Answered

1. An aqueous solution has a [tex]pH = 8.0[/tex]. Determine the [tex]\left[ H _3 O ^{+} \right][/tex], [tex]pOH[/tex], and [tex]\left[ OH ^{-} \right][/tex].



Answer :

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].