Chloride in serum is determined by titration with Hg(NO3)2; 2Cl− + Hg2+ HgCl2. The Hg(NO3)2 is standardized by titrating 2.00 mL of a 0.0108 M NaCl solution, requiring 1.12 mL to reach the diphenylcarbazone end point. A 0.500-mL serum sample is treated with 3.50 mL water, 0.50 mL 10% sodium tungstate solution, and 0.50 mL of 0.33 M H2SO4 solution to precipitate proteins. After the proteins are precipitated, the sample is filtered through a dry filter into a dry flask. A 2.00-mL aliquot of the filtrate is titrated with the Hg(NO3)2 solution, requiring 1.23 mL. Calculate the mg/L chloride in the serum. (Note: mercury is rarely used today due to its toxicity. The problem is illustrative.)
The Correct Answer and Explanation is :
To calculate the chloride concentration in the serum sample, we need to follow several steps, using the information provided in the problem. Let’s break it down into logical steps:
Step 1: Standardization of the Hg(NO3)2 Solution
We start by determining the concentration of the Hg(NO3)2 solution used to titrate the chloride ions in the serum.
- Given information:
- The volume of NaCl solution used for standardization = 2.00 mL.
- The molarity of NaCl solution = 0.0108 M.
- The volume of Hg(NO3)2 solution required to titrate NaCl = 1.12 mL.
- Moles of chloride in the NaCl solution:
[
\text{moles of Cl}^{-} = \text{molarity} \times \text{volume (in L)}
]
[
\text{moles of Cl}^{-} = 0.0108 \, \text{mol/L} \times 2.00 \, \text{mL} \times \left( \frac{1}{1000} \right) \, \text{L/mL} = 0.0000216 \, \text{mol}
] - Moles of Hg(NO3)2 required to titrate this chloride:
According to the reaction, 1 mole of Hg2+ reacts with 2 moles of Cl-. Therefore, the moles of Hg2+ are half the moles of chloride:
[
\text{moles of Hg}^{2+} = \frac{0.0000216}{2} = 0.0000108 \, \text{mol}
] - Concentration of Hg(NO3)2:
The volume of Hg(NO3)2 used is 1.12 mL = 0.00112 L. So, the molarity of Hg(NO3)2 is:
[
M_{\text{Hg(NO3)2}} = \frac{\text{moles of Hg}^{2+}}{\text{volume of Hg(NO3)2 in L}}
]
[
M_{\text{Hg(NO3)2}} = \frac{0.0000108 \, \text{mol}}{0.00112 \, \text{L}} = 0.00964 \, \text{M}
]
Step 2: Titration of the Serum Sample
Next, we calculate the chloride concentration in the serum.
- Given information:
- The volume of the serum sample aliquot used for titration = 2.00 mL.
- The volume of Hg(NO3)2 solution used to titrate this aliquot = 1.23 mL.
- The molarity of Hg(NO3)2 = 0.00964 M.
- Moles of Hg2+ in the titration:
[
\text{moles of Hg}^{2+} = M_{\text{Hg(NO3)2}} \times \text{volume (in L)}
]
[
\text{moles of Hg}^{2+} = 0.00964 \, \text{mol/L} \times 1.23 \, \text{mL} \times \left( \frac{1}{1000} \right) \, \text{L/mL} = 0.0000119 \, \text{mol}
] - Moles of chloride in the serum sample:
Since 1 mole of Hg2+ reacts with 2 moles of Cl-, the moles of chloride are double the moles of Hg2+:
[
\text{moles of Cl}^{-} = 2 \times 0.0000119 \, \text{mol} = 0.0000238 \, \text{mol}
] - Concentration of chloride in the 2.00 mL aliquot:
[
\text{molarity of Cl}^{-} = \frac{\text{moles of Cl}^{-}}{\text{volume of aliquot in L}}
]
[
\text{molarity of Cl}^{-} = \frac{0.0000238 \, \text{mol}}{2.00 \, \text{mL} \times \left( \frac{1}{1000} \right)} = 0.0119 \, \text{M}
]
Step 3: Determining the Chloride Concentration in the Serum
Now we need to calculate the chloride concentration in the entire serum sample, which was diluted before titration.
- The serum sample volume = 0.500 mL. The aliquot used for titration was 2.00 mL, so we need to account for the dilution factor:
[
\text{dilution factor} = \frac{\text{total volume after dilution}}{\text{aliquot volume}} = \frac{0.500 + 3.50}{2.00} = \frac{4.00}{2.00} = 2
] - Final chloride concentration in the serum:
The chloride concentration in the 2.00 mL aliquot is 0.0119 M. Therefore, the chloride concentration in the original 0.500 mL serum sample is:
[
\text{Concentration in serum} = 0.0119 \, \text{M} \times 2 = 0.0238 \, \text{M}
] - Convert molarity to mg/L:
The molar mass of chloride (Cl-) is 35.45 g/mol. To convert molarity to mg/L:
[
\text{Concentration in mg/L} = 0.0238 \, \text{mol/L} \times 35.45 \, \text{g/mol} \times 1000 \, \text{mg/g} = 843.7 \, \text{mg/L}
]
Final Answer:
The chloride concentration in the serum sample is 843.7 mg/L.
Explanation:
In this problem, the key steps involved standardizing the titrant (Hg(NO3)2) using a known chloride solution (NaCl) and then using the standardized titrant to determine the chloride concentration in a diluted serum sample. By using stoichiometry based on the titration reactions, we calculated the chloride concentration in the original serum sample and converted it to mg/L.