(H^+)(OH^-) = Kw = 1E-14
You know OH^-, solve for H^+.
NOTE: (H^+) = (H3O^+)
You know OH^-, solve for H^+.
NOTE: (H^+) = (H3O^+)
Given that the hydroxide ion concentration is 2.50 x 10⁻⁶ M, we can substitute this value into the equation and solve for the hydronium ion concentration.
[H₃O⁺] [OH⁻] = 1.0 x 10⁻¹⁴
[H₃O⁺] = (1.0 x 10⁻¹⁴) / [OH⁻]
[H₃O⁺] = (1.0 x 10⁻¹⁴) / (2.50 x 10⁻⁶)
[H₃O⁺] = 4.0 x 10⁻⁹ M
Therefore, the hydronium ion concentration in the aqueous solution is 4.0 x 10⁻⁹ M.
The equation for the ion product of water is:
Kw = [H3O+][OH-] = 1.0 x 10^-14
Given that the solution is 2.50 x 10^-6 M in hydroxide ions ([OH-]), we can use the ion product of water to solve for the concentration of hydronium ions ([H3O+]).
First, rearrange the equation to solve for [H3O+]:
[H3O+] = Kw / [OH-]
Substituting the given values:
[H3O+] = (1.0 x 10^-14) / (2.50 x 10^-6)
Calculating this expression:
[H3O+] ≈ 4.0 x 10^-9 M
Therefore, the hydronium ion concentration in the aqueous solution is approximately 4.0 x 10^-9 M.