please answer asap

Reproduce the table. Fill in the blanks. The formulas are: [H3O+][OH]=11014pH=log[H3O+][H3O+]=10pHpOH=log[OH][OH]=10pOHpH+pOH=14 \begin{tabular}{|l|l|l|l|l|} {[H3O+](M)} & {[OH](M)} & PH & POH & AcidorBase \\ \hline 3.7106 & & & & \\ \hline & 2.0104 & & & \\ \hline & & 3.7 & & \\ \hline & & 3.7 & \\ \hline \end{tabular}