# 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

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 Click here to see ALL problems on logarithm 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