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Magazine Chapter 15: Problem 41
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 Ca鈧�(PO鈧�)鈧� in a 0.20 M Na鈧働O鈧� solution is approximately \(4.46 \times 10^{-12} \, M\).
Step by step solution
01
Write the balanced chemical equation
First, let's write the balanced chemical equation for the dissolution of calcium phosphate in water:
\[Ca_3(PO_4)_2 (s) \rightleftharpoons 3Ca^{2+} (aq) + 2PO_4^{3-} (aq)\]
Now that we have the balanced chemical equation, we can proceed to write the equilibrium expression.
02
Write the equilibrium expression
The equilibrium expression for the solubility of Ca鈧�(PO鈧�)鈧� can be written as:
\[K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\]
Since we know the value of Ksp (1.3 脳 10鈦宦陈�), we can use it to find the solubility of Ca鈧�(PO鈧�)鈧� in the solution.
03
Determine the common ion effect
When we dissolve Ca鈧�(PO鈧�)鈧� in Na鈧働O鈧�, both solutions will contain phosphate ions (PO鈧劼斥伝). The presence of these common ions in the solution will cause the solubility of the calcium phosphate to decrease due to the common ion effect.
Let the solubility of Ca鈧�(PO鈧�)鈧� be S mol/L. At equilibrium, the concentrations of the ions will be:
\[Ca^{2+} = 3S\]
\[PO_4^{3-} = 2S + 0.20\]
Here we will assume that the concentration of PO鈧劼斥伝 from the Na鈧働O鈧� solution will dominate and the increase in concentration due to the dissolving of Ca鈧�(PO鈧�)鈧� will be negligible.
04
Set up an ICE table
Now, let's set up an ICE table (Initial, Change, Equilibrium).
| | Ca虏鈦� | PO鈧劼斥伝 |
|-------------|------|-------|
| Initial | 0 | 0.20 |
| Change | +3S | +2S |
| Equilibrium | 3S | 0.20 |
As we have assumed that the PO鈧劼斥伝 concentration from the Na鈧働O鈧� solution dominates, we can set up the Ksp expression:
\[K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\]
\[1.3 \times 10^{-32} = (3S)^3 (0.20)^2\]
05
Solve for solubility of Ca鈧�(PO鈧�)鈧�
Now, we will solve this equation for S to determine the solubility of Ca鈧�(PO鈧�)鈧� in the 0.20 M Na鈧働O鈧� solution:
\[1.3 \times 10^{-32} = (3S)^3 (0.04)\]
Divide both sides by 0.04:
\[\frac{1.3 \times 10^{-32}}{0.04} = (3S)^3\]
Now, find the cube root of both sides:
\[\sqrt[3]{\frac{1.3 \times 10^{-32}}{0.04}} = 3S\]
Finally, solve for S:
\[S = \frac{\sqrt[3]{\frac{1.3 \times 10^{-32}}{0.04}}}{3}\]
\[S \approx 4.46 \times 10^{-12} \, M\]
So, the solubility of Ca鈧�(PO鈧�)鈧� in the 0.20 M Na鈧働O鈧� solution is approximately 4.46 脳 10鈦宦孤� mol/L.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Common Ion Effect
The common ion effect is a phenomenon where the presence of an ion in a solution that also derives from a soluble compound decreases the compound's solubility. In simpler terms, when you have two different substances in solution that share a common ion, the one that dissolves later or less readily has its solubility affected by the one that's already present.
For example, in the given exercise, calcium phosphate \(Ca_3(PO_4)_2\) is less soluble in a solution that already contains the phosphate ion \(PO_4^{3-}\) provided by the sodium phosphate \(Na_3PO_4\). The extra phosphate ions from \(Na_3PO_4\) shift the dissolution equilibrium of \(Ca_3(PO_4)_2\) to the left, according to Le Chatelier's principle, resulting in decreased solubility. In essence, the solution is already 'busy' with the common ions, making it harder for the solid to dissociate further. This effect is key to understanding why the solubility of \(Ca_3(PO_4)_2\) drops in the presence of \(Na_3PO_4\).
For example, in the given exercise, calcium phosphate \(Ca_3(PO_4)_2\) is less soluble in a solution that already contains the phosphate ion \(PO_4^{3-}\) provided by the sodium phosphate \(Na_3PO_4\). The extra phosphate ions from \(Na_3PO_4\) shift the dissolution equilibrium of \(Ca_3(PO_4)_2\) to the left, according to Le Chatelier's principle, resulting in decreased solubility. In essence, the solution is already 'busy' with the common ions, making it harder for the solid to dissociate further. This effect is key to understanding why the solubility of \(Ca_3(PO_4)_2\) drops in the presence of \(Na_3PO_4\).
Calculating Solubility Product Constant (Ksp)
The solubility product constant, abbreviated as Ksp, is crucial for quantifying the extent of solubility for sparingly soluble ionic compounds. This equilibrium constant represents the level at which a solid can dissolve in water to form its constituent ions.
To calculate Ksp, you'll need the equilibrium concentrations of the ions formed when the ionic solid dissolves. For the compound \(Ca_3(PO_4)_2\), with Ksp given as \(1.3 \times 10^{-32}\), the formula takes the appearance of \[K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\]. Through this formula, we understand that the solubility equilibrium involves the concentrations of calcium and phosphate ions raised to the power of their stoichiometric coefficients from the balanced dissolution equation. Ksp calculations are vital to predicting whether a precipitate will form in a solution and are fundamental in the study of solubility equilibrium.
To calculate Ksp, you'll need the equilibrium concentrations of the ions formed when the ionic solid dissolves. For the compound \(Ca_3(PO_4)_2\), with Ksp given as \(1.3 \times 10^{-32}\), the formula takes the appearance of \[K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\]. Through this formula, we understand that the solubility equilibrium involves the concentrations of calcium and phosphate ions raised to the power of their stoichiometric coefficients from the balanced dissolution equation. Ksp calculations are vital to predicting whether a precipitate will form in a solution and are fundamental in the study of solubility equilibrium.
Utilizing ICE Table Methodology
The ICE table stands for Initial, Change, Equilibrium; it's a systematic way to consider the concentrations of all species in a solubility equilibrium calculation. The technique involves setting up a table to track the concentration changes that occur when a solute dissolves in a solvent.
In practice, you start with the initial concentrations, account for the change in concentration when the solid starts dissolving, and finally, you determine the equilibrium concentrations. For the problem in question, we initially have \(0.20\,M\) of \(PO_4^{3-}\) from \(Na_3PO_4\) and effectively none of \(Ca^{2+}\) and \(PO_4^{3-}\) from \(Ca_3(PO_4)_2\). As the solid begins to dissolve, the table helps you track the increase in ion concentration and, ultimately, determine their final values at equilibrium.
These calculations allow us to apply the equilibrium constants accurately to find the solubility of the ionic compound in question. The ICE table is a powerful tool for visualizing and solving equilibrium problems in chemistry.
In practice, you start with the initial concentrations, account for the change in concentration when the solid starts dissolving, and finally, you determine the equilibrium concentrations. For the problem in question, we initially have \(0.20\,M\) of \(PO_4^{3-}\) from \(Na_3PO_4\) and effectively none of \(Ca^{2+}\) and \(PO_4^{3-}\) from \(Ca_3(PO_4)_2\). As the solid begins to dissolve, the table helps you track the increase in ion concentration and, ultimately, determine their final values at equilibrium.
These calculations allow us to apply the equilibrium constants accurately to find the solubility of the ionic compound in question. The ICE table is a powerful tool for visualizing and solving equilibrium problems in chemistry.
