A level Chemistry QuestionWatch
At 60° Kw= 9.31x10^-14
At 10° Kw = 2.93 x10^-15
Calculate the pH of water at 60° and 10°
How can I do this question?
But you can assume that [H+]=[OH-] because H2O fully dissociates into H+ and OH- ions
As [H+]=[OH-], [H+][OH-] can be written as [H+]^2
Therefore Kw = [H+]^2
Square root Kw to get [H+], then use -log[H+] to find the pH
At 60°C, the pH is -log(√9.31x10^-14) = 6.52
At 10°C, the pH is -log(√2.93 x10^-15) = 7.27
Hope that helps!