Question

how many moles of na2s2o3 are needed to dissolve 0.075 mol of agbr in a solution volume of 1.0 l, given that ksp for agbr is 3.3×10−13 and kf for the complex ion [ag(s2o3)3−2] is 4.7×1013?

Answer 1

To determine the mass of Na2S2O3 required to dissolve 0.075 mol of AgBr, we use the solubility product and formation constant to find the moles of Na2S2O3, then convert these to mass based on its molar mass of 158.1 g/mol.

Step-by-step explanation:

The question deals with the dissolution of AgBr (silver bromide) in the presence of Na2S2O3 (sodium thiosulfate) to form a complex ion. To find the number of moles of Na2S2O3 needed, we can use the fact that the solubility of AgBr is greatly increased by the formation of the [Ag(S2O3)2]3- complex ion, driven by the large Kf value. The overall equilibrium constant for the dissolution and complex formation is K = Ksp ×Kf.

The amount of AgBr we are dealing with is 0.075 mol. Given the molar mass of AgBr (187.77 g/mol) and the mass of AgBr (1.00 g), we can calculate the concentration of bromide ions in solution. Using the Ksp for AgBr, which is 3.3×10⁻¹³, and the Kf for the complex ion [Ag(S2O3)2]3-, which is 4.7×10¹³, we can determine the mass of Na2S2O3 required to dissolve the AgBr.

To dissolve 0.075 mol of AgBr, a certain number of moles of Na2S2O3 are needed, as indicated by the stoichiometry of the reaction where two moles of S2O32- combine with one mole of Ag+ to form the complex ion. The mass of Na2S2O3 required can be calculated from the number of moles needed and its molar mass, 158.1 g/mol.

Answer 2

0.17 mol Na2S2O3

Step-by-step explanation:

AgBr(s) ⇌ Ag +(aq) + Br−(aq)

Some amount of free Ag+ may also be present in solution, but this amount is limited by the Ksp of AgBr and by the Br− in solution. Keeping in mind that the number of moles of each species will equal its molar concentration in this 1.0 L solution, we can use these values to solve for the maximum concentration of free Ag+:

Ksp = [Ag+][Br−]

3.3E−13 = [Ag+](0.075 M)

[Ag+] = 4.4E−12 M

Now that we know the concentration of free Ag+ in solution, we can calculate the amount of bound Ag+ in the complex (abbreviated here as x) from the original moles of AgBr:

moles of AgBr = moles of free Ag+ + moles of bound Ag+

0.075 mol = 4.4E−12 mol + x

x = 0.075 mol Ag+ in complex

Note that the amount of free Ag+ is so small that virtually all of the silver in solution is part of the [Ag(S2O3)2]3− complex. We also know by the stoichiometry of the complex that the moles of bound Ag+ must be equal to the moles (and concentration) of the complex itself in a 1.0 L solution. Thus, [Ag(S2O3)2]3− = 0.075 M. Given this information, along with the equilibrium concentration of Ag+ found above and the given value of Kf, we can use the expression for Kf to find the equilibrium concentration of [S2O3]2−, which we will abbreviate as y.

Ag+(aq) + 2S2O2−3(aq) ⇌ Ag(S2O3)3−2(aq)

Kf = [Ag(S2O3)2]3− / [Ag+][S2O2−3]2

4.7E13 = 0.075 M / (4.4×10−12 M) × y^2

y = 0.019 M

We must add this amount of free S2O2−3 to the moles of S2O2−3 in the complex to obtain the total moles of S2O2−3 needed to make the solution. Since two equivalents of S2O2−3 are required for each unit of [Ag(S2O3)2]3−, we double the moles of [Ag(S2O3)2]3− to find the moles of S2O2−3 contained therein:

2 × 0.075 moles of [Ag(S2O3)2]3− = 0.15 moles of bound S2O2−3

Adding the moles of bound S2O2−3 to the moles of free S2O2−3 in solution:

0.15 mol + 0.019 mol = 0.17 total moles of S2O2−3

This is the total amount of S2O2−3 that must be added to the solution via the dissolution of Na2S2O3. Luckily, Na2S2O3 is completely soluble, and each equivalent of dissolved Na2S2O3 produces exactly one S2O2−3 ion by the reaction:

Na2S2O3(s) ⟶ 2Na+(aq) + S2O2−3(aq)