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 1013
To find the mass of AgBr that can dissolve in 1.40 L of 0.425 M Na2S2O3 solution, we need to determine the concentration of Ag+ ions in the solution and then use the solubility product constant (Ksp) to calculate the mass.
First, let's find the concentration of Ag+ ions in the solution:
The balanced equation shows that 1 mol of Ag+ reacts with 2 mol of S2O32-. Therefore, the concentration of Ag+ ions is twice that of Na2S2O3.
Concentration of Ag+ = 2 * 0.425 M = 0.85 M
Next, we can use the Ksp expression to find the molar solubility (x) of AgBr in terms of Ag+:
Ksp = [Ag+][Br-]
Ksp = (0.85 M)(x)
Ksp = 5.0 x 10^-13
From the Ksp expression, we can see that the molar solubility of AgBr is x = 5.0 x 10^-13 / 0.85
Now, we can calculate the mass of AgBr that dissolves in 1.40 L of the Na2S2O3 solution.
First, we convert the volume of the solution from liters to moles:
Moles of Na2S2O3 = (0.425 mol/L) * (1.40 L)
Since 1 mol of Na2S2O3 reacts with 1 mol of AgBr, the number of moles of AgBr that dissolves will be the same as the moles of Na2S2O3.
Moles of AgBr = (0.425 mol/L) * (1.40 L)
Finally, we can calculate the mass of AgBr:
Mass of AgBr = Moles of AgBr * Molar mass of AgBr
The molar mass of AgBr can be determined by adding the atomic masses of silver (Ag) and bromine (Br):
Molar mass of AgBr = Atomic mass of Ag + Atomic mass of Br
Now, by substituting the values and performing the calculations, we can find the mass of AgBr that can dissolve in 1.40 L of 0.425 M Na2S2O3 solution.