Problem 54 - Chemistry: The Central Science

Question

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\) Calculate \(K_{\mathrm{sp}}\). (b) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dissolves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{1 p}\) value from Appendix D, calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) -

Step-by-Step Solution

Answer

Answer: (a) The \(K_{sp}\) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). (b) The \(K_{sp}\) for \(\mathrm{AgIO}_3\) is approximately \(8.3 \times 10^{-8}\). (c) The pH of the saturated solution of \(\mathrm{Ca(OH)_2}\) is __(insert calculated pH value)__.

Step 1: Molar solubility equation of \(\mathrm{PbBr}_{2}\) #

First, we need to write the balanced solubility equilibrium equation for \(\mathrm{PbBr}_{2}\): \(\mathrm{PbBr}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{Br}^-_{(aq)}\)

Step 2: Concentrations and \(K_{sp}\) formula #

Now, we will use the given molar solubility of \(1.0 \times 10^{-2}\) mol/L to find the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Br}^-\) ions. Let x be the molar solubility of \(\mathrm{PbBr}_{2}\). Then, the concentration of \(\mathrm{Pb}^{2+}\) is x, and the concentration of \(\mathrm{Br}^-\) is 2x. Plugging these concentrations into the \(K_{sp}\) formula, we have: \(K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{Br}^-]^2\)

Step 3: Calculate \(K_{sp}\) #

Now, we will substitute the values of x and 2x to calculate the \(K_{sp}\) value for \(\mathrm{PbBr}_{2}\). \(K_{sp} = (1.0 \times 10^{-2})(2(1.0 \times 10^{-2}))^2\) \(K_{sp} = 4.0 \times 10^{-6}\) Answer: The \(K_{sp}\) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). ## Part (b) - \(K_{sp}\) for \(\mathrm{AgIO}_{3}\) ##

Step 4: Convert mass solubility to molar solubility #

We are given the mass solubility of \(\mathrm{AgIO}_3\) per liter of solution, which is \(0.0490\) g. We need to convert this to molar solubility by using the molar mass of \(\mathrm{AgIO}_3\). The molar mass of \(\mathrm{AgIO}_3\) is \(169.88\) g/mol. To find the molar solubility, we will use the following conversion formula: Molar solubility = \(\frac{Mass \ solubility}{Molar \ mass}\) Molar solubility = \(\frac{0.0490 \ g}{169.88 \ g/mol} \approx 2.88 \times 10^{-4}\) mol/L

Step 5: Molar solubility equation of \(\mathrm{AgIO}_{3}\) #

Now, we can write the balanced solubility equilibrium equation for \(\mathrm{AgIO}_{3}\): \(\mathrm{AgIO}_{3(s)} \rightleftharpoons \mathrm{Ag}^+_{(aq)} + \mathrm{IO}^-_{3(aq)}\)

Step 6: Concentrations and \(K_{sp}\) formula #

Since the molar solubility of \(\mathrm{AgIO}_{3}\) is \(2.88 \times 10^{-4}\) mol/L, the concentrations of \(\mathrm{Ag}^+\) and \(\mathrm{IO}_3^-\) are equal. Plug these concentrations into the \(K_{sp}\) formula: \(K_{sp} = [\mathrm{Ag}^+] [\mathrm{IO}_3^-]\)

Step 7: Calculate \(K_{sp}\) #

Now, we will substitute the molar solubility to calculate the \(K_{sp}\) value: \(K_{sp} = (2.88 \times 10^{-4})(2.88 \times 10^{-4})\) \(K_{sp} \approx 8.3 \times 10^{-8}\) Answer: The \(K_{sp}\) for \(\mathrm{AgIO}_3\) is approximately \(8.3 \times 10^{-8}\). ## Part (c) - pH of a saturated solution of \(\mathrm{Ca(OH)_2}\) ##

Step 8: Molar solubility equation of \(\mathrm{Ca(OH)_2}\) #

We start by writing the balanced solubility equilibrium equation for \(\mathrm{Ca(OH)}_{2}\): \(\mathrm{Ca(OH)}_{2(s)} \rightleftharpoons \mathrm{Ca}^{2+}_{(aq)} + 2\mathrm{OH}^-_{(aq)}\)

Step 9: Solubility, \(K_{sp}\), and concentration of \(\mathrm{OH}^-\) ions #

We are given the \(K_{sp}\) value of \(\mathrm{Ca(OH)_2}\) in Appendix D. Using it, we can find the concentration of the \(\mathrm{OH}^-\) ions. Let x be the concentration of \(\mathrm{Ca^{2+}_{(aq)}}\). Then, the concentration of \(\mathrm{OH}^-_{(aq)}\) is 2x. Plugging these concentrations into the \(K_{sp}\) formula, we get: \(K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{OH}^-]^2\) We can now solve for the concentration of \(\mathrm{OH}^-\) ions.

Step 10: Calculate \(pOH\) and \(pH\) #

Next, we compute the \(pOH\), which is the negative logarithm of the concentration of \(\mathrm{OH}^-\) ions. With \(pOH\), we can find the \(pH\) using the relation: \(pH + pOH = 14\) Now, we can substitute the values we obtained to calculate the \(pH\) of the saturated solution of \(\mathrm{Ca(OH)_2}\). Answer: The pH of the saturated solution of \(\mathrm{Ca(OH)_2}\) is __(insert calculated pH value)__.

Key Concepts

Molar Solubility

Understanding molar solubility is crucial for solving problems involving solubility products, often shown as \( K_{sp} \). Molar solubility refers to the number of moles of a solute that can dissolve in one liter of solvent until the solution becomes saturated. In our exercise, we calculate the molar solubility of compounds like \( \mathrm{PbBr}_2 \) by setting up a solubility equilibrium.
Molar solubility is often expressed in \( \text{mol/L} \) and can be determined using the relationship between the ions in the solution. For \( \mathrm{PbBr}_2 \) dissolving at equilibrium:\[\mathrm{PbBr}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{Br}^-_{(aq)}\].
You can calculate the \( K_{sp} \) using the concentrations of the ions derived from the molar solubility. Using the expression \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Br}^-]^2 \), replace the concentrations with those based on molar solubility values. Such calculations reveal how sparingly soluble salts behave under equilibrium conditions.

Equilibrium Equations

Equilibrium equations form the foundation for understanding how compounds dissolve and reach a state of balance. They are expressions that represent the reversible reactions of soluble salts dissolving and the resulting concentrations in solutions.
To write these equations, start by identifying the dissociation products: for instance, \( \mathrm{AgIO}_{3(s)} \rightleftharpoons \mathrm{Ag}^+_{(aq)} + \mathrm{IO}_3^-_{(aq)} \).
In this scenario, each component has a one-to-one mole ratio. The equilibrium constant \( K_{sp} \) provides insights into the solubility characteristics of the compound, following the equation \( K_{sp} = [\mathrm{Ag}^+][\mathrm{IO}_3^-] \).

• Write equations that reflect the stoichiometry of the dissociated ions.
• Take into account the states (solid to aqueous) of each species involved.

These equations help students predict the extent of solubility and the related concentrations of ions in saturated solutions.

pH Calculation

Calculating the pH of a solution is an essential skill in chemistry, especially for solutions involving slightly soluble bases like \( \mathrm{Ca(OH)_2} \). The pH is connected to the concentration of hydrogen ions (\( \mathrm{H}^+ \)) in a solution.
First, find the molar solubility of \( \mathrm{Ca(OH)_2} \) from its \( K_{sp} \) value, and remember that upon dissolving, it produces \( \mathrm{Ca}^{2+} \) and \( 2\mathrm{OH}^- \) ions. Use the formula:\[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2\]
Then, compute the \( pOH \) as:\[ pOH = -\log_{10}([\mathrm{OH}^-]) \].Using the relationship \( pH + pOH = 14\), convert the \( pOH \) to \( pH \).

• Ensure accuracy in logarithmic calculations.
• Verify that the pH scale used aligns with the parameters of 0 to 14 for aqueous solutions.

Accurate pH calculation provides insights into the acidity or basicity of the solution.

Concentration Units

Understanding the various concentration units used in chemical calculations is fundamental to grasping how solutions behave. Molarity, which is moles of solute per liter of solution, is one of the most commonly used units, especially in solubility equations and \( K_{sp} \) calculations.
In problems where mass solubility is given, such as the solubility of \( \mathrm{AgIO}_3 \), it is imperative to convert to molarity. Divide the mass of the solute by its molar mass to obtain the molar solubility in \( \text{mol/L} \). This conversion is critical for substituting into \( K_{sp} \) expressions.

• Keep track of units to ensure consistency in calculations.
• Understand the context in which each unit applies, particularly in equilibrium-related calculations.

Recognizing and correctly applying concentration units simplify the analysis of chemical reactions and enhance precise interpretations of solubility products.