Problem 27 - Chemistry Matter and Change

Question

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

Step-by-Step Solution

Answer

a. \(pOH \approx 6\); \(pH \approx 8\) b. \(pOH \approx 3.19\); \(pH \approx 10.81\) c. \(pH \approx 8.44\); \(pOH \approx 5.56\) d. \(pH \approx 1.60\); \(pOH \approx 12.40\)

Step 1: To do this, we will use the formula: pOH = -log\(_{10}[\mathrm{OH}^{-}]\) \(pOH = -\log_{10}\left(1.0 \times 10^{-6}\right)\) #Step 2: Calculate pH# Now that we have the pOH, we can determine the pH using the relationship:

pH + pOH = 14 \(pH = 14 - pOH\) #b. \(\left[\mathrm{OH}^{-}\right]=6.5 \times 10^{-4} M\) #Step 1: Calculate pOH# First, we need to calculate the pOH of the solution.

Step 2: To do this, we will use the formula: pOH = -log\(_{10}[\mathrm{OH}^{-}]\) \(pOH = -\log_{10}\left(6.5 \times 10^{-4}\right)\) #Step 2: Calculate pH# Now, we can determine the pH.

Use the relationship: pH + pOH = 14 \(pH = 14 - pOH\) #c. \(\left[\mathrm{H}^{+}\right]=3.6 \times 10^{-9} M\) #Step 1: Calculate pH# First, we need to calculate the pH of the solution.

Step 3: To do this, we will use the formula: pH = -log\(_{10}[\mathrm{H}^{+}]\) \(pH = -\log_{10}\left(3.6 \times 10^{-9}\right)\) #Step 2: Calculate pOH# Now, we can determine the pOH.

Use the relationship: pH + pOH = 14 \(pOH = 14 - pH\) #d. \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-2} \mathrm{M}\) #Step 1: Calculate pH# First, we need to calculate the pH of the solution.

Step 4: To do this, we will use the formula: pH = -log\(_{10}[\mathrm{H}^{+}]\) \(pH = -\log_{10}\left(2.5 \times 10^{-2}\right)\) #Step 2: Calculate pOH# Now, we can determine the pOH.

Use the relationship: pH + pOH = 14 \(pOH = 14 - pH\) Once we've calculated the pH and pOH for each part, we'll have the answers to the exercise.

Key Concepts

pOH calculation

Understanding pOH is essential when dealing with bases in chemistry. It helps us measure the alkalinity of a solution. To calculate the pOH, you use the formula:

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

This means you're finding the negative logarithm (base 10) of the hydroxide ion concentration (\([\mathrm{OH}^{-}]\)). For example, if you have a concentration\( \left[ \mathrm{OH}^{-} \right] = 1.0 \times 10^{-6} \) M, the calculation becomes:\[ \text{pOH} = -\log_{10} \left( 1.0 \times 10^{-6} \right) = 6 \]The value indicates the strength of the base: a lower pOH means a stronger base.

hydrogen ion concentration

The concentration of hydrogen ions \([\mathrm{H}^{+}]\) determines the acidity of a solution. It's measured in moles per liter (M). A higher concentration of \([\mathrm{H}^{+}]\) indicates a stronger acid.To find the pH of a solution based on \([\mathrm{H}^{+}]\), use:

• \( \text{pH} = -\log_{10} [\mathrm{H}^{+}] \)

For instance, if you have \( [\mathrm{H}^{+}] = 3.6 \times 10^{-9} \) M, the pH is:\[ \text{pH} = -\log_{10}(3.6 \times 10^{-9}) \approx 8.44 \]A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic one.

hydroxide ion concentration

Hydroxide ions \([\mathrm{OH}^{-}]\) determine the basicity of a solution. Just like hydrogen ions, their concentration is measured in moles per liter (M).To better understand a solution's basic nature, you can relate \([\mathrm{OH}^{-}]\) to pOH using the formula:

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

When given a set concentration, like \( [\mathrm{OH}^{-}] = 6.5 \times 10^{-4} \) M, the pOH is calculated as:\[ \text{pOH} = -\log_{10}(6.5 \times 10^{-4}) \approx 3.19 \]This gives insight into how basic the solution is. Lower pOH values indicate stronger bases.

logarithmic scale

A logarithmic scale is used to handle a wide range of values, like ion concentrations. It simplifies numbers by expressing them as powers of 10.Here's how it applies to pH and pOH:

• The equation \( \text{pH} = -\log_{10} [\mathrm{H}^{+}] \) and \( \text{pOH} = -\log_{10} [\mathrm{OH}^{-}] \) both use logarithms.
• This means each unit change reflects a tenfold change in concentration.

For example, if \( [\mathrm{H}^{+}] \) changes from \( 1 \times 10^{-3} \) M to \( 1 \times 10^{-4} \) M, the pH increases by 1.This scale helps in easily comparing the acidity or basicity of different solutions!