I'm doing this stuff for my college chemistry class. Hope I can help...
First, let me tell you the relationships
a. Relationship between [H+] and [OH-]
Kw = [H3O+][OH-] = 1.0 x 10^-14 (mol/L^2)
Now, if like me, that confusion totally messed with your head, here's my dummie way of remembering. With OH and H3O, if the coefficient of the number before the 10^whatever is a 1 (like 1.5, 1.99, 1.2, etc), then you can use the numbers in the ^whatever part to figure it out b/c the ^numbers of the H3O+ and OH- will always add to -14.
b. Relationships between pH and [H+]
pH = -log[H+]
Okay, so your problems...
Problem 1. If pH=3 , then what is [H+] and [OH-] ?
Okay, get your calculator handy!
a. Type in the ph (3)
b. Change the sign (press change-sign button, it looks like a + and - sign together)
c. Take the antilog (press antilog or 10^x button, or press INV or 2nd function and then log button, depending on your calculator).
d. Your calculator will show 0.001, which is 1.0 x 10^-3
e. You now know that your [H+] value is 1.0 x 10^-3
f. Since the first number is a 1, you can use the -14 - (answer) to get 14 - -3, which is -14 + 3 = -11. Your [OH-] value is therefore 1 x 10^-11
Problem 2: 10 ^-8 (ten to power of negative eight), then what is [OH-] and pH?
a. Since your first number (before the x 10^-8) is, assumingly, a 1, you can use that "it equals -14" rule or whatever for [OH-]
b. -14 - -8 = -14 + 8 = -6
c. You know that your [OH-] value is therefore 1 x 10^-6
d. Enter the molarity of 1.0 (since that is what is before the x 10^-8) in your calculator
e. Press the button that activates exponential mode (EE, Exp, etc) and enter your [H+] value - don't forget it's the [H+] and not [OH-] value!!; in this case that is 8
f. Change the sign (that +- button on your calculator to make it -8, matching your ^-8 from earlier).
g. Take the logarithim (press log botton, LOG on my calculator)
h. Change the sign
i. You get the value of 8 on your calculator. 8 is your pH.
Problem 3. I'm not going to do the rationals since it's same as the last problem, but I'll take you through the steps
a. -14 - -7 = -14 + 7 = 7; your [H+] value is 1 x 10^-7
b. Enter 1.0 in your calculator.
c. Press the button that activates exponential mode (EE, Exp, etc) and enter 7
d. Change the sign (that +- button on your calculator to make it -7).
e. Take the logarithim (press log botton, LOG on my calculator)
f. Change the sign
g. You get the pH value of 7.
You probably noticed that when the number before the x 10^whatever of your [H+] value is 1.0 exactly, the pH will be the positive version of the ^whatever number. But it is very important to learn those calculator steps because your H+ value will hardly ever just be 1.0
Good luck!
