A) [H+] = 1 * 10^-6M
B) [H+] = 0.00010M
C) [OH-] = 1 * 10^-2M
D) [OH-] = 1 * 10^-11M
Thanks!
B) [H+] = 0.00010M
C) [OH-] = 1 * 10^-2M
D) [OH-] = 1 * 10^-11M
Thanks!
-
These are all pretty easy. pH = -log[H^+]
Put the value for the [H^+] in your calculator and press the log button. The number you get will be negative. Now press the +/- button and you will obtain your answer.
Example: log(1 * 10^-6) = -6
press +/- and pH = 6
For pH from [OH^-], you must know that [H^+][OH^-] = 1 x 10^-14. Calculate [H^+] from [OH^-] by rearranging the equation to:
[H^+] = (1 x 10^-14)/[OH^-]
In C), for example:
[H^+] = (1 x 10^-14)/(1 x 10^-2) = 1 x 10^-12 Thus,
pH = -log(1 x 10^-12) = 12
You should be able to do the rest without further help.
Hope this is helpful to you. JIL HIR
Put the value for the [H^+] in your calculator and press the log button. The number you get will be negative. Now press the +/- button and you will obtain your answer.
Example: log(1 * 10^-6) = -6
press +/- and pH = 6
For pH from [OH^-], you must know that [H^+][OH^-] = 1 x 10^-14. Calculate [H^+] from [OH^-] by rearranging the equation to:
[H^+] = (1 x 10^-14)/[OH^-]
In C), for example:
[H^+] = (1 x 10^-14)/(1 x 10^-2) = 1 x 10^-12 Thus,
pH = -log(1 x 10^-12) = 12
You should be able to do the rest without further help.
Hope this is helpful to you. JIL HIR