Question

Write an equation that shows what happens when \(\mathrm{Ba}(\mathrm{OH})_{2}\) dissolves in water. What is the \(\mathrm{pH}\) of \(0.01 M\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) ?

Short answer

The equation for the dissolution of \(\mathrm{Ba}(\mathrm{OH})_{2}\) in water is \(\mathrm{Ba}(\mathrm{OH})_{2} \rightarrow Ba^{2+} + 2 OH^-\). The pH of 0.01 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) is 12.30.

Step-by-step solution

Writing the Dissolution Equation

In water, barium hydroxide (\(\mathrm{Ba}(\mathrm{OH})_{2}\)) dissociates into barium (\(Ba^{2+}\)) ions and hydroxide (\(OH^-)\) ions. The balanced chemical equation is: \(\mathrm{Ba}(\mathrm{OH})_{2} \rightarrow Ba^{2+} + 2 OH^-\)

Determine the Concentration of \(OH^-\) ions

The mole ratio of \(\mathrm{Ba}(\mathrm{OH})_{2}\) to \(OH^-\) ions is 1:2. Since the molarity of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is 0.01 M, this makes the \(OH^-\) ion concentration 0.02 M.

Calculate the pOH

pOH is calculated using the formula: \(pOH = -\log [OH^-]\). Substituting our \(OH^-\) concentration into this equation gives: \(pOH = -\log [0.02]\). This calculates to approximately 1.70.

Calculate the pH

The sum of pH and pOH equals 14 at 25°C (the common standard condition). So, to find the pH, subtract the pOH from 14. This results in: \(pH = 14 - pOH = 14 - 1.70 = 12.30\).

Key concepts

Chemical Dissolution Equation

Understanding the chemical dissolution equation is crucial when studying how compounds interact with solvents like water. Dissolution is the process of a solid, liquid, or gaseous solute forming a solution with a solvent. During this process, the solute breaks apart into smaller particles or ions.

For ionic compounds such as barium hydroxide \( \mathrm{Ba}(\mathrm{OH})_{2} \), dissolution occurs with the compound dissociating into its constituent ions. The dissolution of barium hydroxide in water can be represented by the chemical equation:
\[ \mathrm{Ba}(\mathrm{OH})_{2} (s) \rightarrow Ba^{2+} (aq) + 2 OH^{-} (aq) \]
This equation is balanced and shows that one molecule of solid barium hydroxide produces one barium ion and two hydroxide ions once dissolved in water. The \( (s) \) and \( (aq) \) denote the physical states of the substances—solid and aqueous (dissolved in water), respectively. Understanding this process is foundational for grasping how ionic compounds behave in solution.

Molarity and Concentration

Molarity is a measure of concentration in chemistry and is defined as the number of moles of solute per liter of solution. The concentration of a solution can tell us how 'strong' or 'weak' a solution is based on the amount of solute present in a given volume of solvent.

To calculate molarity, you can use the formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the scenario of barium hydroxide dissolving, the original molarity of the \( \mathrm{Ba}(\mathrm{OH})_{2} \) solution was given as 0.01 M. However, because each unit of \( \mathrm{Ba}(\mathrm{OH})_{2} \) produces two \( OH^{-} \) ions, the molarity of the hydroxide ions in the solution is effectively doubled, becoming 0.02 M. Relating molarity to the chemical dissolution equation allows us to understand not only the qualitative but also the quantitative aspects of the dissolution process.

pH and pOH Calculations

Calculating pH and pOH is vital when we want to determine the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with values less than 7 indicating acidity and values greater than 7 indicating basicity. pOH gives a measure of the hydroxide ion concentration and works in tandem with pH.

pOH can be found using the formula:
\[ pOH = -\log [OH^{-}] \]
With the \(OH^{-}\) concentration at 0.02 M, the pOH is calculated to be approximately 1.70. To find the pH from pOH, you can use the relation:
\[ pH = 14 - pOH \]
Given the pOH of 1.70, we can calculate the pH of the \( \mathrm{Ba}(\mathrm{OH})_{2} \) solution to be about 12.30, confirming it as a basic solution. pH and pOH are log scale measures, meaning that each integer change in value represents a tenfold increase or decrease in hydrogen or hydroxide ion concentration, respectively. This makes their use highly effective in comparing the relative strengths of acids and bases.