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 \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Short answer

The solubility-product constant (\(K_{sp}\)) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). The \(K_{sp}\) for \(\mathrm{AgIO}_{3}\) is \(4.44 \times 10^{-8}\). The pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is 12.51.

Step-by-step solution

Write the balanced chemical equation

The dissolution of \(\mathrm{PbBr}_{2}\) in water can be represented by the following balanced equation: \[\mathrm{PbBr}_{2}(s)\rightleftharpoons \mathrm{Pb^{2+}}(aq)+2 \mathrm{Br^-}(aq)\]

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Pb}^{2+}][\mathrm{Br}^{-}]^{2}\]

Calculate the concentrations of the ions

Molar solubility of \(\mathrm{PbBr}_{2}\) is given as \(1.0 \times 10^{-2} \mathrm{M}\). Since the stoichiometry is 1:2 in the balanced equation, this means that the concentration of \(\mathrm{Br}^{-}\) is twice the concentration of \(\mathrm{Pb}^{2+}\). So, the concentrations are: [\(\mathrm{Pb^{2+}}\)] \(= 1.0 \times 10^{-2} \mathrm{M}\) [\(\mathrm{Br^-}\)] \(= 2(1.0 \times 10^{-2}) \mathrm{M} = 2.0 \times 10^{-2} \mathrm{M}\)

Calculate the solubility-product constant

Now, we can plug these values into the \(K_{sp}\) expression: \[K_{sp}=(1.0 \times 10^{-2})(2.0 \times 10^{-2})^{2} = 4.0 \times 10^{-6}\] Thus, the \(K_{sp}\) for \(\mathrm{PbBr}_{2}\) is \(4.0 \times 10^{-6}\). (b) Calculate the solubility-product constant for \(\mathrm{AgIO}_{3}\).

Write the balanced chemical equation

The dissolution of \(\mathrm{AgIO}_{3}\) in water can be represented by the following balanced equation: \[\mathrm{AgIO}_{3}(s)\rightleftharpoons \mathrm{Ag^{+}}(aq)+\mathrm{IO}_{3}^{-}(aq)\]

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Ag}^{+}][\mathrm{IO}_{3}^{-}]\]

Calculate the concentrations of the ions

Given that \(0.049 \mathrm{g}\) of \(\mathrm{AgIO}_{3}\) dissolve per liter of solution, the molar solubility of \(\mathrm{AgIO}_{3}\) can be calculated using the molar mass of \(\mathrm{AgIO}_{3}\), which is approximately \(232 \mathrm{g/mol}\). Molar solubility of \(\mathrm{AgIO}_{3}\): \(\frac{0.049\;\mathrm{g}}{232\;\mathrm{g/mol}} = 2.11 \times 10^{-4} \;\mathrm{M}\) Since the stoichiometry is 1:1, we have: [\(\mathrm{Ag^{+}}\)] \(= 2.11 \times 10^{-4} \;\mathrm{M}\) [\(\mathrm{IO}_{3}^{-}\)] \(= 2.11 \times 10^{-4} \;\mathrm{M}\)

Calculate the solubility-product constant

Now, we can plug these values into the \(K_{sp}\) expression: \[K_{sp}=(2.11 \times 10^{-4})(2.11 \times 10^{-4}) = 4.44 \times 10^{-8}\] Thus, the \(K_{sp}\) for \(\mathrm{AgIO}_{3}\) is \(4.44 \times 10^{-8}\). (c) Calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Write the balanced chemical equation

The dissolution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water can be represented by the following balanced equation: \[\mathrm{Ca}(\mathrm{OH})_{2}(s) \leftrightarrows \mathrm{Ca}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq)\]

Write the expression for solubility-product constant

The solubility-product constant expression is: \[K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{OH}^{-}]^{2}\]

Use the appendix to find the \(K_{sp}\) value for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

According to Appendix D, the \(K_{sp}\) value for \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(25^{\circ} \mathrm{C}\) is \(5.02 \times 10^{-6}\).

Calculate the concentration of \(\mathrm{OH}^-\)

Let the molar solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) be \(s\). Then we have: [\(\mathrm{Ca^{2+}}\)] \(=s\) [\(\mathrm{OH}^{-}\)] \(=2s\) Plugging these into the \(K_{sp}\) expression, we get: \(5.02 \times 10^{-6}=s(2s)^{2}\) Solving for \(s\), we get \(s=1.61 \times 10^{-2} \;\mathrm{M}\). Since \(\mathrm{OH}^{-}\) ion concentration is twice of \(s\), we have: [\(\mathrm{OH}^{-}\)] \(=2(1.61 \times 10^{-2}) = 3.22 \times 10^{-2} \;\mathrm{M}\)

Calculate the \(pOH\) and then find the pH

Now, we can calculate the \(pOH\): \(pOH=-\log([\mathrm{OH}^{-}])=-\log(3.22 \times 10^{-2})=1.49\) Using the relationship between \(pOH\) and \(pH\) (\(pH + pOH = 14\)), we can calculate the pH: \(pH=14-pOH=14-1.49=12.51\) Thus, the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is 12.51.

Key concepts

Molar Solubility

Molar solubility is a measure of how much solute can dissolve in a solvent to form a saturated solution, typically expressed in moles per liter (M). It defines the point at which the solution reaches equilibrium between the dissolved solute and undissolved residue.
For example, for a compound like lead bromide, \(\mathrm{PbBr}_2\), when it dissolves in water, it splits into its constituent ions: \(\mathrm{Pb^{2+}}\) and \(\mathrm{Br^-}\).
The molar solubility indicates how much of this dissociation occurs per liter of water.

Understanding molar solubility helps in calculating the solubility-product constant \(K_{sp}\), which quantifies the solubility equilibrium level of ionic solids.
This value depends on the temperature and the specific solute-solvent combination.
In the exercise provided, determining the molar solubility of \(\mathrm{PbBr}_2\) involves using the formula \(K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Br^-}]^2\), which gives information on how well this salt dissolves under set conditions.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. In the context of solubility, this state is achieved when the rate of dissolving of a compound equals the rate of precipitating.

A key characteristic of chemical equilibrium is that the concentrations of reactants and products remain constant over time, though not necessarily equal in value. They settle at a ratio that is defined by the equilibrium constant, such as \(K_{sp}\) for solubility equilibria.

• This equilibrium constant provides insight into the extent to which a solute will dissolve.
• It helps predict whether a precipitate will form under certain conditions.

To manipulate this equilibrium, conditions like concentration and temperature can be altered, which can either increase or decrease the solubility of a substance.
This concept is critical in understanding reactions in solution and is applicable in various industries, from pharmaceuticals to environmental science.

pH Calculation

Calculating pH is essential for understanding the acidity or basicity of a solution. pH is the negative logarithm of the hydrogen ion concentration, quantifying how acidic or basic a solution is.
For basic solutions, like those involving calcium hydroxide \(\mathrm{Ca(OH)_2}\), pH levels tend towards the higher end of the scale (closer to 14).

To find the pH of a saturated solution, you first need the concentration of hydroxide ions \(\mathrm{OH^-}\). Using the equilibrium expression and \(K_{sp}\), calculate the molar solubility and thereafter, the concentration of \(\mathrm{OH^-}\).
Once the \(\mathrm{OH^-}\) concentration is known, calculate the \(pOH\) using:

• \(pOH = -\log[\mathrm{OH^-}]\)

Finally, convert \(pOH\) to \(pH\) with the equation \(pH + pOH = 14\).
This method was exemplified in calculating the pH of a saturated \(\mathrm{Ca(OH)_2}\) solution as being around \(12.51\), indicating a basic solution.