A solution prepared by mixing 47.1 mL of 0.4 M AgNO3 and 47.1 mL of 0.4 M TlNO3 was titrated with 0.8 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TlBr is Ksp = 3.6 × 10–6 and the solubility constant of AgBr is Ksp = 5.0 × 10–13.
(a) Which of the following expressions shows how cell voltage depends on [Ag ]?
b) What is the cell voltage when the following volumes of 0.8 M NaBr have been added? (b) 1.0 mL (c) 13.5 mL (d) 22.6 mL (e) 23.5 mL (f) 23.9 mL (g) 37.0 mL (h) 47.1 mL (i) 50.5 mL
I found part a) to be:
E = .175-(.799-.05916log(1/[Ag+])
I tried doing part b) over 10 times now and I can't figure it out.
To find the cell voltage when different volumes of 0.8 M NaBr have been added, you need to determine the concentration of Ag+ ions at each stage.
The initial state of the solution can be considered as mixing 47.1 mL of 0.4 M AgNO3 and 47.1 mL of 0.4 M TlNO3.
Since both AgNO3 and TlNO3 are strong electrolytes, they completely dissociate in water. Therefore, the initial concentration of AgNO3 and TlNO3 are both equal to their molarities: 0.4 M.
Now, we need to use the solubility products (Ksp) of TlBr and AgBr to determine the concentrations of Tl+ and Ag+ ions.
TlBr(s) ⇌ Tl+(aq) + Br-(aq), and Ksp(TlBr) = [Tl+][Br-]
For TlBr: Ksp = 3.6 × 10–6
[Tl+] = x (let's assume)
Since the initial concentration of TlNO3 is 0.4 M, the concentration of Tl+ ions contributed by TlNO3 is 0.4 M.
Therefore, [Br-] = 3.6 × 10–6 / 0.4
Next, we consider AgBr:
AgBr(s) ⇌ Ag+(aq) + Br-(aq), and Ksp(AgBr) = [Ag+][Br-]
[Tl+] = 0.4 M (contribution from TlNO3)
[Br-] = 3.6 × 10–6 / 0.4
Since the initial concentration of AgNO3 is also 0.4 M, the concentration of Ag+ ions from AgNO3 is 0.4 M.
The remaining Ag+ concentration can be determined by subtracting the Ag+ concentration contributed by the formation of TlBr from the initial concentration of Ag+:
[Ag+] = 0.4 - [Br-] (contribution from TlBr) - [Br-] (contribution from AgNO3)
Now that you have determined the concentration of Ag+ at each stage, you can substitute the respective values of [Ag+] into the given expression of E:
E = 0.175 - (0.799 - 0.05916 log(1/[Ag+]))
Calculating the value of E for each volume of 0.8 M NaBr solution added will give you the cell voltage.
Repeat these steps for volumes of 1.0 mL, 13.5 mL, 22.6 mL, 23.5 mL, 23.9 mL, 37.0 mL, 47.1 mL, and 50.5 mL of 0.8 M NaBr solution to find the respective cell voltages.