Answer :
Let's solve the problem step-by-step to find the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] in a solution with a pOH of 8.20.
### Step 1: Calculate the pH
Given the formula that relates pH and pOH in water:
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
we can substitute the given pOH into the equation:
[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]
[tex]\[ \text{pH} = 14 - 8.20 \][/tex]
[tex]\[ \text{pH} = 5.80 \][/tex]
### Step 2: Calculate the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex]
To find the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] ions in the solution, use the formula:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 10^{-\text{pH}} \][/tex]
Substitute the calculated pH into the formula:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 10^{-5.80} \][/tex]
The result is:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 1.6 \times 10^{-6} \][/tex]
### Conclusion
Based on the calculations above, the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] in a solution with a pOH of 8.20 is:
[tex]\[ 1.6 \times 10^{-6} \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \quad 1.6 \times 10^{-6}} \][/tex]
### Step 1: Calculate the pH
Given the formula that relates pH and pOH in water:
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
we can substitute the given pOH into the equation:
[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]
[tex]\[ \text{pH} = 14 - 8.20 \][/tex]
[tex]\[ \text{pH} = 5.80 \][/tex]
### Step 2: Calculate the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex]
To find the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] ions in the solution, use the formula:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 10^{-\text{pH}} \][/tex]
Substitute the calculated pH into the formula:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 10^{-5.80} \][/tex]
The result is:
[tex]\[ [ \text{H}_3\text{O}^+ ] = 1.6 \times 10^{-6} \][/tex]
### Conclusion
Based on the calculations above, the concentration of [tex]\( \text{H}_3\text{O}^+ \)[/tex] in a solution with a pOH of 8.20 is:
[tex]\[ 1.6 \times 10^{-6} \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \quad 1.6 \times 10^{-6}} \][/tex]