To solve for the hydroxide ion concentration [tex]\(\left[ OH^- \right]\)[/tex] in an aqueous solution at [tex]\(25^{\circ} C\)[/tex] given the hydronium ion concentration [tex]\(\left[ H_3O^+ \right] = 2.4 \times 10^{-4}\)[/tex] M, we can use the ion-product constant for water ([tex]\(K_w\)[/tex]) at this temperature, which is always [tex]\(1.0 \times 10^{-14}\)[/tex] M².
At [tex]\(25^{\circ} C\)[/tex], the relationship between the concentrations of hydronium ions and hydroxide ions in pure water is given by the ion-product constant of water:
[tex]\[ K_w = \left[ H_3O^+ \right] \left[ OH^- \right] \][/tex]
We can rearrange this equation to solve for the hydroxide ion concentration:
[tex]\[ \left[ OH^- \right] = \frac{K_w}{\left[ H_3O^+ \right]} \][/tex]
Now, substitute the known values into the equation:
[tex]\[ \left[ OH^- \right] = \frac{1.0 \times 10^{-14}}{2.4 \times 10^{-4}} \][/tex]
Perform the division:
[tex]\[ \left[ OH^- \right] \approx \frac{1.0 \times 10^{-14}}{2.4 \times 10^{-4}} \approx 4.1666666666666665 \times 10^{-11} \][/tex]
Therefore, the hydroxide ion concentration [tex]\(\left[ OH^- \right]\)[/tex] in the given solution at [tex]\(25^{\circ} C\)[/tex] is approximately [tex]\(4.17 \times 10^{-11}\)[/tex] M.
