SOLUTION: I have a problem requiring me to find the value of p^H using the value of [H^+] by using the formula {{{ pH = -log(H) }}} Similarly, I have another problem requiring me to find

Question 614192: I have a problem requiring me to find the value of p^H using the value of [H^+] by using the formula
Similarly, I have another problem requiring me to find the value of [H^+] by using pH.
I'll put these two problems to show what I mean.
1. Determine the p^H given the following value:

2. Determine the [H+] from the following pH value:

I figured that the first question's answer was 9, but it was actually 8.8, I'm guessing that's because of the 1.58, however I don't see how to use 1.58 to help find the answer.
Another thing troubling me about these is, that I don't understand logarithms very much, so I don't understand the process used to solve the problems.
Thanks in advance.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
a logarithm is an exponent that some base must be raised to, to equal some value
___ the base for "common" logarithms is 10 (logs are common unless otherwise indicated)

when quantities are multiplied, their logs are added
___ when divided, the logs are subtracted

log(a) = b means 10^b = a

1. log[H+] = log(1.58) + log(10^-9)

log[H+] = .199 + -9 = -8.801

pH = - log[H+] = - -8.801 = 8.8

2. - log[H+] = 3.3

log[H+] = - 3.3 = .7 - 4 (the log of the coefficient is less than one, while the exponent is an integer))

[H+] = 10^.7 * 10^-4 = 5.012 * 10^-4
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