Calculate the molar solubility of barium thiosulfate in 0.0045 M Na2S2O3(aq). (Hint: Check any simplifying assumptions that you make.)

BaS2O3 ==> Ba^+2 + S2O3^-2

Ksp = (Ba^+2)(S2O3^-2)

If you call solubility BaS2O3 = S, then S2O3^-2 = S from BaS2O3 and 0.0045 from Na2S2O3 for a total of S+0.0045M. Substitute into Ksp and solve for S. I assume from the hint that you can not make the assumption that S+0.0045 = 0.0045; i.e., that S is not negligible compared to 0.0045 which probably means a quadratic equation solution.

To calculate the molar solubility of barium thiosulfate (Ba(S2O3)2), we need to set up and solve an equilibrium expression.

The balanced chemical equation for the dissolution of barium thiosulfate is:
Ba(S2O3)2(s) ⇌ Ba2+(aq) + 2S2O3^2-(aq)

Let's assume that the molar solubility of Ba(S2O3)2 is "x" mol/L in the saturated solution. This means, at equilibrium, the concentration of Ba2+(aq) will also be "x" mol/L, and the concentration of S2O3^2-(aq) will be 2x mol/L.

Now, let's set up the equilibrium expression using the balanced chemical equation:
Ksp = [Ba2+][S2O3^2-]^2

Substituting the equilibrium concentrations:
Ksp = x * (2x)^2 = 4x^3

The value of Ksp for Ba(S2O3)2 has been determined to be 1.0 x 10^-12 at room temperature.

Now, we can solve the Ksp expression for x:
1.0 x 10^-12 = 4x^3

To solve this equation, we can take the cube root of both sides:
(x^3) = (1.0 x 10^-12) / 4
x^3 = 2.5 x 10^-13

Taking the cube root of both sides, we get:
x ≈ 6.3 x 10^-5

Therefore, the molar solubility of barium thiosulfate in 0.0045 M Na2S2O3(aq) is approximately 6.3 x 10^-5 mol/L.

To calculate the molar solubility of barium thiosulfate (BaS2O3) in 0.0045 M Na2S2O3(aq), we need to first write and balance the chemical equation for the dissolution of BaS2O3.

The balanced equation for the dissolution of BaS2O3 is:
BaS2O3(s) ⇌ Ba2+(aq) + S2O32-(aq)

According to this equation, 1 mole of BaS2O3 will produce 1 mole of Ba2+ ions and 1 mole of S2O32- ions.

The solubility of BaS2O3 can be represented by the molar solubility, which is defined as the number of moles of BaS2O3 that dissolve per liter of solution.

Let's assume that x moles of BaS2O3 dissolve per liter of solution. Since each mole of BaS2O3 dissociates to produce 1 mole of Ba2+ ions and 1 mole of S2O32- ions, the concentration of Ba2+ and S2O32- ions will also be x M.

Now, let's express the solubility product expression for BaS2O3. The solubility product constant, Ksp, can be calculated as the product of the concentrations of the ions raised to their stoichiometric coefficients.

Ksp = [Ba2+][S2O32-]

Plugging in the concentrations, we get:
Ksp = x * x = x^2

We know that the concentration of Na2S2O3 is 0.0045 M, which means the concentration of S2O32- ions is also 0.0045 M.

Using the stoichiometric ratio from the balanced equation, we know that 1 mole of BaS2O3 produces 1 mole of S2O32- ions. Therefore, the concentration of S2O32- ions is the same as the concentration of Na2S2O3.

Hence, [S2O32-] = 0.0045 M.

Substituting this value back into the solubility product expression, we have:
Ksp = x^2 = [Ba2+][S2O32-] = [Ba2+](0.0045)

We need to find the value of x, which represents the molar solubility of BaS2O3.

Now, considering that the concentration of Ba2+ ions is equal to the concentration of S2O32- ions, we substitute [Ba2+] into the equation:

Ksp = x^2 = [Ba2+][S2O32-] = (x)(0.0045)

Since the concentration of [Ba2+] = concentration of [S2O32-] = 0.0045 M, we can rewrite the equation as:

Ksp = x^2 = (0.0045)(0.0045)

Now, solve for x by taking the square root of both sides:

x = √(Ksp)

To calculate the molar solubility of barium thiosulfate in 0.0045 M Na2S2O3(aq), you need to consult a reference or a reliable source for the value of the solubility product constant (Ksp) for BaS2O3. Then, substitute the value of Ksp into the equation x = √(Ksp) to obtain the molar solubility of BaS2O3.