Titration of a 0.485-g sample by the Mohr method required 36.8 mL of standard 0.1060 M AgNO3 solution. Calculate the percentage of chloride in the sample. 2. Because the yellow color of CrO42- tends to obscure the first appearance of the red Ag2CrO4, it is common practice to hold the CrO42- concentration to about 2.5 x 10-3 M. Calculate the relative titration error (neglecting the volume of AgNO3 needed to produce a detectable amount of Ag2CrO4) in the titration of 50.0 mL of 0.0500 M NaCl with 0.1000 M AgNO3. 3. The Fajans method is to be used in the routine analysis of solids for their chloride content. It is desired that the volume of standard AgNO3 used in these titrations be numerically equal to the percent Cl- when 0.2500-g samples are used. What should the molarity of the AgNO3 solution be?
The Correct Answer and Explanation is :
Let’s work through the three problems step-by-step:
1. Titration of a 0.485-g Sample by the Mohr Method
The Mohr method involves titrating chloride ions (( \text{Cl}^- )) with silver nitrate (( \text{AgNO}_3 )) to form a precipitate of silver chloride (( \text{AgCl} )).
We are given:
- Mass of sample = 0.485 g
- Volume of ( \text{AgNO}_3 ) solution used = 36.8 mL = 0.0368 L
- Molarity of ( \text{AgNO}_3 ) solution = 0.1060 M
To calculate the percentage of chloride in the sample, we need to:
- Determine the moles of ( \text{AgNO}_3 ) used:
[
\text{moles of AgNO}_3 = M \times V = 0.1060 \, \text{mol/L} \times 0.0368 \, \text{L} = 0.0039008 \, \text{mol}
]
- From the reaction, ( \text{AgNO}_3 ) reacts in a 1:1 ratio with ( \text{Cl}^- ), so the moles of chloride in the sample are the same:
[
\text{moles of Cl}^- = 0.0039008 \, \text{mol}
]
- Now calculate the mass of chloride:
[
\text{mass of Cl}^- = \text{moles of Cl}^- \times \text{molar mass of Cl} = 0.0039008 \, \text{mol} \times 35.45 \, \text{g/mol} = 0.1382 \, \text{g}
]
- Finally, calculate the percentage of chloride in the sample:
[
\text{percentage of Cl}^- = \left( \frac{\text{mass of Cl}^-}{\text{mass of sample}} \right) \times 100 = \left( \frac{0.1382}{0.485} \right) \times 100 = 28.5\%
]
2. Relative Titration Error Due to CrO4²⁻
The yellow color of ( \text{CrO}_4^{2-} ) can obscure the red precipitate of ( \text{Ag}_2\text{CrO}_4 ) during titration, leading to errors. The concentration of ( \text{CrO}_4^{2-} ) should be kept low to minimize this effect. We are asked to calculate the relative error when titrating a sample of NaCl with ( \text{AgNO}_3 ).
Given:
- Volume of NaCl solution = 50.0 mL = 0.0500 L
- Molarity of NaCl = 0.0500 M
- Molarity of ( \text{AgNO}_3 ) = 0.1000 M
- ( \text{CrO}_4^{2-} ) concentration = ( 2.5 \times 10^{-3} ) M
To calculate the relative error, the first step is to find the moles of NaCl:
[
\text{moles of NaCl} = 0.0500 \, \text{L} \times 0.0500 \, \text{mol/L} = 0.00250 \, \text{mol}
]
This means that 0.00250 mol of chloride ions are present in the sample.
Next, calculate the volume of ( \text{AgNO}_3 ) needed to titrate these chloride ions:
[
\text{moles of AgNO}_3 = \text{moles of NaCl} = 0.00250 \, \text{mol}
]
Now calculate the volume of ( \text{AgNO}_3 ) solution required:
[
V = \frac{\text{moles of AgNO}_3}{\text{molarity of AgNO}_3} = \frac{0.00250 \, \text{mol}}{0.1000 \, \text{mol/L}} = 0.0250 \, \text{L} = 25.0 \, \text{mL}
]
To find the relative error, we need to account for the volume of ( \text{AgNO}_3 ) required to form detectable ( \text{Ag}_2\text{CrO}_4 ) due to the CrO4²⁻ concentration:
The moles of ( \text{CrO}_4^{2-} ) present are:
[
\text{moles of CrO}_4^{2-} = 0.0500 \, \text{L} \times 2.5 \times 10^{-3} \, \text{mol/L} = 1.25 \times 10^{-4} \, \text{mol}
]
Since ( 2 \, \text{moles of Ag}^+ ) react with ( 1 \, \text{mol of CrO}_4^{2-} ), the moles of ( \text{AgNO}_3 ) required for the detection are:
[
\text{moles of AgNO}_3 = 2 \times 1.25 \times 10^{-4} = 2.50 \times 10^{-4} \, \text{mol}
]
The volume of ( \text{AgNO}_3 ) required to form detectable ( \text{Ag}_2\text{CrO}_4 ) is:
[
V = \frac{2.50 \times 10^{-4}}{0.1000} = 2.50 \times 10^{-3} \, \text{L} = 2.50 \, \text{mL}
]
Thus, the relative error is:
[
\text{relative error} = \frac{\text{volume due to CrO}_4^{2-}}{\text{total volume used}} = \frac{2.50 \, \text{mL}}{25.0 \, \text{mL}} = 0.100 = 10\%
]
3. Required Molarity of AgNO₃ Solution for Fajans Method
The Fajans method involves the titration of chloride with silver nitrate, and the volume of AgNO₃ used should be numerically equal to the percent chloride in the sample.
Given:
- Mass of sample = 0.2500 g
- Desired volume of AgNO₃ = Percentage of chloride in sample
- First, let the percentage of chloride be ( x ). The mass of chloride in the sample is:
[
\text{mass of Cl}^- = \frac{x}{100} \times 0.2500 \, \text{g}
]
- The moles of chloride are:
[
\text{moles of Cl}^- = \frac{\text{mass of Cl}^-}{\text{molar mass of Cl}} = \frac{x}{100} \times \frac{1}{35.45}
]
- For the titration, moles of AgNO₃ are equal to moles of chloride, so:
[
\text{moles of AgNO}_3 = \frac{x}{100} \times \frac{1}{35.45}
]
- The volume of AgNO₃ solution required is:
[
V = \frac{\text{moles of AgNO}_3}{\text{molarity of AgNO}_3} = \frac{x}{100} \times \frac{1}{35.45 \times M}
]
We want the volume of AgNO₃ to be equal to the percentage of chloride ( x ). Therefore:
[
\frac{x}{100} \times \frac{1}{35.45 \times M} = x
]
Solve for ( M ):
[
M = \frac{1}{35.45 \times 100} = 0.0002825 \, \text{M}
]
Thus, the required molarity of AgNO₃ is approximately 0.0002825 M.