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Chapter 15: Problem 44

If 0.150 mol of \(\mathrm{KOH}\) is dissolved in 500 \(\mathrm{mL}\) of water, what are \(\left[\mathrm{OH}^{-}\right]\) and \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) ?

Short Answer

Expert verified
[OH-] = 0.300 M and [H3O+] = 3.33 x 10^-14 M

Step by step solution

01

Calculate the Concentration of OH-

First, calculate the concentration of hydroxide ions [OH-], which is given by the moles of KOH divided by the volume of the solution in liters. Here, the volume of the solution is 500 mL or 0.5 L. So, [OH-] = 0.150 mol / 0.5 L = 0.300 M.
02

Calculate the Concentration of H3O+

Now, calculate the concentration of hydronium ions [H3O+]. We know the ion product for water is \(Kw = [H3O+][OH-] = 1.0 x 10^-14\) at 25 degrees Celsius. Hence, we can use this equation to find [H3O+] = Kw / [OH-] = 1.0 x 10^-14 / 0.300 = 3.33 x 10^-14 M.

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