Solutions of sodium thiosulfate are used to dissolve unexposed AgBr (Ksp = 5.0 10-13) in the developing process for black-and-white film. What mass of AgBr can dissolve in 1.40 L of 0.425 M Na2S2O3? Ag+ reacts with S2O32- to form a complex ion.
Ag+(aq) + 2 S2O32-(aq)--> Ag(S2O3)23-(aq) K = 2.9 10^13
AgBr(s) ==> Ag^+ + Br^-
Ag^+ + 2S2O3^2- ==> Ag(S2O3)2
Add the two equations together for their sum and multiply the two k values (Ksp x Kformation). From the total equation you get Keq for the two. It should look like this.
..AgBr(s)+2S2O3^2- =>Ag(S2O3^2-)2 + Br^-
I.........0.425.........0...........0
C...........-2x.........x...........x
E.........0.425-2x......x...........x
Substitute the E line into the Keq expression (Keq = 14.5 if I punched my calculator right.
Solve x (a quadratic is not necessary) which gives M of the S2O3 solution. That x 1.40L gives mols and that x molar mass gives grams. Post your worki if you get stuck.
To calculate the mass of AgBr that can dissolve in 1.40 L of 0.425 M Na2S2O3 solution, we can use the solubility product constant (Ksp) and the reaction stoichiometry.
The balanced equation shows that 1 mol of AgBr reacts with 2 mol of S2O3^2- to form the complex ion Ag(S2O3)2^3-.
Step 1: Calculate the concentration of Ag+(aq)
Since 1 mol of AgBr produces 1 mol of Ag+, and the concentration of Na2S2O3 is given as 0.425 M, the concentration of Ag+(aq) will be the same.
Ag+(aq) concentration = 0.425 M
Step 2: Calculate the concentration of S2O3^2-(aq)
Since the reaction stoichiometry is 1:2 for Ag+ and S2O3^2-, the concentration of S2O3^2- will be twice the concentration of Ag+.
S2O3^2-(aq) concentration = 2 * (0.425 M) = 0.85 M
Step 3: Calculate the concentration of Ag(S2O3)2^3-(aq)
Since the concentration of Ag+(aq) and S2O3^2-(aq) are the same, and the reaction stoichiometry is 1:1 for Ag+ and Ag(S2O3)2^3-, the concentration of Ag(S2O3)2^3- will also be the same.
Ag(S2O3)2^3-(aq) concentration = 0.425 M
Step 4: Calculate the concentration of AgBr that can dissolve
Using the solubility product constant (Ksp = 5.0 × 10^-13), we can set up an equilibrium expression:
Ksp = [Ag+][Br-]
Given that the concentration of Ag+ is 0.425 M, and assuming the concentration of Br- is x (as AgBr dissolves), we can write:
Ksp = (0.425 M)(x) = 5.0 × 10^-13
Solving for x:
x = (5.0 × 10^-13) / (0.425 M) = 1.18 × 10^-12 M
Step 5: Calculate the mass of AgBr that can dissolve
To determine the mass of AgBr that can dissolve, we need to convert the concentration of AgBr (x) to moles using the volume of the solution and the molar mass of AgBr.
moles of AgBr = concentration (M) × volume (L) = (1.18 × 10^-12 M) × (1.40 L)
Finally, to convert moles to mass, we multiply by the molar mass of AgBr (187.77 g/mol):
mass of AgBr = moles of AgBr × molar mass of AgBr = (1.18 × 10^-12 M) × (1.40 L) × (187.77 g/mol)
Calculate this expression to find the mass of AgBr that can dissolve in the solution.
To determine the mass of AgBr that can dissolve in 1.40 L of 0.425 M Na2S2O3, we need to first find the concentration of Ag+ ions in the solution.
Given the balanced equation:
Ag+(aq) + 2 S2O32-(aq) --> Ag(S2O3)23-(aq)
The stoichiometry shows that 1 mol of Ag+ reacts with 2 mol of S2O32-. Therefore, the concentration of Ag+ ions is twice the concentration of Na2S2O3. Thus, the concentration of Ag+ ions can be calculated as follows:
Ag+ concentration = 2 * Na2S2O3 concentration
Given that the Na2S2O3 concentration is 0.425 M, the Ag+ concentration would be:
Ag+ concentration = 2 * 0.425 M = 0.85 M
Next, to find the mass of AgBr that can dissolve, we need to use the solubility product constant (Ksp) expression for AgBr, which is:
Ksp = [Ag+][Br-]
Rearranging the equation, we have:
[Ag+] = Ksp / [Br-]
Given that the Ksp value for AgBr is 5.0 * 10^-13 and we know the Ag+ concentration is 0.85 M, we can solve for [Br-]:
[Br-] = Ksp / [Ag+] = (5.0 * 10^-13) / (0.85 M) = 5.88 * 10^-13 M
The concentration of Br- ions is now known. To find the moles of Br- ions present in the solution, we need to multiply the concentration by the volume of the solution:
moles of Br- = [Br-] * volume
= (5.88 * 10^-13 M) * (1.40 L)
Finally, we can calculate the mass of AgBr dissolved using stoichiometry:
molar mass of AgBr = atomic mass of Ag + atomic mass of Br
molar mass of AgBr = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol
mass of AgBr = moles of Br- * molar mass of AgBr
= [(5.88 * 10^-13 M) * (1.40 L)] * (187.77 g/mol)
Now we plug in the values to calculate the mass of AgBr.