A solution prepared by mixing 16.7 mL of 0.760 M NaCl and 16.7 mL of 0.760 M KI was titrated with 0.380 M AgNO3 in a cell containing a silver indicator electrode and a saturated calomel reference electrode.
(a) What is [Ag+] when 16.2 mL of 0.380 M AgNO3 have been added? Express your answer as x, where [Ag+] is a quotient having the form Ksp,AgI/x.
(b) What is [Ag+] when 48.0 mL of 0.380 M AgNO3 have been added? Express your answer as y, where [Ag+] is a quotient having the form Ksp,AgCl/y.
d)The difference between voltages measured when 48.0 mL and 16.2 mL of AgNO3 have been added is 0.3857 V. Using this voltage difference and your answers to questions (a), (b), and (c), calculate the numerical value of the quotient Ksp,AgCl/Ksp,AgI.
**answer to C was E = ( 0.799 - 0.05916log(1/[Ag+]) ) - 0.241**
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To solve this problem, we'll follow these steps:
Step 1: Calculate the initial moles of NaCl and KI.
Step 2: Determine the moles of AgNO3 in 16.2 mL and 48.0 mL.
Step 3: Calculate the molar concentration of Ag+ ions in the solution at each point.
Step 4: Use the given formula to find the value of [Ag+].
Step 5: Calculate the voltage difference.
Step 6: Use the voltage difference and the values of [Ag+] to calculate the quotient Ksp,AgCl/Ksp,AgI.
Let's go step-by-step:
Step 1: Calculate the initial moles of NaCl and KI.
Moles of NaCl = concentration × volume = 0.760 M × 16.7 mL = 12.692 mmol
Moles of KI = concentration × volume = 0.760 M × 16.7 mL = 12.692 mmol
Step 2: Determine the moles of AgNO3 in 16.2 mL and 48.0 mL.
Moles of AgNO3 in 16.2 mL = concentration × volume = 0.380 M × 16.2 mL = 6.156 mmol
Moles of AgNO3 in 48.0 mL = concentration × volume = 0.380 M × 48.0 mL = 18.24 mmol
Step 3: Calculate the molar concentration of Ag+ ions in the solution at each point.
The initial volume is 16.7 mL, so the total volume after adding 16.2 mL of AgNO3 is 16.7 mL + 16.2 mL = 32.9 mL.
The concentration of Ag+ ions at this point can be calculated using the moles of AgNO3 and the total volume:
[Ag+] = (moles of AgNO3) / (total volume)
= (6.156 mmol) / (32.9 mL)
= 0.1870 M
Similarly, the concentration of Ag+ ions after adding 48.0 mL of AgNO3 can be calculated:
[Ag+] = (moles of AgNO3) / (total volume)
= (18.24 mmol) / (64.7 mL)
= 0.2820 M
Step 4: Use the given formula to find the value of [Ag+].
For part (a): The formula is [Ag+] = Ksp,AgI / x
where x is the value we need to find.
Substituting the known values:
0.1870 M = Ksp,AgI / x
For part (b): The formula is [Ag+] = Ksp,AgCl / y
where y is the value we need to find.
Substituting the known values:
0.2820 M = Ksp,AgCl / y
Step 5: Calculate the voltage difference.
Voltage difference = E(48.0 mL) - E(16.2 mL) = 0.3857 V
Step 6: Use the voltage difference and the values of [Ag+] to calculate the quotient Ksp,AgCl/Ksp,AgI.
Using the given formula, we have:
0.799 - 0.05916log(1/[Ag+]) - 0.241 = 0.3857
Now, you can solve this equation to find the numerical value of the quotient Ksp,AgCl/Ksp,AgI.
Please note that steps 1-4 were done to obtain the value needed for part (c), which is not mentioned in your question. Let me know if you have any further questions.
To solve this problem, we'll need to use the principles of equilibrium and electrochemistry. Here's how you can approach each part of the question:
(a) To calculate [Ag+] when 16.2 mL of 0.380 M AgNO3 has been added, we need to understand that AgNO3 reacts with the NaCl and KI in the solution to form AgCl and AgI precipitates, respectively. The concentration of Ag+ can be determined using the solubility product constant (Ksp) for AgI and the volumes and concentrations of the initial solution and added AgNO3.
1. Calculate the moles of NaCl and KI in the initial solution:
Moles of NaCl = (0.760 M) * (16.7 mL) / 1000 mL
Moles of KI = (0.760 M) * (16.7 mL) / 1000 mL
2. Determine the limiting reactant (the one that forms the least amount of precipitate):
Compare the moles of NaCl and KI, and identify the smaller value.
3. Use the limiting reactant to calculate the moles of AgI formed:
Moles of AgI = Moles of limiting reactant
4. Calculate the concentration of Ag+ using the volume of AgNO3 added:
Concentration of Ag+ = Moles of AgI / (16.2 mL + 16.7 mL) / 1000 mL
5. Express the concentration of Ag+ as a quotient:
[Ag+] = Ksp,AgI / x
(b) To calculate [Ag+] when 48.0 mL of 0.380 M AgNO3 has been added, follow the same steps as in part (a), but now using the solubility product constant (Ksp) for AgCl instead of AgI:
[Ag+] = Ksp,AgCl / y
(d) To calculate the quotient Ksp,AgCl/Ksp,AgI using the given voltage difference, the Nernst equation can be used:
E = (0.799 - 0.05916log(1/[Ag+])) - 0.241
Substitute the values of [Ag+] obtained in parts (a) and (b) into the equation, and solve for the voltage difference:
E(48.0 mL) - E(16.2 mL) = 0.3857 V
Rearrange the equation and substitute the values:
Ksp,AgCl / Ksp,AgI = 10^((E(48.0 mL) - E(16.2 mL)) / 0.05916)
Calculate the numerical value of the quotient Ksp,AgCl/Ksp,AgI using the given values and the voltage difference obtained in part (d).
Note: Make sure to use the correct values for the solubility product constants (Ksp,AgI and Ksp,AgCl) in your calculations.