Calculating The Solubility Of Silver Phosphate In Aqueous Solution

In this article, we will delve into the process of calculating the solubility of silver phosphate ($Ag_3PO_4$) in a hot aqueous solution. Solubility, a crucial concept in chemistry, refers to the maximum amount of a solute that can dissolve in a given solvent at a specific temperature to form a saturated solution. Understanding solubility is vital in various applications, including pharmaceutical formulation, environmental chemistry, and industrial processes. Silver phosphate ($Ag_3PO_4$), an ionic compound sparingly soluble in water, serves as an excellent example to illustrate the principles of solubility calculations. We will use the solubility product constant, denoted as $K_{sp}$, to determine the solubility of $Ag_3PO_4$ at a given temperature. The $K_{sp}$ value, which is temperature-dependent, represents the equilibrium constant for the dissolution of a solid into its ions in a solution. By utilizing the provided $K_{sp}$ value at the elevated temperature, we will demonstrate how to set up an equilibrium expression and solve for the solubility of $Ag_3PO_4$ in grams per liter (g/L). This calculation involves understanding the stoichiometry of the dissolution process, constructing an ICE (Initial, Change, Equilibrium) table, and applying algebraic principles to solve for the unknown solubility. Through this step-by-step approach, we aim to provide a clear and comprehensive guide for calculating the solubility of sparingly soluble salts, a fundamental skill in chemistry. This article will enhance your understanding of solubility principles and their practical applications in chemical calculations.

Understanding Solubility and $K_{sp}$

Solubility, a key concept in chemistry, dictates the maximum extent to which a substance, known as the solute, dissolves in a solvent to form a homogenous solution. The solubility of a compound is influenced by several factors, including temperature, pressure, and the nature of the solute and solvent. When a solid compound like silver phosphate ($Ag_3PO_4$) is introduced into water, it dissociates into its constituent ions, silver ions ($Ag^+$), and phosphate ions ($PO_4^{3-}$). However, for sparingly soluble salts like $Ag_3PO_4$, this dissolution process reaches an equilibrium where the rate of dissolution equals the rate of precipitation. The equilibrium constant for this process is known as the solubility product constant, denoted as $K_{sp}$. The $K_{sp}$ value is a temperature-dependent constant that represents the product of the ion concentrations raised to the power of their stoichiometric coefficients in a saturated solution. A higher $K_{sp}$ value indicates greater solubility, while a lower value suggests lower solubility. For $Ag_3PO_4$, the dissolution equilibrium can be represented as follows:

$Ag_3PO_4(s) ightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq)$

The solubility product expression for this equilibrium is given by:

$K_{sp} = [Ag^+]^3[PO_4^{3-}]$

This equation shows that the $K_{sp}$ is the product of the concentrations of silver ions cubed and phosphate ions at equilibrium. Understanding the relationship between solubility and $K_{sp}$ is crucial for calculating the solubility of sparingly soluble salts. By utilizing the given $K_{sp}$ value, we can determine the molar solubility (in moles per liter) and subsequently convert it to grams per liter, providing a quantitative measure of how much $Ag_3PO_4$ can dissolve in water at a specific temperature. This section lays the groundwork for the subsequent calculations, emphasizing the importance of equilibrium principles in solubility determinations. Mastering the concepts of solubility and $K_{sp}$ is essential for solving a wide range of chemical problems related to precipitation, dissolution, and complex ion equilibria.

Setting Up the ICE Table

To accurately calculate the solubility of silver phosphate ($Ag_3PO_4$) using the given solubility product constant ($K_{sp}$), we employ the ICE table method. ICE stands for Initial, Change, and Equilibrium, and it is a systematic approach to solving equilibrium problems. The ICE table helps us organize the initial concentrations, the changes in concentrations as the system reaches equilibrium, and the equilibrium concentrations of the ions involved in the dissolution process. For the dissolution of $Ag_3PO_4$ in water, the equilibrium reaction is:

$Ag_3PO_4(s) ightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq)$

Let's define 's' as the molar solubility of $Ag_3PO_4$, which represents the concentration of $Ag_3PO_4$ that dissolves in water to reach equilibrium. According to the stoichiometry of the reaction, for every one mole of $Ag_3PO_4$ that dissolves, three moles of silver ions ($Ag^+$) and one mole of phosphate ions ($PO_4^{3-}$) are produced. We can now construct the ICE table:

In the ICE table, the initial concentrations of $Ag^+$ and $PO_4^{3-}$ are both 0 since we are starting with pure water and solid $Ag_3PO_4$. The change in concentration is represented by '+3s' for $Ag^+$ and '+s' for $PO_4^{3-}$ because, as $Ag_3PO_4$ dissolves, it produces three times as many $Ag^+$ ions as $PO_4^{3-}$ ions. At equilibrium, the concentrations of $Ag^+$ and $PO_4^{3-}$ are 3s and s, respectively. The concentration of the solid $Ag_3PO_4$ does not appear in the $K_{sp}$ expression because it is a pure solid and its activity is considered to be 1. The ICE table is a powerful tool for visualizing the changes in concentrations during a chemical reaction and is essential for setting up the equilibrium expression correctly. By systematically filling out the ICE table, we ensure that we account for the stoichiometry of the reaction and accurately determine the equilibrium concentrations of the ions. This step is crucial for the subsequent calculation of solubility using the $K_{sp}$ value.

Calculating Molar Solubility

With the ICE table set up, we can now use the given solubility product constant ($K_{sp}$) to calculate the molar solubility (s) of silver phosphate ($Ag_3PO_4$). The $K_{sp}$ expression for the dissolution of $Ag_3PO_4$ is:

$K_{sp} = [Ag^+]^3[PO_4^{3-}]$

From the ICE table, we know that at equilibrium, $[Ag^+] = 3s$ and $[PO_4^{3-}] = s$. Substituting these values into the $K_{sp}$ expression, we get:

$K_{sp} = (3s)^3(s)$

Given that the $K_{sp}$ of $Ag_3PO_4$ at the specified temperature is $6.69 imes 10^{-12}$, we can plug this value into the equation:

$6.69 imes 10^{-12} = (3s)^3(s)$

Simplifying the equation:

$6.69 imes 10^{-12} = 27s^3 imes s$

$6.69 imes 10^{-12} = 27s^4$

Now, we solve for s:

s^4 = rac{6.69 imes 10^{-12}}{27}

$s^4 = 2.4778 imes 10^{-13}$

Taking the fourth root of both sides:

s = oot4{2.4778 imes 10^{-13}}

$s imes 10^{-4}$ mol/L

Thus, the molar solubility of $Ag_3PO_4$ is approximately $1.26 imes 10^{-4}$ mol/L. This value represents the concentration of $Ag_3PO_4$ that dissolves in one liter of water at the given temperature. Calculating molar solubility is a critical step in determining the solubility of sparingly soluble salts. By utilizing the $K_{sp}$ expression and the equilibrium concentrations obtained from the ICE table, we can accurately quantify the extent to which a compound dissolves in a solution. This value is essential for further calculations, such as converting molar solubility to grams per liter, which provides a more practical measure of solubility. The molar solubility calculation demonstrates the application of equilibrium principles and algebraic techniques in solving chemical problems.

Converting Molar Solubility to Grams per Liter

Now that we have calculated the molar solubility (s) of silver phosphate ($Ag_3PO_4$) as $1.26 imes 10^{-4}$ mol/L, the next step is to convert this value into grams per liter (g/L), which is a more commonly used unit for expressing solubility. To perform this conversion, we need the molar mass of $Ag_3PO_4$. The molar mass is calculated by summing the atomic masses of each element in the compound, which are:

• Silver (Ag): 107.87 g/mol
• Phosphorus (P): 30.97 g/mol
• Oxygen (O): 16.00 g/mol

The molar mass of $Ag_3PO_4$ is:

$Molar extit{Mass} = (3 imes 107.87 extit{g/mol}) + (1 imes 30.97 extit{g/mol}) + (4 imes 16.00 extit{g/mol})$

$Molar extit{Mass} = 323.61 extit{g/mol} + 30.97 extit{g/mol} + 64.00 extit{g/mol}$

$Molar extit{Mass} = 418.58 extit{g/mol}$

Now, we can convert the molar solubility to grams per liter using the molar mass:

$Solubility extit{(g/L)} = Molar extit{Solubility(mol/L)} imes Molar extit{Mass(g/mol)}$

$Solubility extit{(g/L)} = (1.26 imes 10^{-4} extit{mol/L}) imes (418.58 extit{g/mol})$

$Solubility extit{(g/L)} = 0.0527 extit{g/L}$

Therefore, the solubility of $Ag_3PO_4$ in the hot aqueous solution is approximately 0.0527 grams per liter. This value represents the mass of $Ag_3PO_4$ that can dissolve in one liter of water at the specified temperature, given the $K_{sp}$ value. Converting molar solubility to grams per liter provides a practical understanding of the amount of solute that can dissolve in a given volume of solvent. This conversion is essential for applications where mass concentrations are more relevant, such as in environmental monitoring or chemical formulations. The final solubility value, expressed in grams per liter, gives a clear indication of the extent to which $Ag_3PO_4$ dissolves in water under the given conditions.

Conclusion

In conclusion, we have successfully calculated the solubility of silver phosphate ($Ag_3PO_4$) in a hot aqueous solution using the provided solubility product constant ($K_{sp}$). We began by understanding the fundamental principles of solubility and the significance of the $K_{sp}$ value as an indicator of a compound's solubility. We then constructed an ICE table to systematically determine the equilibrium concentrations of the ions involved in the dissolution process. Using the $K_{sp}$ expression and the equilibrium concentrations from the ICE table, we calculated the molar solubility of $Ag_3PO_4$ to be approximately $1.26 imes 10^{-4}$ mol/L. Finally, we converted the molar solubility to grams per liter, obtaining a solubility of approximately 0.0527 g/L. This value represents the amount of $Ag_3PO_4$ that can dissolve in one liter of water at the specified temperature. The process of calculating solubility involves several key steps, including setting up the equilibrium reaction, constructing the ICE table, using the $K_{sp}$ expression, and converting molar solubility to other units. Each of these steps is crucial for accurately determining the solubility of sparingly soluble salts. Understanding solubility is essential in various fields, including chemistry, environmental science, and pharmaceuticals. By mastering the concepts and techniques presented in this article, one can confidently approach solubility calculations for a wide range of compounds and conditions. The calculated solubility provides valuable insights into the behavior of $Ag_3PO_4$ in aqueous solutions and highlights the importance of solubility principles in chemical problem-solving. This comprehensive guide serves as a valuable resource for students and professionals seeking to enhance their understanding of solubility and its applications.