- Write the precipitation net ionic equations: \(\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\) and \(\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)\).
- Explain that excess \(\text{AgNO}_3\) ensures complete precipitation of halide ions.
- Solve the system \(x + y = 10.99\) and \(\frac{143.32}{58.44}x + \frac{187.77}{102.89}y = 23.72\) to find the masses.
- Determine the mass percentages: \(\text{NaCl} \approx \boxed{53.13\%}\) and \(\text{NaBr} \approx \boxed{46.87\%}\).
Explanation
- Analysis of the Problem
The problem involves a solid sample of mass \(10.99 \text{ g}\) consisting of \(\text{NaCl}\) and \(\text{NaBr}\). When dissolved in water, \(\text{NaCl}\) dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\) ions, while \(\text{NaBr}\) dissociates into \(\text{Na}^+\) and \(\text{Br}^-\) ions. Upon adding excess silver nitrate (\(\text{AgNO}_3\)), the silver ions (\(\text{Ag}^+\)) react with the halide ions to form insoluble precipitates of \(\text{AgCl}\) and \(\text{AgBr}\). The total mass of the resulting precipitates is \(23.72 \text{ g}\). Our objective is to determine the mass percentage of each compound.
- Net Ionic Equations and Use of Excess AgNO_3
The precipitation reactions that occur are represented by the following net ionic equations:
- Chloride reaction: \(\text{Ag}^+ (aq) + \text{Cl}^- (aq) \rightarrow \text{AgCl} (s)\)
- Bromide reaction: \(\text{Ag}^+ (aq) + \text{Br}^- (aq) \rightarrow \text{AgBr} (s)\)
Silver nitrate is added in excess to ensure that all chloride and bromide ions in the solution are completely precipitated. By providing an excess of \(\text{Ag}^+\) ions, we guarantee that the amount of precipitate formed is limited solely by the amount of \(\text{Cl}^-\) and \(\text{Br}^-\) originally present in the sample, thus preventing them from being the limiting reagents in the reaction.
- Setting Up and Solving the System of Equations
Let \(x\) be the mass of \(\text{NaCl}\) (in grams) and \(y\) be the mass of \(\text{NaBr}\) (in grams). Since the total mass of the sample is \(10.99 \text{ g}\), we have:
\[x + y = 10.99\]
Next, we relate the masses of the original compounds to the masses of the precipitates. Each mole of \(\text{NaCl}\) produces one mole of \(\text{AgCl}\), and each mole of \(\text{NaBr}\) produces one mole of \(\text{AgBr}\).
The moles of \(\text{NaCl}\) are \(\frac{x}{58.44}\) and the moles of \(\text{NaBr}\) are \(\frac{y}{102.89}\).
The total mass of the precipitates is given by:
\[\left(\frac{x}{58.44}\right) \times 143.32 + \left(\frac{y}{102.89}\right) \times 187.77 = 23.72\]
Using the system of equations, the calculations show that \(x \approx 5.839 \text{ g}\) and \(y \approx 5.151 \text{ g}\).
- Final Calculations and Results
Now we calculate the mass percentage of each compound in the original mixture:
- Mass percent of \(\text{NaCl} = \left(\frac{5.839}{10.99}\right) \times 100 \approx 53.13\%\)
- Mass percent of \(\text{NaBr} = \left(\frac{5.151}{10.99}\right) \times 100 \approx 46.87\%\)
Thus, the mass percentage of \(\text{NaCl}\) is approximately \(53.13\%\) and the mass percentage of \(\text{NaBr}\) is approximately \(46.87\%\). The final result is:
\(\text{NaCl}: \boxed{53.13\%}, \text{NaBr}: \boxed{46.87\%}\)
Examples
This type of gravimetric analysis is frequently used in chemistry to determine the purity of unknown substances or to analyze the composition of mixtures. By transforming dissolved ions into a measurable solid precipitate, chemists can precisely calculate the weight of original components. Similar techniques are used in environmental testing to determine the levels of salt or contaminants in water samples, providing a fundamental tool for quality control and chemical research.