Question

An aqueous solution of a strong base has a pH of 11.04 at \(25^{\circ} \mathrm{C}\). Calculate the concentration of the base if the base is (a) \(\mathrm{KOH}\) and (b) \(\mathrm{Ba}(\mathrm{OH})_{2}\).

Short answer

For KOH, \([\text{KOH}] \approx 1.096 \times 10^{-3} \text{ M}\), and for Ba(OH)2, \([\text{Ba(OH)}_2] \approx 5.48 \times 10^{-4} \text{ M}\).

Step-by-step solution

Understand pH and pOH concepts

The pH of a solution is a measure of the hydrogen ion concentration and is given by the formula \( \text{pH} = -\log[\text{H}^+] \). The pOH is similarly defined for hydroxide ion concentration by \( \text{pOH} = -\log[\text{OH}^-] \). In pure water or neutral solutions at 25°C, \( \text{pH} + \text{pOH} = 14 \).

Calculate pOH from given pH

Given that the pH of the solution is 11.04, we can find pOH using the relation: \( \text{pOH} = 14 - \text{pH} \). Substitute the given pH value: \( \text{pOH} = 14 - 11.04 = 2.96 \).

Determine hydroxide ion concentration

Once pOH is known, the hydroxide ion concentration can be found using the formula \( [\text{OH}^-] = 10^{-\text{pOH}} \). For a pOH of 2.96: \[ [\text{OH}^-] = 10^{-2.96} \approx 1.096 \times 10^{-3} \text{ M} \].

Calculate the concentration of KOH

Since each molecule of KOH dissociates into one \(\text{OH}^-\) ion and one \(\text{K}^+\) ion, the concentration of KOH is the same as the hydroxide ion concentration: \[ [\text{KOH}] = 1.096 \times 10^{-3} \text{ M} \].

Calculate the concentration of Ba(OH)2

Each molecule of \(\text{Ba(OH)}_2\) dissociates into one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions. Therefore, the base concentration is half of the hydroxide concentration: \[ [\text{Ba(OH)}_2] = \frac{1.096 \times 10^{-3}}{2} = 5.48 \times 10^{-4} \text{ M} \].

Key concepts

Understanding Strong Bases

A strong base is a substance that completely dissociates into its ions in water. This means when a strong base, like potassium hydroxide (KOH) or barium hydroxide \(\text{Ba(OH)}_2\), is dissolved, it breaks apart entirely into its constituent ions.

• Strong bases produce a significant amount of hydroxide ions \(\text{OH}^-\).
• This complete dissociation increases the solution's pH, making it more basic.
• Examples include sodium hydroxide (NaOH), potassium hydroxide (KOH), and barium hydroxide \(\text{Ba(OH)}_2\).

Due to this complete dissociation, calculations involving strong bases often revolve around their hydroxide ion concentration. Understanding this behavior is crucial for hydrating their ionic solutions effectively.

Hydroxide Ion Concentration

Hydroxide ion concentration is a measure of the amount of \(\text{OH}^-\) ions present in a solution. It directly influences the solution's pH and pOH.

The relationship between pH and pOH is such that, for aqueous solutions at \(25^{\circ}\text{C}\), their sum equals 14. Therefore, knowing either pH or pOH allows you to determine hydroxide ion concentration.

The formula used to find hydroxide ion concentration from pOH is:\[ [\text{OH}^-] = 10^{-\text{pOH}} \]This calculation is critical in determining how basic a solution is.

The Process of Dissociation

Dissociation refers to the splitting of a compound into its component ions when dissolved in water. Strong bases completely dissociate, providing hydroxide ions and sometimes metal cations.

Here’s what happens step-by-step:

• KOH: Dissociates into one potassium ion \(\text{K}^+\) and one hydroxide ion \(\text{OH}^-\).
• Ba(OH)\(_2\): Dissociates into one barium ion \(\text{Ba}^{2+}\) and two hydroxide ions \(\text{OH}^-\).

This dissociation impacts the calculation of the concentration of the base needed to achieve a specific pH.

Calculating Ba(OH)2 Concentration

When calculating the concentration of \(\text{Ba(OH)}_2\), understanding its dissociation is key. Each \(\text{Ba(OH)}_2\) molecule produces two hydroxide ions. Hence the concentration of \(\text{Ba(OH)}_2\) is half that of the hydroxide ion concentration.

For instance, given the hydroxide ion concentration as \(1.096 \times 10^{-3} \text{ M}\), the concentration of \(\text{Ba(OH)}_2\) can be calculated by dividing by two:\[ [\text{Ba(OH)}_2] = \frac{1.096 \times 10^{-3}}{2} = 5.48 \times 10^{-4} \text{ M} \]This calculation confirms the stoichiometric relationships resulting from dissociation and highlights the necessary steps to find precise concentrations in a solution.