Problem 26
Question
Challenge Calculate the \(\mathrm{pH}\) of a solution having \(\left[\mathrm{OH}^{-}\right]=8.2 \times 10^{-6} \mathrm{M}\).
Step-by-Step Solution
Verified Answer
The pH of the solution is 8.91.
1Step 1: Understand the relationship between pH and pOH
The pH of a solution is related to its hydrogen ion concentration \( [H^+] \), while the pOH is related to the hydroxide ion concentration \( [OH^-] \). They are connected by the equation \( pH + pOH = 14 \) at 25°C.
2Step 2: Calculate the pOH from hydroxide ion concentration
To find the pOH of the solution, use the formula \( pOH = -\log([OH^-]) \). Plug in the given concentration: \( pOH = -\log(8.2 \times 10^{-6}) = 5.09 \).
3Step 3: Use the relationship between pH and pOH to find pH
With the value of pOH calculated, find the pH by using the relationship \( pH + pOH = 14 \).So, \( pH = 14 - pOH = 14 - 5.09 = 8.91 \).
Key Concepts
pOHhydroxide ion concentrationlogarithmic function
pOH
Understanding pOH helps us link hydroxide ion concentration to the overall acidity or basicity of a solution. **pOH** is a measure of the concentration of hydroxide ions, [OH^-], in a solution. Just like pH, which measures the concentration of hydrogen ions, pOH helps indicate whether a solution is acidic, basic, or neutral.
To find pOH, you use the logarithmic function:
Understanding pOH without knowing pH can already reveal a lot about the nature of the solution. It serves as a crucial component in solving for pH efficiently.
To find pOH, you use the logarithmic function:
- Calculate pOH by taking the negative logarithm (base 10) of the hydroxide ion concentration: \( pOH = -\log([OH^-]) \).
- At 25°C, the sum of pH and pOH is always 14. This relationship helps us convert between pH and pOH when needed.
Understanding pOH without knowing pH can already reveal a lot about the nature of the solution. It serves as a crucial component in solving for pH efficiently.
hydroxide ion concentration
The **hydroxide ion concentration** represents the amount of hydroxide ions ([OH^-]) available in a solution. These ions are pivotal in determining a solution's basicity. A higher concentration of hydroxide ions generally indicates a more basic solution, while lower concentrations suggest a more acidic environment.
When given the hydroxide ion concentration, such as \(8.2 \times 10^{-6} \) M in our original exercise, chemists can use this value to calculate both the pOH and the eventual pH of the solution. The calculation begins by plugging the concentration into the pOH formula: \( pOH = -\log([OH^-]) \).
When given the hydroxide ion concentration, such as \(8.2 \times 10^{-6} \) M in our original exercise, chemists can use this value to calculate both the pOH and the eventual pH of the solution. The calculation begins by plugging the concentration into the pOH formula: \( pOH = -\log([OH^-]) \).
- Solutions with high hydroxide ion concentrations will have low pOH values, indicating a basic solution.
- Conversely, lower concentrations of hydroxide ions will result in higher pOH values, pointing to an acidic solution.
logarithmic function
At the heart of pH and pOH calculations lies the **logarithmic function**. In chemistry, we often deal with very small or very large numbers. Logarithms help simplify these into more manageable numbers, making calculations easier.
The logarithm used in pOH and pH calculations is base 10, often written as \( \log_{10} \) or simply \( \log \). When calculating things like pOH, this designates the expression:
The logarithm used in pOH and pH calculations is base 10, often written as \( \log_{10} \) or simply \( \log \). When calculating things like pOH, this designates the expression:
- For a hydroxide ion concentration \( [OH^-] \), pOH is calculated as \( pOH = -\log([OH^-]) \).
- This function effectively translates small concentrations into straightforward numeric values, such as changing \(8.2 \times 10^{-6} \) M to a pOH of 5.09.
Other exercises in this chapter
Problem 24
Calculate the \(\mathrm{pH}\) of solutions having the following ion concentrations at 298 \(\mathrm{K} .\) \begin{equation} \text { a. }\left[\mathrm{H}^{+}\rig
View solution Problem 25
Calculate the \(\mathrm{pH}\) of aqueous solutions with the following \([\mathrm{H}+]\) at 298 \(\mathrm{K}\). \begin{equation} \text { a. }\left[\mathrm{H}^{+}
View solution Problem 27
Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of aqueous solutions with the following concentrations at 298 \(\mathrm{K}\). \begin{equation} \begin{array}{
View solution Problem 28
Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of aqueous solutions with the following concentration at 298 \(\mathrm{K} .\) \begin{equation} \begin{array}{
View solution