Question

The \(\mathrm{pH}\) of solution made by mixing \(50 \mathrm{~mL}\) of \(0.01 \mathrm{M}\) barium hydroxide solution with \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) is: (a) \(3.0\) (b) \(6.0\) (c) \(12.0\) (d) \(15.0\)

Short answer

The pH of the solution is 12.0.

Step-by-step solution

Understand the Problem Statement

We need to find the \( \mathrm{pH} \) of a solution made by mixing \( 50 \) mL of \( 0.01 \mathrm{M} \) barium hydroxide \( \left(\mathrm{Ba(OH)_2}\right) \) with \( 50 \) mL of water. Barium hydroxide is a strong base, and each molecule provides two hydroxide ions \( \left(\mathrm{OH}^{-}\right) \) to the solution.

Calculate the Total Volume of the Solution

The solution is prepared by mixing \( 50 \) mL of barium hydroxide solution with \( 50 \) mL of water. Therefore, the total volume of the resultant solution is \( 50 + 50 = 100 \) mL.

Determine the Concentration of Hydroxide Ions

Since \( \mathrm{Ba(OH)_2} \) dissociates completely in solution to give \( 2 \mathrm{OH}^- \) ions, the concentration of \( \mathrm{OH}^- \) will be \( 2 \times 0.01 \mathrm{M} \) in the original \( 50 \) mL solution. Thus, \( [\mathrm{OH}^-] = 0.02 \mathrm{M} \).

Adjust Concentration for Dilution

After mixing with \( 50 \) mL water, the solution has doubled in volume, reducing the concentration of \( \mathrm{OH}^- \) by half. Thus, the new concentration is \( \frac{0.02}{2} = 0.01 \mathrm{M} \) in the 100 mL solution.

Calculate the pOH of the Solution

The \( \mathrm{pOH} \) is calculated using the formula \( \mathrm{pOH} = -\log_{10} [\mathrm{OH}^- ] \). Substituting the concentration: \( \mathrm{pOH} = -\log_{10} (0.01) = 2 \).

Determine the pH of the Solution

Since \( \mathrm{pH} + \mathrm{pOH} = 14 \), we can find \( \mathrm{pH} \) by subtracting \( \mathrm{pOH} \) from 14. Thus, \( \mathrm{pH} = 14 - 2 = 12 \).

Key concepts

barium hydroxide

Barium hydroxide, often denoted as \( \mathrm{Ba(OH)_2} \), is an inorganic compound commonly used for titration in chemistry labs due to its complete dissociation in water. It is categorized as a strong base, meaning it fully ionizes in solution to release hydroxide ions \( \mathrm{OH^-} \). This characteristic makes barium hydroxide highly effective in altering the pH of a solution, raising it significantly when mixed with water. The chemical dissociation of barium hydroxide in water can be represented by the reaction:

• \( \mathrm{Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-} \)

This reaction indicates that each mole of \( \mathrm{Ba(OH)_2} \) yields two moles of hydroxide ions, which significantly impacts the hydroxide ion concentration of the solution. Understanding the dissociation process and relative strength of the base is crucial for correct pH calculations in various chemical applications.

hydroxide ion concentration

The concentration of hydroxide ions in a solution is pivotal for calculating the solution's pH and pOH. In the context of barium hydroxide, the concentration of \( \mathrm{OH^-} \) ions can initially be determined prior to any dilution. For example, in a \( 0.01 \mathrm{M} \) barium hydroxide solution, the dissociation step informs us that:

• \( [\mathrm{OH^-}] = 2 \times 0.01 \mathrm{M} = 0.02 \mathrm{M} \)

This calculation follows from knowing that each molecule of barium hydroxide releases two hydroxide ions. It is important to remember that any dilution, such as adding water, will affect this concentration. Calculating the hydroxide ion concentration accurately is a critical step for further calculating the pOH and eventually the pH of the solution.

pOH calculation

Calculating pOH is an essential skill when dealing with basic solutions. The pOH is related to the concentration of hydroxide ions \([\mathrm{OH^-}]\) and is calculated using the formula:

• \( \mathrm{pOH} = -\log_{10} [\mathrm{OH^-}] \)

For example, with a hydroxide ion concentration of \( 0.01 \mathrm{M} \), the calculation would be:

• \( \mathrm{pOH} = -\log_{10} (0.01) = 2 \)

This numerical value of pOH further connects to the pH of the solution through the fundamental relationship:

• \( \mathrm{pH} + \mathrm{pOH} = 14 \)

Thus, knowing the pOH helps complete our understanding of the solution's acidity or basicity by easily determining the pH.

dilution effect in solutions

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. In the context of this exercise, understanding the dilution effect is crucial because the mixing of \( 50 \mathrm{~mL} \) of barium hydroxide solution with \( 50 \mathrm{~mL} \) of water effectively doubles the total volume of the solution. This results in a new total volume of \( 100 \mathrm{~mL} \), reducing the concentration of hydroxide ions by half:

• New concentration of \( \mathrm{OH^-} = \frac{0.02 \mathrm{M}}{2} = 0.01 \mathrm{M} \)

Understanding dilution is vital as it directly affects any subsequent pOH and pH calculations. Chemical solutions often require precise concentrations for accurate reactions and analysis, making understanding dilution effects fundamental in chemical calculations and experiments.