ok then,
1) amount of H2=4.68L = 4.68/24.47 moles = 0.19 moles of H2
therefore 0.38 moles of OH- have been added into the solution from loss of H+.
originally in water there is 1*10^-7 molL^-1 concentration of OH- and so
in 1.2 L there was 1.2*10^-7 moles of OH-.
therefore now there are...