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Chemistry 13 Online
OpenStudy (haleigh0922):

have to calculate the pH of [H+]=1.1x10^-8 M and also the pH of [H+]=6.0x10^-5 M for a study guide for my college final can anyone help me with this?

OpenStudy (zavier10):

pH = -lg[H+] = 7.96 for [H+] = 1.1x10^-8 =4.22 for [H+] = 6.0x10^-5

OpenStudy (anonymous):

^^^^^ so basically what the guy above did. When they ask for the pH or pOH, you have to do -log[H3O+] for pH, or if you're finding the POH, -log[OH-]. If you need to find the [OH+] or [H3O+], you do antilog-pH, or antilog-pOH

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