Question

A saturated solution of zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2},\) has a \(\mathrm{pH}\) of \(8.44 .\) Find the hydroxide ion concentration and calculate the equilibrium constant.

Short answer

The hydroxide ion concentration is \(2.75 \times 10^{-6}\) M and the \(K_{sp}\) is \(1.04 \times 10^{-17}\).

Step-by-step solution

Understanding the pH and pOH Relationship

The pH scale is used to measure the acidity or basicity of a solution. Since the pH of the solution is given as 8.44, we can find the pOH using the relationship: \[ \text{pH} + \text{pOH} = 14 \]Substituting the given value:\[ \text{pOH} = 14 - 8.44 \]

Calculate the pOH Value

Now, perform the calculation to find the pOH:\[ \text{pOH} = 14 - 8.44 = 5.56 \]

Finding the Hydroxide Ion Concentration

The hydroxide ion concentration, \([\mathrm{OH}^-]\), can be found using the formula:\[ [\mathrm{OH^-}] = 10^{-\text{pOH}} \]Substituting the calculated pOH:\[ [\mathrm{OH^-}] = 10^{-5.56} \]

Calculate the Hydroxide Ion Concentration

Calculate the hydroxide ion concentration:\[ [\mathrm{OH^-}] = 2.75 \times 10^{-6} \text{ mol/L} \]

Setting Up the Equilibrium Expression

Zinc hydroxide dissolves in water according to the equation:\[ \mathrm{Zn(OH)_2} (s) \rightleftharpoons \mathrm{Zn^{2+}} (aq) + 2\mathrm{OH^-} (aq) \]The solubility product constant expression (\(K_{sp}\)) for this reaction is:\[ K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{OH^-}]^2 \]

Solving for the Zinc Ion Concentration

We know from stoichiometry that for every mole of \(\mathrm{Zn^{2+}}\), there should be twice as much \(\mathrm{OH^-}\). Therefore,\[ [\mathrm{Zn^{2+}}] = \frac{[\mathrm{OH^-}]}{2} = \frac{2.75 \times 10^{-6}}{2} \]Perform the calculation:\[ [\mathrm{Zn^{2+}}] = 1.375 \times 10^{-6} \text{ mol/L} \]

Calculate the Equilibrium Constant (Ksp)

Finally, calculate the solubility product constant using:\[ K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{OH^-}]^2 \]Substitute the values:\[ K_{sp} = (1.375 \times 10^{-6})(2.75 \times 10^{-6})^2 \]

Solve for Ksp Value

Perform the calculation:\[ K_{sp} = 1.04 \times 10^{-17} \]

Key concepts

pH and pOH relationship

The pH and pOH are crucial concepts in understanding the acidity and basicity of solutions. They are inversely related and sum up to 14. This relationship is fundamental when working with aqueous solutions. So, when you know the pH value, you can easily find the pOH by using the equation: \[ ext{pH} + ext{pOH} = 14 \] For instance, if the pH of a solution is 8.44, the pOH can be calculated as: \[ ext{pOH} = 14 - 8.44 = 5.56 \] This tells us about the basic nature of the solution. A lower pH indicates a more acidic solution, while a higher pH indicates a more basic solution. Understanding this balance helps chemists determine the concentration of hydrogen and hydroxide ions in a solution, which is fundamental to many chemistry calculations.

• pH measures the concentration of hydrogen ions \( ext{[H}^{+} ext{]} \) in a solution.
• pOH measures the concentration of hydroxide ions \( ext{[OH}^{-} ext{]} \) in a solution.

hydroxide ion concentration

Hydroxide ion concentration is a measure of the number of hydroxide ions \( ext{[OH}^{-} ext{]} \) present in a solution. To find this concentration, we use the formula: \[ [\text{OH}^{-}] = 10^{-\text{pOH}} \] In the context of our example, where the pOH is 5.56, the calculation would be: \[ [\text{OH}^{-}] = 10^{-5.56} \] This results in a hydroxide ion concentration of approximately \( 2.75 \times 10^{-6} \text{ mol/L} \). Such calculations assist in determining how basic the solution is.

• High hydroxide ion concentration indicates a strong base.
• Calculating hydroxide ion concentration helps predict reactions in the solution.

equilibrium constant calculation

In the context of solubility, the equilibrium constant is often referred to as the solubility product constant (\(K_{sp}\)). For \(Zn(OH)_2\), the dissolution in water can be represented as: \[ \mathrm{Zn(OH)_2} (s) \rightleftharpoons \mathrm{Zn^{2+}} (aq) + 2\mathrm{OH^-} (aq) \] The\( K_{sp} \)expression for this reaction is: \[ K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{OH^-}]^2 \] To calculate\( K_{sp} \), we use the concentrations of the ions. For our zn(OH)_2 example:

• The hydroxide ion concentration is\( 2.75 \times 10^{-6} \text{ mol/L} \).
• Solve for\( [\mathrm{Zn^{2+}}] \) using stoichiometry: \( \frac{[\mathrm{OH^-}]}{2} \) which gives \( 1.375 \times 10^{-6} \text{ mol/L} \).

Finally, substituting into the\( K_{sp} \)formula, we calculate: \[ K_{sp} = (1.375 \times 10^{-6})(2.75 \times 10^{-6})^2 \] Giving a final value of about\( K_{sp} = 1.04 \times 10^{-17} \). Understanding these calculations allows one to predict solubility and reaction direction in chemical systems.