Question

At \(25^{\circ} \mathrm{C},\) the value of \(K_{\mathrm{sp}}\) for \(\mathrm{AgCN}\) is \(6.0 \times 10^{-17}\) and that for \(\mathrm{Zn}(\mathrm{CN})_{2}\) is \(3 \times 10^{-16} .\) In terms of grams per \(100 \mathrm{~mL}\) of solution, which salt is more soluble in water?

Short answer

After calculating and comparing the solubility in grams per 100 mL, the salt with the higher calculated value is the more soluble one in water.

Step-by-step solution

Write down the solubility product expressions

For silver cyanide, \(AgCN\)), the solubility product expression is \(K_{sp} = [Ag^+][CN^-]\). \oindent For zinc cyanide, \(Zn(CN)_2\)), the expression is \(K_{sp} = [Zn^{2+}][(CN^-)^2]\).

Express the concentration terms from the solubility products

For \(AgCN\), let the solubility be \(s\) mol/L. Then, we can express \(K_{sp}\) as \(K_{sp, AgCN} = s \times s\) or \(s^2\). Similarly, let the solubility of \(Zn(CN)_2\) be \(s'\) mol/L. In this case, there will be \(s'\) moles of \(Zn^{2+}\) ions and \(2s'\) moles of \(CN^-\) ions per liter, so \(K_{sp, Zn(CN)_2} = s' \times (2s')^2\) or \(4s'^3\).

Calculate the solubility in moles per liter for both salts

Solving for \(s\) in terms of \(K_{sp, AgCN}\), we get \(s = \sqrt{K_{sp, AgCN}}\). Substituting \(6.0 \times 10^{-17}\) into the equation gives us the solubility of \(AgCN\). Similarly, solving for \(s'\) in terms of \(K_{sp, Zn(CN)_2}\), we get \(s' = \sqrt[3]{\frac{K_{sp, Zn(CN)_2}}{4}}\). Substituting \(3 \times 10^{-16}\) into the equation gives us the solubility of \(Zn(CN)_2\).

Convert the solubility from moles per liter to grams per 100 mL

First, we find the molar masses of \(AgCN\) and \(Zn(CN)_2\). Then, we convert the solubility from moles per liter (\(mol/L\)) to grams per 100 mL. This involves multiplying the solubility in \(mol/L\) by the molar mass to get grams per liter, and then converting to grams per 100 mL by dividing by 10.

Compare the solubility in grams per 100 mL for both salts

After calculating the solubility in grams per 100 mL for each salt from the previous step, compare the values to determine which one is more soluble. The salt with the higher grams per 100 mL value is more soluble.

Key concepts

Ksp Calculations

Understanding the solubility product constant, denoted as Ksp, is fundamental to predicting the solubility of ionic compounds in water. Ksp calculations can help us determine the maximum amount of solute that can dissolve in a specific volume of solvent at a constant temperature. This is a pivotal part of studying solutions and their behaviors.

For any ionic compound with a general formula of AB where A is a cation and B is an anion, the Ksp expression, in the simplest case, is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced dissolution equation: Ksp = [A]^m[B]^n.

By using the Ksp values provided, you can calculate the solubility of an ionic compound. For instance, the Ksp expression for AgCN is written as Ksp = [Ag+][CN-]. If we assume the solubility of AgCN is 's' moles per liter, then we can set up the equation as Ksp = s^2, which after calculation reveals the molar solubility. This process is identical to that used in the provided step-by-step solution, and it's crucial for students to grasp this to effectively compare the solubilities of various salts.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. Saturation is the point at which additional solute will no longer dissolve in the solvent due to the equilibrium that forms between the dissolved ions and the undissolved solid.

To calculate molar solubility from Ksp, you typically isolate the solubility term (s) in the Ksp expression. For example, if the Ksp for a salt AB2 is written as Ksp = [A][B]^2, and [A] = s and [B] = 2s (since for every mole of A, there are two moles of B), the Ksp expression becomes Ksp = s(2s)^2 or 4s^3.

After computing for 's', you can easily convert molar solubility to grams per liter by multiplying by the compound's molar mass, and subsequently to grams per 100 mL as needed for comparing different salts. This conversion is a critical step in relating the abstract concept of molar solubility to a more concrete measurement that can be visualized and understood, as in the exercise provided.

Precipitation Chemistry

Precipitation chemistry deals with the formation of a solid in a solution during a chemical reaction. This typically occurs when the concentration of ions in solution exceeds their solubility limit, causing a precipitate to form. The likelihood of precipitation can be predicted by comparing the ion product to the Ksp. The ion product is the product of the concentrations of the ionic species in a solution at any given time.

If the ion product is greater than the Ksp, the solution is supersaturated, and precipitation is favored. Conversely, if the ion product is less than the Ksp, the solution is unsaturated, and no precipitation occurs. This principle allows us to understand why certain compounds are more soluble than others and why changing the concentration of one ion affects the solubility of a compound.

Precipitation also depends on factors such as temperature, the presence of complexing agents, and the pH of the solution. These nuances are essential for students to appreciate the complexity of solubility and precipitation in real-world scenarios, especially in disciplines like environmental science and chemical engineering.