Question

Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\) . Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\) b. \(\left[\mathrm{H}^{+}\right]=8.3 \times 10^{-16} M\) c. \(\left[\mathrm{H}^{+}\right]=12 M\) d. \(\left[\mathrm{H}^{+}\right]=5.4 \times 10^{-5} M\) Also calculate the pH and pOH of each of these solutions.

Short answer

a. \(\left[\mathrm{OH}^{-}\right] = 1.0 \times 10^{-7} M\), neutral, pH = 7.0, pOH = 7.0 b. \(\left[\mathrm{OH}^{-}\right] = 1.2 \times 10^{-2} M\), basic, pH = 15.08, pOH = 1.92 c. \(\left[\mathrm{OH}^{-}\right] \approx 8.3 \times 10^{-16} M\), acidic, pH ≈ -1.08, pOH = 15.08 d. \(\left[\mathrm{OH}^{-}\right] \approx 1.9 \times 10^{-10} M\), acidic, pH = 4.27, pOH = 9.73

Step-by-step solution

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}} = 1.0 \times 10^{-7} M\)

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] = \left[\mathrm{OH}^{-}\right]\), this solution is neutral.

Calculate pH and pOH

pH = \(-\log_{10}(1.0 \times 10^{-7}) = 7.0\) pOH = \(-\log_{10}(1.0 \times 10^{-7}) = 7.0\) b.

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{8.3 \times 10^{-16}} = 1.2 \times 10^{-2} M\)

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] < \left[\mathrm{OH}^{-}\right]\), this solution is basic.

Calculate pH and pOH

pH = \(-\log_{10}(8.3 \times 10^{-16}) = 15.08\) pOH = \(-\log_{10}(1.2 \times 10^{-2}) = 1.92\) c.

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{12} \approx 8.3 \times 10^{-16} M\)

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), this solution is acidic.

Calculate pH and pOH

pH = −\(\log_{10}(12)\) ≈ −1.08 pOH = \(-\log_{10}(8.3 \times 10^{-16}) = 15.08\) d.

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{5.4 \times 10^{-5}} \approx 1.9 \times 10^{-10} M\)

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), this solution is acidic.

Calculate pH and pOH

pH = \(-\log_{10}(5.4 \times 10^{-5}) = 4.27\) pOH = \(-\log_{10}(1.9 \times 10^{-10}) = 9.73\)

Key concepts

acidic and basic solutions

Solutions can be categorized as neutral, acidic, or basic based on their hydrogen ion concentration \\([H^+ ]\). A neutral solution has equal concentrations of hydrogen ions, \\([H^+ ]\), and hydroxide ions, \\([OH^- ]\). At 25°C, pure water is neutral because it has equal \\([H^+ ]\) and \\([OH^- ]\) concentrations of \\(1.0 \times 10^{-7} \, \text{M}\).
For an acidic solution, the \\([H^+ ]\) concentration is greater than \\([OH^- ]\), making \\([H^+ ]\) more impactful. Such solutions have a pH of less than 7, which indicates an acidic nature. A pH below 7 suggests higher \\([H^+ ]\) ions.
Conversely, in basic or alkaline solutions, the \\([OH^- ]\) concentration exceeds that of \\([H^+ ]\). Here, the solution will exhibit a pH higher than 7. The higher pH values suggest a dominance of hydroxide ions. Generally, when \\([H^+ ]\) is less than \\([OH^- ]\), the solution is basic or alkaline.

ion product of water

The ion product of water, denoted as \\(K_w\), is fundamental to understanding aqueous solutions. This constant arises from the self-ionization of water, where water molecules produce equal numbers of hydrogen ions \\([H^+ ]\) and hydroxide ions \\([OH^- ]\). The equation representing this balance is:
\[K_w = [H^+ ][OH^- ] = 1.0 \times 10^{-14} \, \text{at 25°C}\]
This relation means that if you know the concentration of hydrogen ions, you can determine the concentration of hydroxide ions, and vice versa.
For example, if you have a solution with \\([H^+ ] = 1.0 \times 10^{-7} \, \text{M}\), then by substituting into the equation, you can solve for \\([OH^- ]\). This constant is pivotal in calculating whether a solution is neutral, acidic, or basic, as it delinearly connects \\([H^+ ]\) and \\([OH^- ]\) concentrations.

hydrogen ion concentration

The hydrogen ion concentration \\([H^+ ]\) is a measure that plays a crucial role in understanding the acidity of a solution. It is determined by the amount of hydrogen ions present in a solution, usually measured in moles per liter (M).
This concentration can directly indicate how acidic a solution is, as more hydrogen ions equate to stronger acidity. The pH scale, which runs from 0 to 14, expresses this concentration in a more digestible logarithmic manner, where a lower pH corresponds to higher \\([H^+ ]\) concentration.
Calculating this concentration from pH involves inverting the logarithm:

• \[ [H^+ ] = 10^{-\text{pH}} \]

For instance, if a solution has a pH of 3, \\([H^+ ]\) would be calculated as \\(10^{-3}\), or \\(0.001 \, \text{M}\). This method highlights how even small changes in pH denote significant changes in acidity.

pOH calculation

pOH is a measure similar to pH but focuses on hydroxide ion concentration \\([OH^- ]\). This measure helps characterize the basicity of a solution and is calculated from the concentration of hydroxide ions.
Just like pH, pOH is expressed on a logarithmic scale:

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

Knowing either pH or pOH enables easy calculation of the other using the relationship:

• \[ \text{pH} + \text{pOH} = 14 \, \text{at 25°C} \]

For example, if you determine \\([OH^- ]\) to be \\(1.0 \times 10^{-7} \, \text{M}\), you can calculate pOH by substituting into the formula to get a pOH of 7.
Determining pOH is useful in titrations and other scenarios where understanding a solution's basicity is crucial.