We know that in water the following equilibrium operates:
#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-# #K_"eq"=10^-14#
#[H_3O^+][HO^-]=10^-14#
We take #log_10# of both sides:
#log_10[H_3O^+]+log_10[HO^-]=log_10[10^-14]#
#log_10[H_3O^+]+log_10[HO^-]=-14#, and multiply both sides by #-1#,
#-log_10[H_3O^+]-log_10[HO^-]=14#
But #-log_10[H_3O^+]=pH#, and #pOH=-log_10[HO^-]# by definition.
And thus #pH+pOH=14# (for water under standard conditions). This is something that you have to know.
You have #pH=0.7#, and thus #pOH=13.3#. What is #[H_3O^+]#?