molar mass HA = 87.8 g/mole
Given 0.140g HA + 14.5ml(0.110M NaOH) => NaA + H₂O
Rxn is a 1:1 rxn ration => moles HA neutralized = moles NaOH used
=> 0.140g/molar mass of HA = (0.110M)(0.0145L)
=> molar mass of HA = (0.140g)/[(0.110moles/L)(0.0145L)] = 87.8 g/mole
- both b and d.
-since b and d are the closest coastlines then that makes it the most affected.