Water undergoes a measurable autoprotolysis...
#H_2O(l) rightleftharpoons H^+ +HO^-#
For which #K_w=[H^+][HO^-]=10^-14# under standard conditions of temperature and pressure...
We take #logs# of both sides, and rearrange...
#log_10K_w=log_10[H^+] +log_10[HO^-]#
or.....#-log_10K_w=underbrace(-log_10[H^+])_(pH)underbrace( -log_10[HO^-])_(pOH)#
And so #pK_w=pH+pOH#
And here #pOH=-log_10(1.0xx10^-5)=5#
And thus #pH=14-5=9#...the solution is SLIGHTLY basic...
And why do we use the #pH# scale? Well, it is throwback to the old pre-electronic calculator days, when logarithmic tables were used for multiplication and division of very large and very small numbers. Students of mathematics or physics or chemists, or engineers would use log tables to calculate expressions that a 50p calculator would EAT today....
