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?

Question

zavier10 · Answer

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

anonymous · Answer

^^^^^ 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