Least is NaBr (100% dissolved so no NaBr remains, only Na^+1 and Br^-1 H2O yields 10^-7 M H3O^+1 and 10^-7 M OH^-1 (Kw = 1x10^-14 = [H3O+][OH-] Na^+1 and Br^-1 will bothe be 0.1 M H2O is slightly less that 1000 g / L in a 0.1 M NaBr solution, so its concentration is about 55.5 M
NaBr is a strong electrolyte. Hence it dissociates completely in aqueous solution.
Dissociation of NaBr: [tex]NaBr\rightarrow Na^{+}+Br^{-}[/tex]. Hence 1 mol of NaBr produces 1 mol of [tex]Na^{+}[/tex] and 1 mol of [tex]Br^{-}[/tex] upon complete dissociation
So, concentration of NaBr is 0 (M), concentration of [tex]Na^{+}[/tex] is 0.10 (M) and concentration of [tex]Br^{-}[/tex] is 0.10 (M)
Concentrations of [tex]H_{3}O^{+}[/tex] and [tex]OH^{-}[/tex] depends upon solely on autoionization of water.
Hence concentration of [tex]H_{3}O^{+}[/tex] = concentration of [tex]OH^{-}[/tex] = [tex]1.0\times 10^{-7}(M)[/tex]
Concentration of [tex]H_{2}O[/tex] is calculated from it's density (1 g/mL) and molar mass of [tex]H_{2}O[/tex] (18 g/mol). hence concentration of [tex]H_{2}O[/tex] is 55.5 (M)
