To calculate the concentration of hydroxide ions [tex]\([\text{OH}^-]\)[/tex] in a solution where the hydronium ion concentration [tex]\([\text{H} _3 \text{O} ^{+}]\)[/tex] is given, we can use the relationship between [tex]\([\text{H} _3 \text{O} ^{+}]\)[/tex] and [tex]\([\text{OH}^-]\)[/tex]. This relationship is derived from the ion-product constant for water ([tex]\(K_w\)[/tex]) at 25°C, which is:
[tex]\[K_w = [\text{H} _3 \text{O} ^{+}][\text{OH}^-] = 1.0 \times 10^{-14}\][/tex]
Given:
[tex]\[ [\text{H} _3 \text{O} ^{+}] = 3.77 \times 10^{-12} \text{ M} \][/tex]
We need to find [tex]\([\text{OH}^-]\)[/tex]. Re-arrange the ion-product constant equation to solve for [tex]\([\text{OH}^-]\)[/tex]:
[tex]\[ [\text{OH}^-] = \frac{K_w}{[\text{H} _3 \text{O} ^{+}]} \][/tex]
Substitute the given values into the equation:
[tex]\[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{3.77 \times 10^{-12}} \][/tex]
Now, calculate the value:
[tex]\[ [\text{OH}^-] \approx 0.002652519893899204 \text{ M} \][/tex]
So, the concentration of hydroxide ions [tex]\([\text{OH}^-]\)[/tex] in the KOH solution is approximately [tex]\(0.002652519893899204 \text{ M}\)[/tex].
