RbOH is a strong base, so it dissociates in water completely. As 1 mole of RbOH produces, 1 mol of hydroxide ions.
Thus, "[OH^-]=[RbOH]=7.5*10^{-2}M"
As this concentration is much greater than the concentration of the concentration of hydroxide ions contributed by water.
"pOH=-log([OH^-])=2-log(7.5)=2-0.875=1.125"
As "pH+pOH=14 \\implies pH=14-1.125=12.875"
And thus, "[H^+]=10^{-12.875}=0.1333*10^-12=1.333*10^{-13}M"
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