Can someone clarify where I'm going wrong with this?
Kw for pure water is 1.2 x 10-15 at 5°C. The pH of pure water at 5°C?
I thought Kw=[H3O][OH]...
I've tried a few different things, but none seem to work.
Thanks!
I figured it out. :)
Wolframalpha Failed to understand what i meant and have me the wrong answer even though I did everything correctly.
Sure, let's break it down step by step.
First, let's start with the equation Kw = [H3O+][OH-], which represents the autoionization of water. At 5°C, the value of Kw is given as 1.2 x 10^-15.
To find the pH of pure water at 5°C, we need to find the concentration of H3O+ ions ([H3O+]). But in neutral water, the concentration of H3O+ ions is equal to the concentration of OH- ions ([OH-]) since the solution is neither acidic nor basic.
So, in pure water, we can assume that [H3O+] = [OH-]. Let's denote their concentration as x.
Now we can substitute this into the Kw equation:
Kw = [H3O+][OH-]
1.2 x 10^-15 = x * x
Next, let's solve this equation:
x^2 = 1.2 x 10^-15
x = √(1.2 x 10^-15)
Using a calculator, we can find that x ≈ 1.097 x 10^-8.
Since [H3O+] = [OH-] = x, the concentration of H3O+ ions in pure water at 5°C is approximately 1.097 x 10^-8 M.
Now, to find the pH, we can use the formula:
pH = -log[H3O+]
Calculating the pH using the concentration of H3O+ ions we just found:
pH = -log(1.097 x 10^-8)
pH ≈ 7.96
Therefore, the pH of pure water at 5°C is approximately 7.96.
I hope this clarifies where you may have gone wrong and how to properly calculate the pH of pure water at 5°C. If you have any further questions, feel free to ask!