Alright guys.. so question is at 37 deg Celcius, Kw for H2O is 5 x 10 ^-14, what is Ph at this temperature.
So i solved and got the correct answer by setting the Kw = (OH-)(H+)... then going to 5x10^-14 = x^2. Then i proceeded to solve for x and got ~2.5 x 1o^-7.. plugged this into pH= -log (H3O+) and got the pH is equal to around 6.7 or so... ok so thats all good.
But wouldn't that also mean that i could have used the same value of x (2.5 x 10^-7) for (OH-) and used that in pOH = -log(OH-)?? but then I would get the same pOH as pH and that doesn't make sense... I feel like I am missing something here? Could anyone explain?
So i solved and got the correct answer by setting the Kw = (OH-)(H+)... then going to 5x10^-14 = x^2. Then i proceeded to solve for x and got ~2.5 x 1o^-7.. plugged this into pH= -log (H3O+) and got the pH is equal to around 6.7 or so... ok so thats all good.
But wouldn't that also mean that i could have used the same value of x (2.5 x 10^-7) for (OH-) and used that in pOH = -log(OH-)?? but then I would get the same pOH as pH and that doesn't make sense... I feel like I am missing something here? Could anyone explain?