QuestionJune 5, 2025

At 25^circ C the ion -product constant of wate is K_(w)=1.0times 10^-14 In an a queous s solut ion, the hydroxide ion conce tration is [OH^-]=2.5times 10^-4M c alcula te the hy dronium ion col ncen tration [H_(3)O^+] in the solution. selec tone [H_(3)O^+]=2.5times 10^-11M [H_(3)O^+]=2.5times 10^-9M [H_(3)O^+]=4.0times 10^-10M [H_(3)O^+]=4.0times 10^-11M

At 25^circ C the ion -product constant of wate is K_(w)=1.0times 10^-14 In an a queous s solut ion, the hydroxide ion conce tration is [OH^-]=2.5times 10^-4M c alcula te the hy dronium ion col ncen tration [H_(3)O^+] in the solution. selec tone [H_(3)O^+]=2.5times 10^-11M [H_(3)O^+]=2.5times 10^-9M [H_(3)O^+]=4.0times 10^-10M [H_(3)O^+]=4.0times 10^-11M
At 25^circ C the ion -product constant of wate is K_(w)=1.0times 10^-14 In an a queous s solut ion, the hydroxide ion
conce tration is [OH^-]=2.5times 10^-4M c alcula te the hy dronium ion col ncen tration [H_(3)O^+] in the solution.
selec tone
[H_(3)O^+]=2.5times 10^-11M
[H_(3)O^+]=2.5times 10^-9M
[H_(3)O^+]=4.0times 10^-10M
[H_(3)O^+]=4.0times 10^-11M

Solution
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Answer

[H_{3}O^{+}] = 4.0 \times 10^{-11}M Explanation 1. Apply the ion-product constant formula Use K_w = [H_3O^+][OH^-]. Given K_w = 1.0 \times 10^{-14} and [OH^-] = 2.5 \times 10^{-4}M, solve for [H_3O^+]. 2. Solve for hydronium ion concentration Rearrange to find [H_3O^+] = \frac{K_w} OH^- = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}}.

Explanation

1. Apply the ion-product constant formula<br /> Use $K_w = [H_3O^+][OH^-]$. Given $K_w = 1.0 \times 10^{-14}$ and $[OH^-] = 2.5 \times 10^{-4}M$, solve for $[H_3O^+]$.<br />2. Solve for hydronium ion concentration<br /> Rearrange to find $[H_3O^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}}$.
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