pH=-log_10[H_3O^+]pH=−log10[H3O+] by definition.
And thus, if pH=10pH=10, [H_3O^+]=10^(-10)*mol*L^-1[H3O+]=10−10⋅mol⋅L−1, (i.e. pH=-log_10(10^(-10))=-(-10)=10pH=−log10(10−10)=−(−10)=10
And if pH=8pH=8, [H_3O^+]=10^(-8)*mol*L^-1[H3O+]=10−8⋅mol⋅L−1, which is one hundred times more concentrated than the first instance, as required. In other words, if the DeltapH=2, there is a 10^2, i.e. one hundredfold difference in [H_3O^+].
Do not be intimidated by the log function. When we write log_ab=c, we ask to what power we raise the base a to get c. Here, a^c=b. And thus log_(10)10=1, , log_(10)100=2, log_(10)10^(-1)=-1 . And log_(10)1=0.
I acknowledge that I have hit you with a lot of facts. But back in the day A level students routinely used log tables before the advent of electronic calculators. If you can get your head round the logarithmic function you will get it.
