Question

For each hydrogen or hydroxide ion concentration listed, calculate the concentration of the complementary ion and the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the solution. a. \(\left[\mathrm{H}^{+}\right]=5.72 \times 10^{-4} M\) b. \(\left[\mathrm{OH}^{-}\right]=8.91 \times 10^{-5} M\) c. \(\left[\mathrm{H}^{+}\right]=2.87 \times 10^{-12} M\) d. \(\left[\mathrm{OH}^{-}\right]=7.22 \times 10^{-8} M\)

Short answer

a. \(\mathrm{pH} = 3.24\), \(\mathrm{pOH} = 10.76\), \(\left[\mathrm{OH}^{-}\right] = 1.74 \times 10^{-11} M\)\newline b. \(\mathrm{pH} = 4.05\), \(\mathrm{pOH} = 9.95\), \(\left[\mathrm{H}^{+}\right] = 8.91 \times 10^{-5} M\)\newline c. \(\mathrm{pH} = 11.54\), \(\mathrm{pOH} = 2.46\), \(\left[\mathrm{OH}^{-}\right] = 2.87 \times 10^{-12} M\)\newline d. \(\mathrm{pH} = 7.14\), \(\mathrm{pOH} = 6.86\), \(\left[\mathrm{H}^{+}\right] = 7.22 \times 10^{-8} M\)

Step-by-step solution

Calculate \(\mathrm{pH}\)#

Given \(\left[\mathrm{H}^{+}\right] = 5.72 \times 10^{-4} M\), we will calculate \(\mathrm{pH}\) using the formula: \(\mathrm{pH} = -\log(\left[\mathrm{H}^{+}\right])\) \(\mathrm{pH} = -\log(5.72 \times 10^{-4})\) Now calculate the value for \(\mathrm{pH}\).

Calculate \(\mathrm{pOH}\)#

We will use the relationship: \(\mathrm{pOH} = 14 - \mathrm{pH}\) Now calculate the value for \(\mathrm{pOH}\).

Calculate \(\left[\mathrm{OH}^{−}\right]\)#

As we have calculated \(\mathrm{pOH}\), we can find \(\left[\mathrm{OH}^{-}\right]\) using the formula: \(\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\) Now calculate the value for \(\left[\mathrm{OH}^{-}\right]\). ##Part b:##

Calculate \(\mathrm{pOH}\)#

Given \(\left[\mathrm{OH}^{-}\right] = 8.91 \times 10^{-5} M\), we will calculate \(\mathrm{pOH}\) using the formula: \(\mathrm{pOH} = -\log(\left[\mathrm{OH}^{-}\right])\) \(\mathrm{pOH} = -\log(8.91 \times 10^{-5})\) Now calculate the value for \(\mathrm{pOH}\).

Calculate \(\mathrm{pH}\)#

We will use the relationship: \(\mathrm{pH} = 14 - \mathrm{pOH}\) Now calculate the value for \(\mathrm{pH}\).

Calculate \(\left[\mathrm{H}^{+}\right]\)#

As we have calculated \(\mathrm{pH}\), we can find \(\left[\mathrm{H}^{+}\right]\) using the formula: \(\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\) Now calculate the value for \(\left[\mathrm{H}^{+}\right]\). ##Part c:##

Calculate \(\mathrm{pH}\)#

Given \(\left[\mathrm{H}^{+}\right] = 2.87 \times 10^{-12} M\), we will calculate \(\mathrm{pH}\) using the formula: \(\mathrm{pH} = -\log(\left[\mathrm{H}^{+}\right])\) \(\mathrm{pH} = -\log(2.87 \times 10^{-12})\) Now calculate the value for \(\mathrm{pH}\).

Calculate \(\mathrm{pOH}\)#

We will use the relationship: \(\mathrm{pOH} = 14 - \mathrm{pH}\) Now calculate the value for \(\mathrm{pOH}\).

Calculate \(\left[\mathrm{OH}^{−}\right]\)#

As we have calculated \(\mathrm{pOH}\), we can find \(\left[\mathrm{OH}^{-}\right]\) using the formula: \(\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\) Now calculate the value for \(\left[\mathrm{OH}^{-}\right]\). ##Part d:##

Calculate \(\mathrm{pOH}\)#

Given \(\left[\mathrm{OH}^{-}\right] = 7.22 \times 10^{-8} M\), we will calculate \(\mathrm{pOH}\) using the formula: \(\mathrm{pOH} = -\log(\left[\mathrm{OH}^{-}\right])\) \(\mathrm{pOH} = -\log(7.22 \times 10^{-8})\) Now calculate the value for \(\mathrm{pOH}\).

Calculate \(\mathrm{pH}\)#

We will use the relationship: \(\mathrm{pH} = 14 - \mathrm{pOH}\) Now calculate the value for \(\mathrm{pH}\).

Calculate \(\left[\mathrm{H}^{+}\right]\)#

As we have calculated \(\mathrm{pH}\), we can find \(\left[\mathrm{H}^{+}\right]\) using the formula: \(\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\) Now calculate the value for \(\left[\mathrm{H}^{+}\right]\).

Key concepts

Hydrogen Ion Concentration

Understanding the hydrogen ion concentration (\( [\text{H}^+] \)) in a solution is crucial for many areas of chemistry, especially when dealing with acids and bases. This concentration refers to the amount of hydrogen ions present per unit volume in a solution. It's often measured in moles per liter (M). The more hydrogen ions present, the more acidic the solution is.

Calculating the hydrogen ion concentration is typically done by taking the negative logarithm (base 10) of the concentration, which gives you the pH value. To reverse the process, and find the hydrogen ion concentration from a known pH, one would use the following formula: \( [\text{H}^+] = 10^{-\text{pH}} \). This calculation helps in quantifying the acidity and is fundamental for solving pH problems in chemistry.

Hydroxide Ion Concentration

In contrast to hydrogen ions, hydroxide ions (\( [\text{OH}^-] \) indicate the basicity of a solution. The hydroxide ion concentration is the amount of hydroxide ions in a liter of solution. Similar to the hydrogen ion concentration, pOH is a measure of the hydroxide ion concentration on a logarithmic scale, defined as \( \text{pOH} = -\text{log}([\text{OH}^-]) \).

To determine the hydroxide concentration from a known pOH, you would use the formula \( [\text{OH}^-] = 10^{-\text{pOH}} \). This measurement is equally important when balancing the acidic and basic nature in chemical reactions and is complementary to the pH.

pOH Calculation

Calculating the pOH of a solution is a direct process once the hydroxide ion concentration is known. You can calculate pOH by taking the negative logarithm of the hydroxide ion concentration: \( \text{pOH} = -\text{log}([\text{OH}^-]) \).

This calculation is essential for understanding the basicity of a solution. It's especially relevant in the context of the pH scale as it provides a complete picture of a solution's acidity or alkalinity when combined with its complementary pH.

pH and pOH Relationship

Constant Sum of pH and pOH

One of the fundamental relationships in acid-base chemistry is between pH and pOH. These two measures are inversely related and always add up to 14 at 25°C. The relationship is described by the equation \( \text{pH} + \text{pOH} = 14 \).

From One to the Other

This means if you know one value, you can easily find the other by subtracting from 14. For example, if a solution has a pH of 3, its pOH would be 14 - 3 = 11. This relationship allows chemists to quickly balance equations involving acids and bases and understand the neutralization process.