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
To find out how much AgBr can dissolve in the given sodium thiosulfate solution, we need to determine the concentration of Ag+ ions in the solution.
First, we need to use the balanced equation to find the mole ratio between Ag+ and Na2S2O3:
Ag+(aq) + 2 S2O32-(aq) --> Ag(S2O3)23-(aq)
The stoichiometric ratio tells us that for every 1 mole of Ag+, we need 2 moles of S2O32-. This means that the concentration of Ag+ will be half of the concentration of Na2S2O3.
Given:
Volume of Na2S2O3 solution (V) = 1.40 L
Concentration of Na2S2O3 (C) = 0.425 M (moles per liter)
The moles of Na2S2O3 can be calculated using the formula:
moles = concentration (C) * volume (V)
moles of Na2S2O3 = 0.425 M * 1.40 L = 0.595 mol
Since the ratio between Ag+ and Na2S2O3 is 1:2, the moles of Ag+ will be half of the moles of Na2S2O3:
moles of Ag+ = 0.595 mol / 2 = 0.2975 mol
Now that we know the moles of Ag+, we can use the Ksp expression for AgBr to determine the maximum amount of AgBr that can dissolve.
The Ksp expression for AgBr is:
Ksp = [Ag+][Br-]
Given the Ksp value for AgBr is 5.0 * 10^(-13), we can rearrange the equation to solve for [Ag+], which is the concentration of Ag+ ions in the solution:
[Ag+] = Ksp / [Br-]
Since Ag+ reacts with S2O32- to form a complex ion, we can assume that the Ag+ concentration remains the same throughout the reaction, even after forming the complex ion.
Therefore, the concentration of Ag+ is 0.2975 mol / 1.40 L = 0.2125 M.
Now, we can plug this concentration value into the Ksp expression for AgBr:
5.0 * 10^(-13) = (0.2125 M) * [Br-]
Rearranging the equation, we can solve for [Br-]:
[Br-] = Ksp / [Ag+]
[Br-] = (5.0 * 10^(-13)) / (0.2125 M)
[Br-] ≈ 2.35 * 10^(-12) M
Finally, we can calculate the mass of AgBr that can dissolve in the solution by multiplying the concentration of Br- ions by the molar mass of AgBr.
Molar mass of AgBr = atomic mass of Ag + atomic mass of Br = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol
Mass of AgBr = molar mass * volume * concentration
Mass of AgBr = 187.77 g/mol * 1.40 L * 2.35 * 10^(-12) M
Mass of AgBr ≈ 6.95 * 10^(-9) g
Therefore, approximately 6.95 * 10^(-9) grams of AgBr can dissolve in 1.40 L of 0.425 M Na2S2O3 solution.