Question

To precipitate as Ag2S(s), all the Ag+ present in 338 mL of a saturated solution of AgBrO3 requires 30.4 mL of H2S(g) measured at 23 ∘C and 748 mmHg. What is Ksp for AgBrO3?

**This solution is posted under the text book solutions (General Chemistry, 10th ED. Chapter 18, question 13), but I can't quite follow the process. I'm hoping someone can break apart the steps a bit more.

Answer 1

We have, Ksp = [Ag⁺][BrO₃^-]

Here, AgBrO₃ solution is saturated. Therefore, first, the moles of H₂S can be calculated. From the moles of H₂S, moles of Ag and BrO₃^- can be easily determined.

From the ideal gas law,

PV= nRT

Here, water at 23°C is used. Therefore, its vapor pressure = 21.1 mm Hg

In addition, R (gas constant) can be taken as 62.36 L mmHg K^-1 mol^-1

Therefore,

Pressure of the gas = Total pressure + Vapor pressure

                                = 748 mm Hg - 21.1 mm Hg = 726.9 mm Hg

Now, moles of H₂S = n = PV/RT = (726.9 x 0.0304 L)/(62.36 x 296) = 1.196 x 10^-3 moles

Moles of H₂S = 1.196 x 10^-3 moles = Limiting reagent, since AgBrO₃ is saturated

AgBrO₃ and H₂S react in 1:1 ratio

Ag⁺ + H₂S    Ag₂S

Therefore, moles of Ag⁺= Moles of BrO₃^- = 2 x (1.196 x 10^-3 moles) = 2.392 x 10^-3 moles

[Ag⁺] = [BrO₃^-] = (2.392 x 10^-3 moles) / 0.338 L = 7.0769 x 10^-3 M

Ksp = [Ag⁺][BrO₃^-] = 5.008 x 10^-5