To determine the concentration of [tex]\([ \text{OH}^- ]\)[/tex] in a solution given the concentration of [tex]\([ \text{H}_3 \text{O}^+ ]\)[/tex], we need to use the water dissociation constant, [tex]\(K_w\)[/tex].
At 25°C, the value of [tex]\(K_w\)[/tex] is [tex]\(1.0 \times 10^{-14}\)[/tex]. This constant represents the product of the molar concentrations of hydrogen ions [tex]\([ \text{H}_3 \text{O}^+ ]\)[/tex] and hydroxide ions [tex]\([ \text{OH}^- ]\)[/tex] in water:
[tex]\[ K_w = [ \text{H}_3 \text{O}^+ ] \times [ \text{OH}^- ] \][/tex]
Given:
[tex]\[ [ \text{H}_3 \text{O}^+ ] = 1.0 \times 10^{-5} \, \text{M} \][/tex]
We need to find:
[tex]\[ [ \text{OH}^- ] \][/tex]
Rearrange the equation:
[tex]\[ [ \text{OH}^- ] = \frac{K_w}{ [ \text{H}_3 \text{O}^+ ] } \][/tex]
Substitute the known values:
[tex]\[ [ \text{OH}^- ] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} \][/tex]
Perform the division:
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{-14} \div 1.0 \times 10^{-5} \][/tex]
Since we are dividing powers of 10, we subtract the exponents:
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{(-14) - (-5)} \][/tex]
[tex]\[ [ \text{OH}^- ] = 1.0 \times 10^{-9} \][/tex]
Thus, the concentration of [tex]\([ \text{OH}^- ]\)[/tex] in the solution is:
[tex]\[ 1.0 \times 10^{-9} \, \text{M} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{1 \times 10^{-9} \, \text{M}} \][/tex]
