Absolutely! Let's solve this step by step.
### Step 1: Determine the [tex]\(\left[ H_3O^+ \right]\)[/tex]
The pH of a solution is defined as:
[tex]\[ \text{pH} = -\log \left( \left[ H_3O^+ \right] \right) \][/tex]
Given that pH = 8.0, we can find the [tex]\(\left[ H_3O^+ \right]\)[/tex] concentration by rearranging the equation:
[tex]\[ \left[ H_3O^+ \right] = 10^{-\text{pH}} \][/tex]
Plugging in the value of pH:
[tex]\[ \left[ H_3O^+ \right] = 10^{-8.0} \][/tex]
Thus,
[tex]\[ \left[ H_3O^+ \right] = 1.0 \times 10^{-8} \, \text{M} \][/tex]
### Step 2: Determine the [tex]\(pOH\)[/tex]
The relationship between pH and pOH in water at 25°C is given by:
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
Given that pH = 8.0, we can solve for pOH:
[tex]\[ \text{pOH} = 14 - \text{pH} \][/tex]
Plugging in the pH value:
[tex]\[ \text{pOH} = 14 - 8.0 \][/tex]
Thus,
[tex]\[ \text{pOH} = 6.0 \][/tex]
### Step 3: Determine the [tex]\(\left[ OH^- \right]\)[/tex]
The pOH of a solution is defined as:
[tex]\[ \text{pOH} = -\log \left( \left[ OH^- \right] \right) \][/tex]
Given that pOH = 6.0, we can find the [tex]\(\left[ OH^- \right]\)[/tex] concentration by rearranging the equation:
[tex]\[ \left[ OH^- \right] = 10^{-\text{pOH}} \][/tex]
Plugging in the value of pOH:
[tex]\[ \left[ OH^- \right] = 10^{-6.0} \][/tex]
Thus,
[tex]\[ \left[ OH^- \right] = 1.0 \times 10^{-6} \, \text{M} \][/tex]
### Summary
To summarize, for an aqueous solution with a pH of 8.0:
- The concentration of [tex]\(\left[ H_3O^+ \right]\)[/tex] is:
[tex]\[ 1.0 \times 10^{-8} \, \text{M} \][/tex]
- The pOH is:
[tex]\[ 6.0 \][/tex]
- The concentration of [tex]\(\left[ OH^- \right]\)[/tex] is:
[tex]\[ 1.0 \times 10^{-6} \, \text{M} \][/tex]
These steps yield the desired values of [tex]\(\left[ H_3O^+ \right]\)[/tex], pOH, and [tex]\(\left[ OH^- \right]\)[/tex].
