- #1
omni
- 192
- 1
i need to Calculate the pH of 1x10^-2 NaOH
so i did like this: [OH-]=10^-2
POH=-log[OH-]=-log[10^-2]=2
and if i work with the Formula :pH+POH=14 so i will get pH+2=14
so the for the 1x10^-2 NaOH the ph=12 do i correct?
thnaks.
so i did like this: [OH-]=10^-2
POH=-log[OH-]=-log[10^-2]=2
and if i work with the Formula :pH+POH=14 so i will get pH+2=14
so the for the 1x10^-2 NaOH the ph=12 do i correct?
thnaks.