To determine the hydroxide ion concentration [tex]\(\left[ \text{OH}^- \right]\)[/tex] for a solution at [tex]\( 25^{\circ} \text{C} \)[/tex] with a given hydronium ion concentration [tex]\(\left[ \text{H}_3\text{O}^+ \right]=2.35 \times 10^{-3} \, \text{M}\)[/tex], we use the relationship between the concentrations of [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] and [tex]\(\left[ \text{OH}^- \right]\)[/tex] governed by the ion product constant of water [tex]\( K_w \)[/tex].
At [tex]\( 25^{\circ} \text{C} \)[/tex], the ion product constant of water [tex]\( K_w \)[/tex] is:
[tex]\[ K_w = \left[ \text{H}_3\text{O}^+ \right] \left[ \text{OH}^- \right] = 1.0 \times 10^{-14} \, \text{M}^2 \][/tex]
Given:
[tex]\[\left[ \text{H}_3\text{O}^+ \right] = 2.35 \times 10^{-3} \, \text{M}\][/tex]
We need to find [tex]\(\left[ \text{OH}^- \right]\)[/tex]. Rearrange the equation for [tex]\( K_w \)[/tex]:
[tex]\[ \left[ \text{OH}^- \right] = \frac{K_w}{\left[ \text{H}_3\text{O}^+ \right]} \][/tex]
Substitute [tex]\( K_w = 1.0 \times 10^{-14} \)[/tex] and [tex]\(\left[ \text{H}_3\text{O}^+ \right]=2.35 \times 10^{-3} \)[/tex]:
[tex]\[ \left[ \text{OH}^- \right] = \frac{1.0 \times 10^{-14}}{2.35 \times 10^{-3}} \][/tex]
Perform the division:
[tex]\[ \left[ \text{OH}^- \right] = 4.26 \times 10^{-12} \, \text{M} \][/tex]
Therefore, the [tex]\(\left[ \text{OH}^- \right]\)[/tex] for the given solution is:
[tex]\[ \boxed{4.26 \times 10^{-12} \, \text{M}} \][/tex]
From the provided options, the correct answer is E. [tex]\( 4.26 \times 10^{-12} \, \text{M} \)[/tex].
