Question

Base Solutions For each strong base solution, determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\) and \(\mathrm{pOH} .\) $$\begin{array}{l}{\text { a. } 0.15 \times 10^{-3} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}} \\ {\text { b. } 1.5 \times 10^{-3} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}} \\ {\text { c. } 4.8 \times 10^{-4} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}} \\ {\text { d. } 8.7 \times 10^{-5} \mathrm{M} \mathrm{KOH}}\end{array}$$

Short answer

For a: \(\left[\mathrm{OH}^{-}\right]=0.3 \times 10^{-3} \mathrm{M}\), \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 3.3 \times 10^{-12} \mathrm{M}\), \(\mathrm{pH} = 11.48\), \(\mathrm{pOH} = 2.52\). Repeat the calculations similarly for b, c, and d to get their respective concentrations and pH values.

Step-by-step solution

Determine \(\left[\mathrm{OH}^{-}\right]\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Since \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissociates into \(\mathrm{Ca}^{2+}\) and two \(\mathrm{OH}^{-}\) per formula unit, multiply the molarity of \(\mathrm{Ca}(\mathrm{OH})_{2}\) by 2 to get \(\left[\mathrm{OH}^{-}\right]\). For part a, it will be \(0.15 \times 10^{-3}\)M times 2 = \(0.3 \times 10^{-3}\)M. Do the same for part b.

Determine \(\left[\mathrm{OH}^{-}\right]\) for \(\mathrm{Sr}(\mathrm{OH})_{2}\) and \(\mathrm{KOH}\)

Repeat the process for \(\mathrm{Sr}(\mathrm{OH})_{2}\), which also produces 2 moles of \(\mathrm{OH}^{-}\) per mole of compound. For \(\mathrm{KOH}\), it produces 1 mole of \(\mathrm{OH}^{-}\) and the molarity remains the same.

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)

Use the formula \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = \frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}\), where \(K_{w} = 1.0 \times 10^{-14}\) at 25°C. Divide \(K_{w}\) by each \(\left[\mathrm{OH}^{-}\right]\) calculated in the previous steps.

Calculate \(\mathrm{pOH}\) and \(\mathrm{pH}\)

Use the formulas \(\mathrm{pOH} = -\log(\left[\mathrm{OH}^{-}\right])\) and \(\mathrm{pH} = 14 - \mathrm{pOH}\). Calculate \(\mathrm{pOH}\) and then use it to find \(\mathrm{pH}\) for each solution.

Compile the results

List the \(\left[\mathrm{OH}^{-}\right]\), \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for each solution (a, b, c, and d).

Key concepts

Hydroxide Ion Concentration

Understanding the concentration of hydroxide ions \( [OH^-] \) in a solution is crucial for analyzing the properties of bases. A strong base like calcium hydroxide \(Ca(OH)_2\) will dissociate completely in water, releasing hydroxide ions into the solution. If you have a concentration of the base, you would need to consider the number of \(OH^-\) ions produced per formula unit to find the actual hydroxide ion concentration. For example, \(Ca(OH)_2\) releases two \(OH^-\) ions, so you'd multiply the molarity of \(Ca(OH)_2\) by two.
When working with other strong bases, such as potassium hydroxide \(KOH\), which only releases one \(OH^-\) ion per molecule, the concentration of \(KOH\) is equal to the concentration of \(OH^-\) in the solution. To recap, always factor in the number of hydroxide ions produced by the dissociation of one unit of the base when calculating \( [OH^-] \).

Hydronium Ion Concentration

Conversely, the hydronium ion concentration \( [H_3O^+] \) indicates the acidity of a solution, but it can also be derived in a basic solution through the self-ionization equilibrium of water. In pure water at 25°C, the product of the hydronium ion concentration and the hydroxide ion concentration is always constant, denoted as \(K_w = 1.0 \times 10^{-14} \).
To find the hydronium ion concentration in a strong base solution, which possesses a known \( [OH^-] \), you can apply the formula \( [H_3O^+] = \frac{K_w}{[OH^-]} \). This inversely proportional relationship allows you to calculate the \( [H_3O^+] \) by dividing the ion product constant of water by the hydroxide ion concentration, connecting the two measures of ions in the solution.

pOH Calculation

The pOH of a solution is a measure of its basicity on a logarithmic scale. Calculating pOH is straightforward once you have the hydroxide ion concentration. You use the formula \( pOH = -\log([OH^-]) \). The negative sign ensures that a higher concentration of hydroxide ions (more basic solution) results in a lower pOH value. It's important to remember that as the pOH decreases, the basicity of the solution increases. This is a key concept when comparing the strengths of base solutions numerically. A lower pOH value indicates a stronger base.

pH Calculation

Lastly, understanding pH is essential for analyzing 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. To calculate pH from pOH for a strong base solution, you can utilize the simple relationship \( pH + pOH = 14 \) at 25°C.
By subtracting the pOH from 14, you obtain the pH, revealing the solution's degree of basicity. For example, a solution with a pOH of 2 will have a corresponding pH of 12, classifying it as a strongly basic solution. Understanding how to convert between pOH and pH allows students to have a complete picture of the solution's chemical properties.