To find the hydroxide ion concentration in an aqueous solution at [tex]\(25^{\circ} C\)[/tex] with a given hydronium ion concentration, we can use the relationship between hydronium ions [tex]\([\text{H}_3\text{O}^+]\)[/tex], hydroxide ions [tex]\([\text{OH}^-]\)[/tex], and the ionic product of water [tex]\(K_w\)[/tex], which is constant at [tex]\(25^{\circ} C\)[/tex].
The ionic product of water [tex]\(K_w\)[/tex] at [tex]\(25^{\circ} C\)[/tex] is given by:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \, \text{M}^2 \][/tex]
Given the hydronium ion concentration [tex]\([\text{H}_3\text{O}^+]\)[/tex] is:
[tex]\[ [\text{H}_3\text{O}^+] = 9.7 \times 10^{-5} \, \text{M} \][/tex]
We need to find the hydroxide ion concentration [tex]\([\text{OH}^-]\)[/tex].
Rearrange the [tex]\(K_w\)[/tex] expression to solve for [tex]\([\text{OH}^-]\)[/tex]:
[tex]\[ [\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]} \][/tex]
Substitute the given values into the equation:
[tex]\[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{9.7 \times 10^{-5}} \][/tex]
Calculate the result:
[tex]\[ [\text{OH}^-] \approx 1.0309278350515464 \times 10^{-10} \, \text{M} \][/tex]
Thus, the hydroxide ion concentration of the solution is approximately:
[tex]\[ [\text{OH}^-] \approx 1.0 \times 10^{-10} \, \text{M} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{1.0 \times 10^{-10} \, M} \][/tex]
