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 a \(3.0 \times 10^{-7}\,\mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Step-by-step solution

Write the dissociation equation for calcium hydroxide

Calcium hydroxide is a strong base that dissociates completely in water. The dissociation equation for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is given by: \[\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\] This equation shows that for every one molecule of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissociates, one \(\mathrm{Ca}^{2+}\) ion and two \(\mathrm{OH}^{-}\) ions are formed.

Calculate the concentration of hydroxide ions using stoichiometry

Since we are given the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) as \(3.0 \times 10^{-7}\,\mathrm{M}\), we can use the stoichiometry from the dissociation equation to find the concentration of hydroxide ions. As per the equation, the formation of hydroxide ions happens in a 1:2 ratio with respect to \(\mathrm{Ca}(\mathrm{OH})_{2}\), so we can write: \[\left[\mathrm{OH}^{-}\right] = 2 \times \left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\] Now, substitute the given concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) into the equation: \[\left[\mathrm{OH}^{-}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M})\]

Calculate the final concentration of hydroxide ions

Now, simply multiply the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) by the stoichiometric coefficient (2) to obtain the concentration of hydroxide ions: \[\left[\mathrm{OH}^{-}\right] = 6.0 \times 10^{-7}\,\mathrm{M}\] Thus, the concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in the given solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, dealing with the quantitative relationships between reactants and products in a chemical reaction. It helps to determine how much of each substance is involved and produced in a reaction. In the context of calcium hydroxide dissociation, stoichiometry plays a crucial role in understanding how many hydroxide ions are formed.

When we look at the dissociation equation of calcium hydroxide:

• \[\mathrm{Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^{-}}\]

This equation tells us that one molecule of \(\mathrm{Ca(OH)_2}\) produces two hydroxide ions \(\mathrm{OH}^{-}\).

Therefore, for a given concentration of calcium hydroxide in the solution, the concentration of hydroxide ions will be twice that of calcium hydroxide. Stoichiometry allows us to set up this 1:2 ratio correctly, giving us the foundational step to move forward with the calculation of \(\mathrm{OH}^{-}\) concentration.

Hydroxide Ion Concentration

Calculating the hydroxide ion concentration in a solution requires one to first understand how a compound dissociates in water. For calcium hydroxide, which is a strong base, it dissociates completely, releasing hydroxide ions into the solution.

The dissociation equation for \(\mathrm{Ca(OH)_2}\) shows that two hydroxide ions are produced for every molecule that dissolves. Given the concentration of \(\mathrm{Ca(OH)_2}\) is \(3.0 \times 10^{-7}\,\mathrm{M}\), we can calculate the \(\mathrm{OH}^{-}\) concentration by simply applying the stoichiometric factor of 2:

• \(\left[\mathrm{OH}^{-}\right] = 2 \times \left[\mathrm{Ca(OH)_2}\right]\)
• \(\left[\mathrm{OH}^{-}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M})\)
• \(\left[\mathrm{OH}^{-}\right] = 6.0 \times 10^{-7}\,\mathrm{M}\)

Thus, the concentration of hydroxide ions is \(6.0 \times 10^{-7}\,\mathrm{M}\). This value indicates how strongly basic the solution is, which is crucial for understanding chemical equilibria and reactions involving acids and bases.

Chemical Equilibrium

Chemical equilibrium refers to a state where the forward and reverse reactions occur at the same rate, leading to no net change in the concentration of reactants and products over time. However, in the context of strong bases like calcium hydroxide, the emphasis is on the fact that it dissociates completely.

This means that when \(\mathrm{Ca(OH)_2}\) is dissolved in water, it yields its maximum production of \(\mathrm{OH}^{-}\) ions instantly, pushing the equilibrium towards more product formation.

For a solution at equilibrium, the concentration of products and reactants is constant, but in a case of complete dissociation, the products dominate.

This full dissociation is key to maintaining a stable level of hydroxide ions in solution, which impacts the pH and influences subsequent chemical processes. Hence, understanding chemical equilibrium in dissociation reactions aids in predicting how solutions will behave in different chemical environments.

In practical applications, grasping these concepts helps in tasks like titrations and reaction predictions, forming a backbone in many analytical chemistry methods.