pH and Kw question: A student dissolves 1.75g of a drain cleaner (based on NaOH) in water and makes the solution up to 100cm3. The student measures the solution pH as 13.60. Determine the percentage of NaOH in the drain cleaner, in terms of mass (g).

[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.K_w = [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^-3K_w 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, K_a = [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, K_a x 55.6 moldm^-3 = [H⁺] x [OH^-]‘K_a x 55.6moldm^-3’ = K_w. Although it can change with temperature, at 298K/25^oC (room temperature) it is 1.00 x 10^-14. K_w 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