Final answer:
To find [OH-], divide Kw by [H3O+]. At 25 °C, where Kw=1.0×10⁻¹⁴, if [H3O+] is given, rearrange the equation as [OH-] = Kw / [H3O+]. Alternatively, if you have pH, calculate pOH as 14 minus pH and then use [OH-] = 10⁻ᵐ₀₈.
Step-by-step explanation:
To find the hydroxide ion concentration ([OH-]) given the hydronium ion concentration ([H3O+]) and the ionic product of water (Kw), you can use the equilibrium constant expression for water. This expression states that Kw is equal to the product of the concentrations of the hydronium ions ([H3O+]) and hydroxide ions ([OH-]). At 25 °C, Kw is 1.0 × 10⁻¹⁴.
You can rearrange the equation to solve for [OH-] by dividing Kw by the given [H3O+] concentration:
[OH-] = Kw / [H3O+]
For example, if [H3O+] is 1 × 10⁻⁴ M, you would divide the constant Kw (1.0 × 10⁻¹⁴) by the given [H3O+] to calculate [OH-]. It implications that an increase in [H3O+] results in a decrease in [OH-] to maintain the constant value of Kw.
If you desire to use the pH and pOH scales, you can calculate [OH-] by first determining the pOH:
pOH = 14 - pH
Once you have pOH, you can find [OH-] using the inverse logarithmic relationship:
[OH-] = 10⁻ᵐ₀₈