What is the [H3O+] for a neutral solution at 50 degrees celsius?
I know that kw=[H3O+][OH-] but I am not sure what to do to get the H3O. The value I have for kw=5.76e-14 and I have the pH 6.62 but I do not know what the OH would be.
A neutral solution has H3O^+ = OH^- so if the pH = 6.62 then the pOH 6.62
pH = 6.62 = -log(H3O^+).
Solve for (H3O^+)
so I would take the - log of 6.62?
I don't know if I put this in my calculator correctly but I got -0.82.
To find the [H3O+] for a neutral solution at 50 degrees Celsius, you need to consider the autoionization of water and the relationship between [H3O+] and [OH-].
The autoionization of water is represented by the equation: H2O ⇌ H3O+ + OH-
In a neutral solution, the concentration of H3O+ is equal to the concentration of OH-. Thus, [H3O+] and [OH-] will be the same.
You mentioned that you have the value of Kw (the equilibrium constant for the autoionization of water) as 5.76e-14. This value is constant at all temperatures.
At 50 degrees Celsius, Kw remains the same, but the concentration of H3O+ and OH- can be different. To find the concentrations, you can use the equation Kw = [H3O+][OH-] and the fact that [H3O+] = [OH-] in a neutral solution.
So, let's solve for [H3O+] or [OH-] using the given Kw value.
Since [H3O+] = [OH-] in a neutral solution, we can substitute [OH-] for [H3O+] in the equation Kw = [H3O+][OH-]. This gives us Kw = [OH-][OH-] = [OH-]^2.
Taking the square root of both sides to isolate [OH-], we get [OH-] = √Kw.
Now, substitute the value of Kw (5.76e-14) into the equation and calculate the square root. [OH-] = √(5.76e-14) ≈ 7.58e-8 M.
Thus, at 50 degrees Celsius, in a neutral solution, both [H3O+] and [OH-] concentrations are approximately 7.58e-8 M.