Question

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

Answer 1

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)

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