We use the following equilibrium..........
#H_2OrightleftharpoonsH^+ +HO^(-)#, and further that we know the equilibrium constant for this expression is......
#[H^+]xx[HO^-]=10^-14# under standard conditions.
Alternatively, #[H_3O^+][HO^-]=10^-14#. We can take #log_10# of both sides and get.........
#log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)#,
or......#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#,
or......#-log_10(10^-14)=pH+pOH=14#
But the expression on the left hand side is simply #14#, and this goes back to definition of the log function, the #log# of something is the exponent to which you raise the #"base"# (normally #10# or #e#) to get the bracketed term.
or......#-log_10(10^-14)=-(-14)=14#
And so finally #pH+pOH=14#, you must know this for A-level....
Here #[HO^-]=10^-3*mol*L^-1#, so #pOH=-log_10(10^-3)=-(-3)=3#
#pH=11#, and #pOH=3#, i.e. taking antilogs.......
#[H_3O^+]=10^-11*mol*L^-1.#
