Problem 27 - Chemistry Matter and Change

Question

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of aqueous solutions with the following concentrations at 298 \(\mathrm{K}\). \begin{equation} \begin{array}{l}{\text { a. }\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-6} M} \\ {\text { b. }\left[\mathrm{OH}^{-}\right]=6.5 \times 10^{-4} \mathrm{M}} \\ {\text { c. }\left[\mathrm{H}^{+}\right]=3.6 \times 10^{-9} \mathrm{M}} \\ {\text { d. }\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-2} \mathrm{M}}\end{array} \end{equation}

Step-by-Step Solution

Answer

a) pH = 8, pOH = 6; b) pH = 10.81, pOH = 3.19; c) pH = 8.44, pOH = 5.56; d) pH = 1.60, pOH = 12.40.

Step 1: Concept of pH and pOH

The pH of a solution is defined as \( \text{pH} = -\log[ \text{H}^+ ] \) and the pOH is defined as \( \text{pOH} = -\log[ \text{OH}^- ] \). At 298 K, the sum of pH and pOH is 14, i.e., \( \text{pH} + \text{pOH} = 14 \).

Step 2a: Calculate pOH for a = \([\text{OH}^-]=1.0 \times 10^{-6}\, M\)

To find pOH, use the formula \( \text{pOH} = -\log[ \text{OH}^- ] = -\log(1.0 \times 10^{-6}) \). This simplifies to \( \text{pOH} = 6 \).

Step 3a: Calculate pH for a

Using the relationship \( \text{pH} + \text{pOH} = 14 \), we find \( \text{pH} = 14 - 6 = 8 \).

Step 2b: Calculate pOH for b = \([\text{OH}^-]=6.5 \times 10^{-4}\, M\)

Apply the formula \( \text{pOH} = -\log[ \text{OH}^- ] = -\log(6.5 \times 10^{-4}) \). This calculates to approximately \( \text{pOH} = 3.19 \).

Step 3b: Calculate pH for b

With \( \text{pH} + \text{pOH} = 14 \), we have \( \text{pH} = 14 - 3.19 = 10.81 \).

Step 2c: Calculate pH for c = \([\text{H}^+]=3.6 \times 10^{-9}\, M\)

Use \( \text{pH} = -\log[ \text{H}^+ ] = -\log(3.6 \times 10^{-9}) \), which is approximately \( \text{pH} = 8.44 \).

Step 3c: Calculate pOH for c

With \( \text{pH} + \text{pOH} = 14 \), \( \text{pOH} = 14 - 8.44 = 5.56 \).

Step 2d: Calculate pH for d = \([\text{H}^+]=2.5 \times 10^{-2}\, M\)

Find \( \text{pH} = -\log[ \text{H}^+ ] = -\log(2.5 \times 10^{-2}) \), which calculates to approximately \( \text{pH} = 1.60 \).

Step 3d: Calculate pOH for d

Using \( \text{pH} + \text{pOH} = 14 \), \( \text{pOH} = 14 - 1.60 = 12.40 \).

Key Concepts

Acid-Base Chemistry

Acid-base chemistry revolves around the behavior and properties of acids and bases, which are two types of substances that play a crucial role in chemical reactions.
An acid is any substance that increases the hydrogen ion (\( ext{H}^+\)) concentration in a solution, while a base increases the hydroxide ion (\( ext{OH}^-\)) concentration.
These changes in ion concentration significantly impact a solution's properties. For example:

• Acids tend to have a sour taste and can neutralize bases.
• Bases generally have a bitter taste and slippery feel, and they can neutralize acids.

At 298K, which is a standard room temperature, a unique relationship exists between \( ext{H}^+\) and \( ext{OH}^-\) concentrations. The product of the \( ext{H}^+\) and \( ext{OH}^-\) concentrations in water is always \(1.0 \times 10^{-14}\). This is a central concept in understanding how equilibria work in aqueous solutions and forms the backbone of how to calculate pH and pOH.

Logarithmic Calculations

Logarithmic calculations are central to determining the pH and pOH of solutions in chemistry.
These calculations use logarithms because they simplify the processes involved with very large or very small numbers.
The pH and pOH calculations employ a base-10 logarithm.
The formulas are:

• \( ext{pH} = -\log[\text{H}^+]\)
• \( ext{pOH} = -\log[\text{OH}^-]\)
• \(\text{pH} + \text{pOH} = 14\)

The key benefit of using logarithmic calculations is that they transform complicated multiplication and division of ion concentrations into manageable addition and subtraction.
This logarithmic scale is particularly useful because it turns the little differences in concentration into visible differences in pH or pOH, making it easier to interpret and understand their effect on a solution's properties.

Solution Concentration

Solution concentration refers to the amount of a substance, such as \( ext{H}^+\) or \( ext{OH}^-\) ions, present in a solution.
It is usually expressed in moles per liter (Molar, M) and dictates the strength and reactivity of the solution.
For acids, you measure the \( ext{H}^+\) concentration, and for bases, you measure \( ext{OH}^-\) concentration.
Concentration profoundly influences chemical behavior in solutions.
For example:

• Increasing \( ext{H}^+\) concentration decreases pH, making the solution more acidic.
• Increasing \( ext{OH}^-\) concentration decreases the pOH, making the solution more basic.

When one works with these quantities, such as in the original exercise, one starts with the provided concentrations and employs the pH or pOH formulas to determine the solution’s acidity or basicity.

Hydrogen Ion Concentration

Hydrogen ion concentration (\([\text{H}^+]\)) is a crucial measure in determining the acidity of a solution. A higher \([\text{H}^+]\) corresponds to a more acidic solution, while a lower \([\text{H}^+]\) characterizes a more basic (or less acidic) solution.
The relationship between \([\text{H}^+]\) and pH is logarithmic, making accurate determination essential for understanding a solution’s chemical nature.
For instance, if \([\text{H}^+]\) increases from \(1.0 \times 10^{-7}\) M to \(1.0 \times 10^{-5}\) M, the pH decreases, indicating increased acidity.
When making these calculations, keep in mind:

• The pH scale ranges from 0 to 14, where values less than 7 indicate acidic conditions, 7 is neutral, and values above 7 indicate basic conditions.
• Understanding and calculating \([\text{H}^+]\) can help predict the outcome of chemical reactions and interactions within the solution.