Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7} M\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Short answer

The concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in the \(3.0 \times 10^{-7} M\) solution of Calcium Hydroxide is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Step-by-step solution

STEP 1: Write the dissociation equation

When Calcium Hydroxide \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissolves in water, it dissociates into its individual ions: \[\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^-\]

STEP 2: Set up the stoichiometry

According to the dissociation equation, the stoichiometry of the reaction states that for each mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissociates, 2 moles of \(\mathrm{OH}^-\) ions are produced. Let's use this ratio to find the concentration of \(\mathrm{OH}^-\) ions.

STEP 3: Calculate the concentration of \(\mathrm{OH}^-\) ions

We know the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (\(3.0 \times 10^{-7} M\)). Since 1 mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) produces 2 moles of \(\mathrm{OH^-}\) ions, we'll simply multiply the given concentration by 2 to get the concentration of \(\mathrm{OH}^-\) ions: \[\left[\mathrm{OH}^{-}\right] = 2 \times \left[\mathrm{Ca}(\mathrm{OH})_{2}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M})\]

STEP 4: Solve for the concentration of \(\mathrm{OH}^-\) ions

By multiplying the given concentration by 2, we can find the concentration of \(\mathrm{OH}^-\) ions in the solution: \[\left[\mathrm{OH}^{-}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M}) = 6.0 \times 10^{-7}\,\mathrm{M}\] Thus, the concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in the \(3.0 \times 10^{-7} M\) solution of Calcium Hydroxide is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Key concepts

Dissociation Equation

Calcium Hydroxide, or \(\text{Ca(OH)}_2\), is an ionic compound that is highly soluble in water. Once it dissolves, the dissociation equation helps us understand how it separates into individual ions. Specifically, when \(\text{Ca(OH)}_2\) dissolves in water, it breaks down according to the following equation:\[\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-\]
In this equation, \(\text{Ca(OH)}_2\) dissociates into one calcium ion \(\text{Ca}^{2+}\) and two hydroxide ions \(\text{OH}^-\). This process is essential because it releases hydroxide ions into the solution, which affects the solution's properties, such as pH. Understanding the dissociation equation is the first step in calculating the concentration of these ions in a solution, which is crucial for various chemical calculations.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. In the dissociation of \(\text{Ca(OH)}_2\), stoichiometry tells us the ratio of each substance involved. For every mole of \(\text{Ca(OH)}_2\) that dissociates, two moles of \(\text{OH}^-\) ions are formed.When calculating concentrations in solution, this stoichiometric ratio is used to determine how much of each ion is present based on what we begin with. In our example, if our starting concentration of \(\text{Ca(OH)}_2\) is \(3.0 \times 10^{-7} \, \text{M}\), stoichiometry informs us that the concentration of \(\text{OH}^-\) ions will be twice that amount:

• This is because 1 mole of \(\text{Ca(OH)}_2\) yields 2 moles of \(\text{OH}^-\).
• Thus, \(\text{OH}^-\) concentration = \(2 \times 3.0 \times 10^{-7} = 6.0 \times 10^{-7} \, \text{M}\).

Understanding stoichiometry ensures that we can accurately calculate the concentrations of all substances in a chemical equilibrium.

Hydroxide Ion Concentration

The concentration of hydroxide ions \(\left[\text{OH}^-\right]\) in a solution is a crucial factor in determining the solution's basicity or alkalinity. Here, we aimed to find \(\left[\text{OH}^-\right]\) in a \(3.0 \times 10^{-7} \, \text{M}\) solution of \(\text{Ca(OH)}_2\).By using the dissociation equation, we know one mole of \(\text{Ca(OH)}_2\) splits into 2 moles of \(\text{OH}^-\). Applying this to our given concentration, we multiply this number by 2:\[\left[\text{OH}^-\right] = 2 \times \left[\text{Ca(OH)}_2\right] = 2 \times 3.0 \times 10^{-7} = 6.0 \times 10^{-7} \, \text{M}\]
This calculated concentration of \(\left[\text{OH}^-\right]\) can help predict the pH of the solution since the presence of \(\text{OH}^-\) ions increases the solution's pH, making it more basic. Such calculations are essential in chemistry for predicting how solutions behave and react with other substances.