(4.95 \times 10^{-11} \mathrm{M}\).

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Key Concepts

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

Solubility Product (Ksp)
The solubility product, denoted as \( K_{sp} \), is a constant that provides insight into the solubility of a sparingly soluble salt in a solution. It is essentially the equilibrium constant specific for a solvation reaction, expressing the product of the concentrations of the ions that the substance dissociates into when in a saturated solution. For example, in the dissolution of calcium phosphate \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \):
  • The chemical reaction is
    \[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Ca^{2+}}(aq) + 2\mathrm{PO^{3-}_{4}}(aq)\]
  • The expression for \( K_{sp} \) for this reaction is
    \[K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO^{3-}_{4}}]^2\]
In practice, a smaller \( K_{sp} \) value means the compound is less soluble. This plays a significant role in predicting whether a precipitate will form in a solution. If the ionic concentrations product exceeds \( K_{sp} \), the solution forms a precipitate.
ICE Table
An ICE table is a simple yet powerful tool used in chemistry to organize and calculate the changes in concentrations of species participating in a chemical reaction as they achieve equilibrium. ICE stands for Initial, Change, Equilibrium. Let's break down how it works for this example:
  • Initial: Begin with the initial concentrations of ions before any dissolution. In our case, \([\mathrm{Ca^{2+}}] = 0\) and \([\mathrm{PO^{3-}_{4}}] = 0.20\) M from the sodium phosphate.
  • Change: This represents the change in concentration as the system approaches equilibrium. For every "s" moles of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) that dissolves, \([\mathrm{Ca^{2+}}]\) increases by "3s" and \([\mathrm{PO^{3-}_{4}}]\) by "2s".
  • Equilibrium: The total concentration at equilibrium is found by combining initial and change. Hence, \([\mathrm{Ca^{2+}}] = 3s\) and \([\mathrm{PO^{3-}_{4}}] = 0.20 + 2s\).
The ICE table simplifies complex equilibrium problems, allowing for systematic solving to find unknown concentrations.
Chemical Equilibrium
Chemical equilibrium exists when a chemical reaction's forward and backward reactions occur at an equal rate, leading to constant concentrations of reactants and products over time. Although it may seem reactions have stopped, they are dynamically balanced.For dissolution, equilibrium is reached when the rate of the solid dissolving equals the rate of its recrystallization from the solution. In our example of \( \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \):
  • Dynamic Balance: At equilibrium, soluble ions of calcium and phosphate form and dissolve from the solid at the same rate.
  • Concentration Stability: The concentrations of \([\mathrm{Ca^{2+}}]\) and \([\mathrm{PO^{3-}_{4}}]\) remain steady, signifying equilibrium.
Calculating \( K_{sp} \) and using an ICE table assists in predicting these equilibrium concentrations efficiently. This knowledge is crucial not only in theoretical calculations but also in practical applications like drug formulation and wastewater treatment, where solubility matters.

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Most popular questions from this chapter

Will a precipitate form when 100.0 \(\mathrm{mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to 100.0 \(\mathrm{mL}\) of \(2.0 \times 10^{-4} \mathrm{MNaOH}\)?

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2},\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

A friend tells you: "The constant \(K_{\mathrm{sp}}\) of a salt is called the solubility product constant and is calculated from the concentrations of ions in the solution. Thus, if salt A dissolves to a greater extent than salt \(\mathrm{B}\) , salt \(\mathrm{A}\) must have a higher \(K_{\mathrm{sp}}\) than salt \(\mathrm{B}\) ." Do you agree with your friend? Explain.

Tooth enamel is composed of the mineral hydroxyapatite. The \(K_{\mathrm{sp}}\) of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) is \(6.8 \times 10^{-37}\) . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\) , forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times 10^{-60}\) . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

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