I did this in lesson today... I have used the method my teacher gave me but I'm not sure whether the pH is given as 7 or if you have to find that?
Hopefully this is somewhat useful anyway
1. Calculate the pH of water at 50 degrees when Kw= 5.48 x 10^-14. Is the water neutral? Explain your answer.
I am assuming the water pH would be 7. So you do 10 to the power of -7 which gives you 1x10^-7
you then rearrage the Kw equation (Kw= [H+] [OH-]) to OH- = 5.48x10^-14 / 1x10^-7
This equals, 5.48x 10^-7 which would make the water slightly acidic.