The solubility of the salt M_xA_y is 1.0 × 10^-2 mol/L at 25°C. The osmotic pressure exhibited by a solution saturated with M_xA_y at 25°C is 1.22 atm. Determine the values of x and y by assuming ideal behavior. A) x = 1, y =3 B) x = 2, y = 5 C) x = 1, y = 1 D) x = 2, y = 3 E) none of these

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First, we use the osmotic pressure equation: π = MRT Where π is osmotic pressure, M is molarity, R is the gas constant, and T is temperature in kelvins. We can rearrange this equation to solve for molarity: M = π/RT Plugging in the given values: M = 1.22 atm / (0.0821 L·atm/mol·K × 298 K) = 0.0492 mol/L Now we need to determine the formula of the salt. We can do this by using the fact that the solubility is 1.0 × 10-2 mol/L. Let's assume that the salt dissociates completely into its ions: M_xA_y ⇌ M^+ + _xA^- + _y^- The solubility product expression for this salt is: Ksp = [M^+][_xA^-][_y^-] Since the solubility is 1.0 × 10-2 mol/L, we know that: [M^+] = 1.0 × 10-2 mol/L [_xA^-] = 1.0 × 10-2 mol/L [_y^-] = 1.0 × 10-2 - x mol/L (since we don't know the stoichiometry yet) Now we can substitute these values into the solubility product expression: Ksp = (1.0 × 10-2 mol/L) × (1.0 × 10-2 mol/L) × (1.0 × 10-2 - x mol/L) = 1.0 × 10-14 Simplifying: 1.0 × 10-6 - 1.0 × 10-4x + x2 = 1.0 × 10-14 Since the solubility is small, we can assume that x is also small, and therefore x2 is negligible compared to 1.0 × 10-4x. Therefore: 1.0 × 10-6 - 1.0 × 10-4x = 1.0 × 10-14 Solving for x: 1.0 × 10-4x = 1.0 × 10-6 x = 2 Now we can use the fact that MS2AS is the formula of the salt, with divalent cation M^+ and monovalent anions _{A^- and}^-: MS2AS ⇌ M^+ + 2_{A^- +}^- The values of x and y are therefore: x = 2, y = 3 Therefore, the answer is D.

The Solubility of the Salt MxAy Is 1

The Solubility of the Salt M_xA_y Is 1