Question

Strong bases are bases that completely ionize in water to produce hydroxide ion, \(\mathrm{OH}^{-}\). The strong bases include the hydroxides of the Group 1 elements. For example, if 1.0 mole of \(\mathrm{NaOH}\) is dissolved per liter, the concentration of \(\mathrm{OH}^{-}\) ion is \(1.0 \mathrm{M}\). Calculate the \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH},\) and \(\mathrm{pH}\) for each of the following strong base solutions. a. \(0.10 M \mathrm{NaOH}\) b. \(2.0 \times 10^{-4} M \mathrm{KOH}\) c. \(6.2 \times 10^{-3} M \mathrm{CsOH}\) d. \(0.0001 M \mathrm{NaOH}\)

Short answer

a. \([\mathrm{OH}^{-}] = 0.10 M\), \(\mathrm{pOH} = 1\), \(\mathrm{pH} = 13\) b. \([\mathrm{OH}^{-}] = 2.0 \times 10^{-4} M\), \(\mathrm{pOH} = 3.7\), \(\mathrm{pH} = 10.3\) c. \([\mathrm{OH}^{-}] = 6.2 \times 10^{-3} M\), \(\mathrm{pOH} = 2.21\), \(\mathrm{pH} = 11.79\) d. \([\mathrm{OH}^{-}] = 0.0001 M\), \(\mathrm{pOH} = 4\), \(\mathrm{pH} = 10\)

Step-by-step solution

a. Calculate \([\mathrm{OH}^{-}]\) for \(0.10 M \mathrm{NaOH}\)

Since strong bases completely ionize in water, the concentration of the \(\mathrm{OH}^{-}\) ions will be the same as the concentration of the given base. In this case: \([\mathrm{OH}^{-}] = 0.10 M\)

a. Calculate \(\mathrm{pOH}\) for \(0.10 M \mathrm{NaOH}\)

We'll use the formula \(\mathrm{pOH} = -\log_{10}\left[\mathrm{OH}^{-}\right]\). Applying the formula, we get: \(\mathrm{pOH} = -\log_{10}(0.10) = 1\)

a. Calculate \(\mathrm{pH}\) for \(0.10 M \mathrm{NaOH}\)

We'll use the formula \(\mathrm{pH} + \mathrm{pOH} = 14\). Since we calculated \(\mathrm{pOH} = 1\), we can find \(\mathrm{pH}\) as follows: \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1 = 13\)

b. Calculate \([\mathrm{OH}^{-}]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for \(2.0 \times 10^{-4} M \mathrm{KOH}\)

Using the same steps as in part (a), we can calculate \([\mathrm{OH}^{-}]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for \(2.0 \times 10^{-4} M \mathrm{KOH}\): \([\mathrm{OH}^{-}] = 2.0 \times 10^{-4}M\) \(\mathrm{pOH} = -\log_{10}(2.0 \times 10^{-4}) = 3.7\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.7 = 10.3\)

c. Calculate \([\mathrm{OH}^{-}]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for \(6.2 \times 10^{-3} M \mathrm{CsOH}\)

Using the same steps as above: \([\mathrm{OH}^{-}] = 6.2 \times 10^{-3}M\) \(\mathrm{pOH} = -\log_{10}(6.2 \times 10^{-3}) = 2.21\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.21 = 11.79\)

d. Calculate \([\mathrm{OH}^{-}]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for \(0.0001 M \mathrm{NaOH}\)

Using the same steps as in the previous parts, we can calculate \([\mathrm{OH}^{-}]\), \(\mathrm{pOH}\), and \(\mathrm{pH}\) for \(0.0001 M \mathrm{NaOH}\): \([\mathrm{OH}^{-}] = 0.0001 M\) \(\mathrm{pOH} = -\log_{10}(0.0001) = 4\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4 = 10\)

Key concepts

Ionization

When we talk about ionization in the context of strong bases, like NaOH, KOH, and CsOH, we're referring to their ability to completely dissociate into ions when dissolved in water. This dissociation results in the formation of hydroxide ions, \( \text{OH}^- \). It's important to understand that strong bases, due to their full ionization, will have the same hydroxide ion concentration as the base's initial concentration.

• Complete Ionization: Strong bases completely ionize, which means that essentially all the molecules of the base become ions. This attribute is different from weak bases, which only partially ionize in solution.
• Relevance of Group 1 Elements: Many strong bases are hydroxides of Group 1 elements on the periodic table, such as sodium (Na), potassium (K), and cesium (Cs).

When you dissolve 1.0 mole of NaOH in water, you'll get 1.0 mole of \( \text{OH}^- \) ions per liter. The ability to calculate the hydroxide concentration directly from the base concentration makes dealing with strong bases fairly straightforward.

Hydroxide Ion Concentration

Hydroxide ion concentration, denoted as \([\text{OH}^-]\), is a measure of the amount of hydroxide ions present in a solution. For strong bases, this concentration is directly equal to the base's molarity. Why is this important?
Understanding the \([\text{OH}^-]\) helps us determine the basicity of the solution because hydroxide ions give bases their characteristic properties. The higher the concentration of \( \text{OH}^- \), the stronger or more basic the solution is.

• Example For NaOH: A 0.10 M NaOH solution will have a hydroxide ion concentration of 0.10 M, because NaOH completely dissociates in water.
• Calculating Hydroxide Concentration: Since strong bases fully dissociate, determining \([\text{OH}^-]\) involves simply acknowledging the base's concentration.

This straightforward relationship is what allows us to easily construct calculations involving \([\text{OH}^-]\), \( \text{pOH} \), and \( \text{pH} \) for these solutions.

pH and pOH Calculations

Calculating pH and pOH helps us understand the acidity or basicity of a solution. The pOH is directly related to the hydroxide ion concentration using the formula:
\[ \text{pOH} = -\log_{10} \left( [\text{OH}^-] \right) \] This logarithmic scale simplifies representing wide-ranging hydroxide concentrations.
Once we know the pOH, we can easily find the pH of the solution with the formula: \[ \text{pH} + \text{pOH} = 14 \] It's like solving a puzzle where knowing one angle opens up the entire angle sum.

• pOH Calculation: If a solution has \([\text{OH}^-] = 0.10\) M, the pOH will be \( \text{pOH} = 1 \).
• Using the 14 Rule: Given the pOH, you find the pH by subtracting the pOH from 14. This relationship results from the constant sum of pH and pOH in pure water at 25°C.

Remember, a low pH indicates an acidic solution, while a high pH means the solution is basic. A strong base solution like NaOH with a high \([\text{OH}^-]\) typically has a low pOH and a correspondingly high pH, reflecting its strong basic nature.