Question

Part A

Use the molar solubility 1.08×10−5M in pure water to calculate Ksp for BaCrO4.

Ksp =

Part B

Use the molar solubility 1.55×10−5M in pure water to calculate Ksp for Ag2SO3.

Ksp =

Part C

Use the molar solubility 2.22×10−8M in pure water to calculate Ksp for Pd(SCN)2.

Ksp =

Answer 1

Part A

The K_sp expression is:

Ksp = [Ba²⁺][CrO₄^2-]

There is a 1:1 molar ratio between the BaCrO4 that dissolves and Ba²⁺ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved BaCrO4 and CrO₄^2- in solution. This means that, when 1.08 x 10¯⁵ mole per liter of BaCrO4 dissolves, it produces 1.08 x 10¯⁵ mole per liter of Ba²⁺ and 1.08 x 10¯⁵ mole per liter of CrO₄^2- in solution.

Putting the values into the K_sp expression, we obtain:

K_sp = [1.08 x 10¯⁵] [1.08 x 10¯⁵]

K_sp = 1.1664 x 10^-10

PART B

The K_sp expression is:

K_sp = [Ag²⁺]² [SO₃²¯]

We know the following:

These is a 2:1 ratio between the concentration of the Ag²⁺ and the molar solubility of theAg₂SO₃.

There is a 1:1 ratio between the concentation of the SO₃²¯ ion and the molar solubility of the Ag₂SO₃.

K_sp = [Ag²⁺]² [SO₃²¯]

K_sp = [3.1×10⁻⁵]² [1.55×10⁻⁵]

K_sp = 1.489 x 10^-14

PART C

The K_sp expression is:

K_sp = [Pd²⁺] [SCN^-1]²

K_sp = [ 2.22×10⁻⁸] [4.44×10⁻⁸]²

K_sp = 4.376 x 10^-23