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Chapter 7: Problem 81

In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. \(2 \mathrm{Cl}^{-}(a q)+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow 2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{HgCl}_{2}(s)\) What is the \(\mathrm{Cl}^{-}\) concentration in a 0.25 -mL sample of normal serum that requires \(1.46 \mathrm{mL}\) of \(8.25 \times 10^{-4} \mathrm{M}\) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) to reach the end point?

Short Answer

Expert verified
The concentration of Cl⁻ in the serum sample is calculated using the moles of titrant and the serum sample volume.

Step by step solution

01

Write down the balanced chemical equation

The balanced chemical equation for the reaction is: \[2 \mathrm{Cl}^{-}(aq) + \mathrm{Hg}(\mathrm{NO}_{3})_{2}(aq) \rightarrow 2 \mathrm{NO}_{3}^{-}(aq) + \mathrm{HgCl}_{2}(s)\].This shows that 2 moles of Cl⁻ react with 1 mole of Hg(NO₃)₂.
02

Calculate moles of Hg(NO₃)₂ used in titration

Use the concentration and volume of the titrant to find out the moles of Hg(NO₃)₂ used:\[\text{Moles of Hg(NO₃)₂} = \text{Concentration of Hg(NO₃)₂} \times \text{Volume of Hg(NO₃)₂}\]\[\text{Moles of Hg(NO₃)₂} = 8.25 \times 10^{-4} \text{M} \times 1.46 \times 10^{-3} \text{L}\]Calculate this value.
03

Determine moles of Cl⁻ reacted

Using the stoichiometry from the balanced equation, calculate how many moles of Cl⁻ reacted. Since 2 moles of Cl⁻ react with 1 mole of Hg(NO₃)₂:\[\text{Moles of Cl⁻} = 2 \times \text{Moles of Hg(NO₃)₂}\].
04

Calculate concentration of Cl⁻

The concentration of Cl⁻ is found by dividing the moles of Cl⁻ by the volume of the serum sample in liters:\[\text{Concentration of Cl⁻} = \frac{\text{Moles of Cl⁻}}{\text{Volume of serum sample in L}}\].Remember to convert the volume of the serum sample from mL to L before the calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Titration Calculation
Understanding titration calculations is crucial for analyzing substances such as blood serum. In titration, a solution of known concentration, called the titrant, is used to determine the concentration of an analyte in a sample. The point at which the reaction is completed is known as the endpoint, which is determined by a color change or another indicator. For titration calculations:
  • Start with the balanced chemical equation to identify the reaction ratio of the reactants.
  • Using the titrant's concentration and volume, calculate the moles of titrant used.
  • Apply stoichiometry to find the moles of the analyte reacted based on the reaction ratio.
  • Finally, calculate the analyte's concentration by dividing its moles by the sample's volume.
This step-by-step approach allows students to work systematically through titration problems.
Chemical Stoichiometry
Chemical stoichiometry is a section of chemistry that involves using the coefficients from balanced chemical equations to calculate relative quantities of reactants and products in chemical reactions. In the context of our problem:
  • We have a balanced equation with a 2:1 ratio of Cl⁻ to Hg(NO₃)₂.
  • Stoichiometry tells us that for every mole of Hg(NO₃)₂, two moles of Cl⁻ will react.
The ability to convert from moles of one substance to moles of another using these ratios is fundamental for understanding chemical reactions and is especially important for titrations.
Molarity and Volume Relationship
The molarity and volume relationship is an integral part of titration and stoichiometry. Molarity (M) is defined as the number of moles of solute per liter of solution. For a given solution:
  • Multiplying molarity by volume (in liters) yields the number of moles of solute.
  • In titration, we use this relationship to find the moles of the titrant and, by extension, the moles of the analyte.
Note the importance of units here; while molarity is given in moles per liter, volumes are often measured in milliliters and must be converted to liters before performing calculations.
Balancing Chemical Equations
Balancing chemical equations ensures that the law of conservation of mass is respected, stating that matter cannot be created or destroyed in a chemical reaction. A balanced equation has equal numbers of each type of atom on both sides of the reaction. For instance, our titration problem involved a reaction with a balanced equation indicating a 2:1 mole ratio between Cl⁻ ions and Hg(NO₃)₂. It is this balanced equation that serves as the foundation of stoichiometry and allows us to perform accurate calculations in titration problems.

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