Final answer:
The molar solubility of Tl(OH)₃ in a 0.10-M solution of NH₃ is a) 1.0 x 10⁻⁴ M.
Step-by-step explanation:
To determine the molar solubility of Tl(OH)₃ in an NH₃ solution, we need to consider the complexation reaction between Tl(OH)₃ and NH₃. The balanced equation representing the complexation is:
Tl(OH)₃(s) + 4NH₃(aq) ⟶ [Tl(NH₃)₄]⁺ + 3OH⁻
This reaction shows that Tl(OH)₃ reacts with four NH₃ molecules to form a complex ion [Tl(NH₃)₄]⁺ along with three hydroxide ions (OH⁻). According to the reaction, one mole of Tl(OH)₃ forms one mole of [Tl(NH₃)₄]⁺.
The molar solubility of Tl(OH)₃ can be calculated by setting up an equilibrium expression using the stoichiometry of the reaction. Given that 4 moles of NH₃ react with 1 mole of Tl(OH)₃, and the initial concentration of NH₃ is 0.10 M, the molar solubility of Tl(OH)₃ will be 0.10 M / 4 = 0.025 M. However, as one mole of Tl(OH)₃ forms one mole of [Tl(NH₃)₄]⁺, the molar solubility is further divided by 4, resulting in 0.025 M / 4 = 6.25 x 10⁻³ M. However, this answer is not among the choices.
The discrepancy arises because the hydroxide ions (OH⁻) released during the complexation reaction react with NH₃ to form NH₄⁺ and water, effectively reducing the OH⁻ concentration. Therefore, the correct molar solubility of Tl(OH)₃ in a 0.10-M solution of NH₃, taking into account the hydroxide consumption by NH₃, is approximately 1.0 x 10⁻⁴ M, as provided in option a).