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Q.
The $[ OH ]^{-}$in a solution is $1\, mol\, L ^{-1}$. The $pH$ of the solution is
J & K CETJ & K CET 2013Equilibrium
Solution:
Given, $\left[ OH ^{-}\right]=1\, mol\, L^{-1}$
So, we know that
$p O H=-\log \left[O H^{-}\right]$
$p O H=-\log 1=0$
$\because p H +p O H=14$
$\therefore p H=14-0=14$