If performing a potentiometric halide titration with Ag+:
1. Is siver nitrate a primary standard? Explain.
2. What is the purpose of using a bisulfate buffer?
3. Have 1.0000 grams of unknown, containing .3000g KCl and .7000g KI. You dissolve in 100 mL of water then use an aliquot of 25mL for titration. How many milliliters of 0.700 M AgNO3 would you expect the mixture to consume?
1. I don't know the details of your experiment but I suspect AgNO3 is, indeed, a primary standard. How you explain that is up to you.
2. Too basic and you have problems with Ag(OH)2.
3. moles KCl = grams/molar mass
moles KI = grams/molar mass
Total Cl and I = sum of the two.
Amount titrated is sum x (25/100) = ?
moles AgNO3 used = moles in the 25 mL aliquot.
M AgNO3 = moles AgNO3/L AgNO3. Solve for L and convert to mL.
1. To determine if silver nitrate (AgNO3) is a primary standard for a potentiometric halide titration with Ag+, one needs to consider the criteria for a substance to be considered a primary standard.
A primary standard is a highly pure compound that can be used to directly standardize a solution of known concentration. It should have the following characteristics:
- High purity: The compound should be easily obtainable in a highly pure form.
- Stability: The compound should be stable, meaning it does not undergo degradation over time or form impurities that could affect its concentration.
- Accurate stoichiometry: The compound should have a known and consistent stoichiometry, allowing for precise calculations of its concentration.
- High molar mass: A higher molar mass reduces the relative impact of weighing errors during preparation, resulting in more accurate concentration determination.
In the case of silver nitrate (AgNO3), it fulfills these criteria and is commonly used as a primary standard. It is highly pure, stable, and has a known stoichiometry (Ag+ ions). Its molar mass is also relatively high, which aids in reducing weighing errors during preparation. Therefore, silver nitrate can be considered a primary standard for the potentiometric halide titration.
2. The purpose of using a bisulfate buffer in a potentiometric halide titration with Ag+ is to maintain a constant pH. The bisulfate buffer consists of a mixture of sodium bisulfate (NaHSO4) and sodium sulfate (Na2SO4) in a specific ratio.
Maintaining a constant pH is crucial in a potentiometric titration because it helps ensure the reaction between the analyte and the titrant proceeds accurately and rapidly. In the case of halide titrations, the bisulfate buffer helps maintain a slightly acidic pH (around pH 4-5) that is suitable for the reaction between Ag+ and the halide ions (e.g., Cl-, Br-, or I-) present in the sample solution.
By maintaining a constant pH, the bisulfate buffer minimizes the influence of pH changes on the electrochemical potential, allowing for accurate and reproducible measurements during the titration process. Therefore, the purpose of using a bisulfate buffer is to provide a stable pH environment for the halide titration.
3. To calculate the expected amount of 0.700 M AgNO3 solution that the mixture would consume during titration, you need to consider the stoichiometry of the reaction between silver ions (Ag+) and the halide ions (KCl and KI) present in the unknown sample.
The balanced equation for the reaction between Ag+ and halide (X-) ions is as follows:
Ag+ + X- -> AgX
where X- represents Cl-, Br-, or I-.
From the given information, you have 0.3000 g of KCl and 0.7000 g of KI in a 1.0000 g unknown sample. To calculate the amount of KCl and KI in moles, you need to divide the mass of each compound by its molar mass.
- KCl molar mass: 39.10 g/mol (atomic mass of K) + 35.45 g/mol (atomic mass of Cl)
- KI molar mass: 39.10 g/mol (atomic mass of K) + 126.90 g/mol (atomic mass of I)
Calculate the number of moles for each compound:
- Moles of KCl = 0.3000 g / (39.10 g/mol + 35.45 g/mol)
- Moles of KI = 0.7000 g / (39.10 g/mol + 126.90 g/mol)
Next, you need to determine which halide ion (Cl-, Br-, or I-) is present in excess in the aliquot of the 25 mL sample solution. To do this, you can compare the moles of KCl and KI present in the aliquot. The one with the excess moles will be the limiting reagent.
To calculate the excess moles, multiply the moles of each compound by the ratio of the halide ion present in the compound:
- Moles of Cl- from KCl = Moles of KCl
- Moles of I- from KI = Moles of KI
If Cl- is in excess, the reaction will consume exactly 1 mole of Ag+ per mole of Cl-. Similarly, if I- is in excess, the reaction will consume exactly 1 mole of Ag+ per mole of I-.
Now that you know the limiting reagent, you can calculate the moles of Ag+ consumed during the titration reaction based on the stoichiometry of the reaction. The number of moles of Ag+ consumed will be equal to the moles of Cl- or I- (depending on the limiting reagent).
Finally, to determine the volume of the 0.700 M AgNO3 solution required, you can use the equation:
Volume (in L) = Moles of Ag+ consumed / Molarity of AgNO3 solution
Note: Make sure to convert grams to moles and milliliters to liters when performing the calculations.
By following these calculations, you can determine the expected volume of the 0.700 M AgNO3 solution that the mixture would consume during titration.