To calculate the $\text{OH}^-$ concentration from the given $\text{H}_3\text{O}^+$ concentration at $25^\circ \text{C}$, we use the water ion product constant:

[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] ]

At $25^\circ \text{C}$, $K_w = 1.0 \times 10^{-14}$.

Given:

$[\text{H}_3\text{O}^+] = 2.0 \times 10^{-13} \, \text{M}$

Calculation:

Rearranging the formula for $[\text{OH}^-]$: [ [\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]} ]

Substitute the values: $[\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-13}}$

Simplify: $[\text{OH}^-] = 5.0 \times 10^{-2} \, \text{M}$

Final Answer:

$[\text{OH}^-] = 5.0 \times 10^{-2} \, \text{M}$

Let me know if you'd like further clarification or additional examples!

Related Questions:

1. What happens to $[\text{H}_3\text{O}^+]$ if the $[\text{OH}^-]$ concentration increases?
2. How does temperature affect the value of $K_w$?
3. Why does the sum of the pH and pOH always equal 14 at $25^\circ \text{C}$?
4. What is the relationship between $[\text{H}_3\text{O}^+]$ and acidity?
5. How can the pH be determined from the $[\text{H}_3\text{O}^+]$ concentration?

Tip:

Always double-check significant figures in your calculations—they should match the precision of the given data!