We use the following equilibrium..........
H_2OrightleftharpoonsH^+ +HO^(-)H2O⇌H++HO−, and further that we know the equilibrium constant for this expression is......
[H^+]xx[HO^-]=10^-14[H+]×[HO−]=10−14 under standard conditions.
Alternatively, [H_3O^+][HO^-]=10^-14[H3O+][HO−]=10−14. We can take log_10log10 of both sides and get.........
log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)log10[H3O+]+log10[HO−]=log10(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.
