Question

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

Answer 1

Answer has been given below

Step-by-step 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* 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)

Answer 2

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