[H+] = 10-pH, so [H+] = 10-13.6 = 2.5 x 10-14 moldm-3. pH=-log10([H+]), so we have rearranged this to make [H+] the subject of the formula.Kw = [H+] x [OH-] = 1.0 x 10-14 moldm-3. This means that [OH-] = (1.0 x 10-14) / (2.5 x 10-14) [OH-] = 0.4 moldm-3Kw is the ionic product of water. Water, although neutral, is able to ionise slightly or partially dissociate, like weak acids do. This is shown in the equation: H2O + H2O H3O+ + OH-For weak acids, Ka = [H+] x [OH-] / [HA]. If we treat water as a weak acid, then [H2O] = 1000/18 = 55.6 moldm-3 (for 1dm3 or 1000g of water; concentration = volume/moles)Now, Ka x 55.6 moldm-3 = [H+] x [OH-]‘Ka x 55.6moldm-3’ = Kw. Although it can change with temperature, at 298K/25oC (room temperature) it is 1.00 x 10-14. Kw can be used to work out the relative concentrations of H+ and OH- ions in an aqueous solution: you will always have both types of ions present, just that in basic solutions you have more OH- ions than H+ ions (and vice versa for acidic solutions). n (moles of NaOH, in mol) = c (concentration of NaOH, in moldm-3) x v (volume of solution, in dm3); n = c x vVolume (dm3) = 100cm3/1000 = 0.1dm3n (NaOH) = 0.4 moldm-3 x 0.1dm3 = 0.04molm (mass of NaOH, in g) = n (moles of NaOH, in mol) x M (molecular mass of NaOH, in g/mol); m = n x MMolecular mass = 23 + 16 + 1 = 40m (NaOH) = 0.04 mol x 40 = 1.6g% mass = (mass of NaOH, g) / (total mass of drain cleaner added, g)% mass of NaOH = 1.6g / 1.75g x 100% = 91.428… = 91.4% to 3SF