Calculate the Ksp for silver sulfite if the solubility of Ag2SO3 in pure water is 4.6 × 10-3 g/L.Group of answer choices1.5 × 10-144.8 × 10-102.4 × 10-103.8 × 10-15

To calculate the Ksp (solubility product constant) for silver sulfite (Ag2SO3), we first need to understand the dissociation of Ag2SO3 in water:

Ag2SO3(s) ⇌ 2Ag+(aq) + SO3^2-(aq)

From the balanced chemical equation, we can see that for each formula unit of Ag2SO3 that dissolves, 2 moles of Ag+ ions and 1 mole of SO3^2- ions are produced.

Given that the solubility of Ag2SO3 is 4.6 × 10^-3 g/L, we first need to convert this to moles/L (Molarity, M) because Ksp is usually expressed in terms of molar concentrations.

The molar mass of Ag2SO3 = 2*(molar mass of Ag) + molar mass of S + 3*(molar mass of O) = 2107.87 g/mol + 32.06 g/mol + 316.00 g/mol = 247.8 g/mol

So, the molar solubility of Ag2SO3 = (4.6 × 10^-3 g/L) / (247.8 g/mol) = 1.86 × 10^-5 M

Now, we can set up the expression for Ksp. According to the balanced chemical equation, the Ksp is [Ag+]^2[SO3^2-]. But since the dissolution of one formula unit of Ag2SO3 produces two Ag+ ions, the concentration of Ag+ is 2 times the molar solubility of Ag2SO3, and the concentration of SO3^2- is equal to the molar solubility of Ag2SO3.

So, Ksp = (2*1.86 × 10^-5)^2 * (1.86 × 10^-5) = 1.3 × 10^-14

Therefore, the Ksp for silver sulfite is approximately 1.3 × 10^-14. This value is not exactly the same as any of the answer choices you provided, but it is closest to 1.5 × 10^-14.