To find the hydroxide ion ([tex]\(OH^{-}\)[/tex]) concentration in an aqueous solution at [tex]\(25^{\circ} C\)[/tex] given the hydronium ion ([tex]\(H_3O^+\)[/tex]) concentration, you can use the ion-product constant of water ([tex]\(K_w\)[/tex]). At [tex]\(25^{\circ} C\)[/tex], [tex]\(K_w\)[/tex] is [tex]\(1.0 \times 10^{-14} \, \text{mol}^2/\text{L}^2\)[/tex].
Given:
- [tex]\( [H_3O^+] = 0.085 \, \text{M} \)[/tex]
The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the ion-product constant of water:
[tex]\[
K_w = [H_3O^+][OH^-]
\][/tex]
Rearranging this equation to solve for the hydroxide ion concentration:
[tex]\[
[OH^-] = \frac{K_w}{[H_3O^+]}
\][/tex]
Substitute the given values into the equation:
[tex]\[
[OH^-] = \frac{1.0 \times 10^{-14} \, \text{mol}^2/\text{L}^2}{0.085 \, \text{mol/L}}
\][/tex]
Calculate the hydroxide ion concentration:
[tex]\[
[OH^-] = \frac{1.0 \times 10^{-14}}{0.085} \approx 1.176 \times 10^{-13} \, \text{mol/L}
\][/tex]
To express this result with 2 significant digits, round the value:
[tex]\[
[OH^-] \approx 0.0 \, \text{mol/L}
\][/tex]
Thus, the [tex]\(OH^-\)[/tex] concentration in the solution is approximately [tex]\(0.0 \, \text{mol/L}\)[/tex].
