We have that the the resulting pH is mathematically given as
PH=2
From the question we are told
If a solution at pH 5 undergoes a 1000-fold increase in [OH-], what is the resulting pH?
Generally the equation for the solution is is mathematically given as
5=-log[H+]
H+=10e-5*1e3
H+=10e-2
Therefore
-log[H+]=-log[10e-2]
PH=2
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