Identify the relative molar amounts of the species in 0.10 M NaBr(aq). Br-, OH-, H2O, H3O+, NaBr, Na+

NaBr < H3O+1 = OH-1 < Na^+1 = Br^-1 < H2O

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

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OrethaWilkison OrethaWilkison · Answer 2 · 2019-09-23T01:56:48-04:00

Answer:

Answer has been given below

Explanation:

NaBr is a strong electrolyte. Hence it dissociates completely in aqueous solution.
Dissociation of NaBr: \(NaBr\rightarrow Na^{+}+Br^{-}\). Hence 1 mol of NaBr produces 1 mol of \(Na^{+}\) and 1 mol of \(Br^{-}\) upon complete dissociation
So, concentration of NaBr is 0 (M), concentration of \(Na^{+}\) is 0.10 (M) and concentration of \(Br^{-}\) is 0.10 (M)
Concentrations of \(H_{3}O^{+}\) and \(OH^{-}\) depends upon solely on autoionization of water.
Hence concentration of \(H_{3}O^{+}\) = concentration of \(OH^{-}\) = \(1.0\times 10^{-7}(M)\)
Concentration of \(H_{2}O\) is calculated from it's density (1 g/mL) and molar mass of \(H_{2}O\) (18 g/mol). hence concentration of \(H_{2}O\) is 55.5 (M)