Question

What mass of AgBr will dissolve in 500.0 mL of 2.50 M NH₃? (Ksp for AgBr is 5.0 × 10⁻13. For the reaction: Ag(aq) + 2NH₃(aq) = Ag(NH₃)₂+(aq) K = 1.7 x 10⁷)

A) 1.05 x 10⁻4 grams
B) 1.25 x 10^-3 grams
C) 2.0 x 10^-4 grams
D) 2.5 x 10^-3 grams

Answer 1

The mass of AgBr that will dissolve in a solution of 500.0 mL of 2.50 M NH₃ can be calculated using the Ksp equation for AgBr and the K value for the formation of Ag(NH₃)₂⁺. By substituting the given values and using stoichiometry, the mass is determined to be 1.825 grams.

Step-by-step explanation:

To determine the mass of AgBr that will dissolve in 500.0 mL of 2.50 M NH₃, we can use the Ksp equation for AgBr and the K value for the formation of Ag(NH₃)₂⁺. The balanced equation for the formation of Ag(NH₃)₂⁺ is Ag⁺ (aq) + 2NH₃(aq) ⇒ Ag(NH₃)₂⁺(aq). From the Ksp equation for AgBr, we know that the Ksp value is 5.0 × 10⁻¹³. Therefore, we can set up the following equation:

[Ag⁺][Br⁻] = Ksp

Substituting the given values, we have:

(2.50 M)[Br⁻] = 5.0 × 10⁻¹³

[Br⁻] = (5.0 × 10⁻¹³) / (2.50 M)

[Br⁻] = 2.0 × 10⁻¹³ M

Since each AgBr dissociates into one Ag⁺ ion and one Br⁻ ion, we can say that the concentration of Ag⁺ ions is also 2.0 × 10⁻¹³ M.

Now, using the K value for the formation of Ag(NH₃)₂⁺, we can set up the following equation:

K = [Ag(NH₃)₂⁺] / ([Ag⁺][NH₃]²)

Substituting the known values:

(1.7 × 10⁷) = [Ag(NH₃)₂⁺] / ((2.0 × 10⁻¹³ M)(2.50 M)²)

[Ag(NH₃)₂⁺] = (1.7 × 10⁷)(2.0 × 10⁻¹³ M)(2.50 M)²

Now, we can calculate the moles of Ag(NH₃)₂⁺ formed:

Moles = (1.7 × 10⁷)(2.0 × 10⁻¹³ M)(2.50 M)²

Moles = 0.017

Since there is a 1:1 stoichiometric ratio between AgBr and Ag(NH₃)₂⁺, the moles of AgBr that dissolve is also 0.017.

Now, we can calculate the mass of AgBr using its molar mass:

Mass = Moles × Molar mass

Mass = 0.017 × (107.87 g/mol)

Mass = 1.825 g

Therefore, the mass of AgBr that will dissolve in 500.0 mL of 2.50 M NH₃ is 1.825 grams.