pH is a symbolic representation of:
[tex]pH=-\log \lbrack H^+\rbrack[/tex]
and for water we now that:
[tex]\lbrack H^+\rbrack\lbrack OH^-\rbrack=10^{-14}[/tex]
that means the text give us the information to calculate the concentration of protons and therefore the concentration of ions OH-:
[tex]pH=10.2\rightarrow\lbrack H^+\rbrack=10^{-10.2}^{}[/tex]
therefore the concentration of OH- can be calculated:
[tex]\lbrack OH^-\rbrack=\frac{10^{-14}}{10^{-10.2}}=10^{-3.8}=1.58\times10^{-4}\text{ M}[/tex]
but they are not asking about the concetration of OH- they ask about concentration of Ba(OH)2 and for that we need to know the disolution stechiometry
[tex]Ba(OH)_2\rightarrow Ba^++2\times OH^-[/tex]
Whic mean the concentration of Ba(OH)2 is half the concentration of OH-:
[tex]\lbrack Ba(OH)_2\rbrack=\frac{\lbrack OH^-\rbrack}{2}=7.92\times10^{-5}M[/tex]
