Calculate both [H30] and [OH-] for a solution that is:
pH=5.50
ph=7.00
pH = -log(H^+)
Plug and chug for (H^+).
Then Kw = (H^+)(OH^-). Substitute (H^+) and solve for (OH^-).
pH = 7.00
pH = -log(H3O^+)
7.00 = -log(H3O^+)
-7.00 = log(H3O^+)
Now you take the antilog of both sides of the equation. Do you know how to do that on your calculator. Punch in 7.00, change the sign to -7.00, then punch the 10x button on your calculator. The answer should come up as 1 x 10^-7. So 1 x 10^-7 M = (H3O^+). Now to calculate (OH^-).
(H3O^+)(OH^-) = Kw. I'm sure you have had in class that Kw is the ion product for water and that equals 1 x 10^-14 at room T. So
(H3O^+)(OH^-) = 1 x 10^-14
Substitute 1 x 10^-7 for (H3O^+) and solve for (OH^-). You should get 1 x 10^-7 M.
Repost above if you have trouble with any of the others but show your work.
To calculate the concentrations of both [H3O+] and [OH-], we need to use the relationship between pH and pOH, which is given by the equation:
pH + pOH = 14
For the first solution with pH = 5.50:
1. Calculate the pOH:
pOH = 14 - pH = 14 - 5.50 = 8.50
2. Calculate the concentration of [OH-]:
[OH-] = 10^(-pOH) = 10^(-8.50) = 3.16 x 10^(-9) M
3. Calculate the concentration of [H3O+] using the equation for the ion product of water:
[H3O+] = Kw / [OH-]
where Kw is the ionization constant of water (1.0 x 10^(-14) M^2):
[H3O+] = (1.0 x 10^(-14) M^2) / (3.16 x 10^(-9) M) = 3.16 x 10^(-6) M
So, for the solution with pH = 5.50, [H3O+] is 3.16 x 10^(-6) M and [OH-] is 3.16 x 10^(-9) M.
For the second solution with pH = 7.00:
1. Calculate the pOH:
pOH = 14 - pH = 14 - 7.00 = 7.00
2. Calculate the concentration of [OH-]:
[OH-] = 10^(-pOH) = 10^(-7.00) = 1.00 x 10^(-7) M
3. Calculate the concentration of [H3O+] using the equation for the ion product of water:
[H3O+] = Kw / [OH-]
where Kw is the ionization constant of water (1.0 x 10^(-14) M^2):
[H3O+] = (1.0 x 10^(-14) M^2) / (1.00 x 10^(-7) M) = 1.00 x 10^(-7) M
So, for the solution with pH = 7.00, [H3O+] is 1.00 x 10^(-7) M and [OH-] is 1.00 x 10^(-7) M.