A dilute solution of Ca(OH)2 has a concentration of 5.2x10^-4M
Calculate PH POH and H+
and another problem I don't understand is
What is the [OH-] of a solution with a PH of 3.95?
i'm failing chemistry
and I don't know how to do these problems
if you can help me thank you so much
i'm more looking for an explanation of how to do these problems, not just the answers.
Calculate PH POH and H+
and another problem I don't understand is
What is the [OH-] of a solution with a PH of 3.95?
i'm failing chemistry
and I don't know how to do these problems
if you can help me thank you so much
i'm more looking for an explanation of how to do these problems, not just the answers.
-
Ok, quick crash course in calculating these things.
Ca(OH2) is a strong base. That means it will dissociate completely, so: Ca(OH)2 --> Ca2+ + 2OH-
Therefore, [OH]- is actually twice the Ca(OH)2 cause every calcium hydroxide molecule gives rise to 2 OH- ions. Make sense so far? That means that the [OH-] is actually 10.4x10^-4, which can be rewritten as 1.04 x 10^-3.
pOH= --log[OH-]. Simply take the negative logarithm of that concentration and that will tell you the pOH.
Now let's calculate pH. Here's an important equation to know:
pH + pOH = 14. Whatever you got for pOH, simply subtract that number from 14 and that tells you the pH. For example, when pH is 2, pOH will be 12.
(This may be beyond the scope of your class but I'll give you the background for why the equation is like that. Don't worry if you don't understand it. You might not have covered this stuff yet. Water spontaneously dissociates into a proton and a hydroxide molecule. The vast majority of the water is still H2O, but a few water molecules will split into those two ions. The equation goes like this: H2O--> H+ + OH-. The equilibrium constant for the equation is Kw = [H+][OH-], and at 25 degrees Celsius, the value of Kw is 10^-14. Therefore, when you take the negative logarithm of both sides, you get pH + pOH = pKw, and pKw is 14 when the water is room temperature).
Now let's figure out [H+]. pH= --log[H+]. Divide both sides by the negative sign. Now you have --pH = log[H+]. Simply use both of those as exponents of base 10, and you find that 10^(--pH) = [H+].
(Since 10^(log X) = X).
For your second question, 14 - 3.95 = 10.05. That's the pOH. Now, pOH= --log[OH-], so divide both sides by -1, you get -10.05=log[OH-]. Raise both sides to base 10, and you get 10^(-10.05) = [OH-].
Good luck! Study hard!
Ca(OH2) is a strong base. That means it will dissociate completely, so: Ca(OH)2 --> Ca2+ + 2OH-
Therefore, [OH]- is actually twice the Ca(OH)2 cause every calcium hydroxide molecule gives rise to 2 OH- ions. Make sense so far? That means that the [OH-] is actually 10.4x10^-4, which can be rewritten as 1.04 x 10^-3.
pOH= --log[OH-]. Simply take the negative logarithm of that concentration and that will tell you the pOH.
Now let's calculate pH. Here's an important equation to know:
pH + pOH = 14. Whatever you got for pOH, simply subtract that number from 14 and that tells you the pH. For example, when pH is 2, pOH will be 12.
(This may be beyond the scope of your class but I'll give you the background for why the equation is like that. Don't worry if you don't understand it. You might not have covered this stuff yet. Water spontaneously dissociates into a proton and a hydroxide molecule. The vast majority of the water is still H2O, but a few water molecules will split into those two ions. The equation goes like this: H2O--> H+ + OH-. The equilibrium constant for the equation is Kw = [H+][OH-], and at 25 degrees Celsius, the value of Kw is 10^-14. Therefore, when you take the negative logarithm of both sides, you get pH + pOH = pKw, and pKw is 14 when the water is room temperature).
Now let's figure out [H+]. pH= --log[H+]. Divide both sides by the negative sign. Now you have --pH = log[H+]. Simply use both of those as exponents of base 10, and you find that 10^(--pH) = [H+].
(Since 10^(log X) = X).
For your second question, 14 - 3.95 = 10.05. That's the pOH. Now, pOH= --log[OH-], so divide both sides by -1, you get -10.05=log[OH-]. Raise both sides to base 10, and you get 10^(-10.05) = [OH-].
Good luck! Study hard!