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A \(0.5131-\mathrm{g}\) sample that contains \(\mathrm{KBr}\) is dissolved in \(50 \mathrm{~mL}\) of distilled water. Titrating with \(0.04614 \mathrm{M} \mathrm{AgNO}_{3}\) requires \(25.13 \mathrm{~mL}\) to reach the Mohr end point. A blank titration requires \(0.65 \mathrm{~mL}\) to reach the same end point. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{KBr}\) in the sample.

Short Answer

Expert verified
The sample contains 43.8% w/w KBr.

Step by step solution

01

Calculate the net volume of AgNO3 used

Subtract the volume of the blank titration from the total volume of AgNO3 used: \( V_{net} = 25.13 \, \text{mL} - 0.65 \, \text{mL} = 24.48 \, \text{mL} \).
02

Convert the net volume to liters

Convert the net volume from mL to liters to use it in the molarity calculation:\( V_{net} = 24.48 \, \text{mL} = 0.02448 \, \text{L} \).
03

Calculate moles of AgNO3 used in titration

Use the molarity formula to calculate moles of AgNO3: \( \text{moles} = M \times V = 0.04614 \, \text{mol/L} \times 0.02448 \, \text{L} = 0.001129\, \text{moles} \).
04

Determine moles of KBr in sample

According to the reaction, 1 mole of AgNO3 reacts with 1 mole of KBr. Therefore, the moles of KBr are equal to the moles of AgNO3 used: \( \text{moles of KBr} = 0.001129 \).
05

Calculate the mass of KBr

Calculate the mass of KBr using its molar mass (119.0 g/mol for K and 79.9 g/mol for Br, totaling 198.9 g/mol): \( \text{mass of KBr} = 0.001129 \, \text{mol} \times 198.9 \, \text{g/mol} = 0.2247 \, \text{g} \).
06

Calculate the weight percentage of KBr in the sample

Use the formula for weight percentage:\(\% \text{w/w KBr} = \left( \frac{\text{mass of KBr}}{\text{mass of sample}} \right) \times 100 \%= \left( \frac{0.2247 \text{ g}}{0.5131 \text{ g}} \right) \times 100 \% = 43.8 \% \).

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Key Concepts

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

Titration
Titration is a method used in chemistry to determine the concentration of a solute in a solution. In this process, a solution of known concentration, called the titrant, is added to a solution with an unknown concentration until the reaction reaches its endpoint. This point often involves a noticeable change, such as a color shift, indicating the completion of the chemical reaction. In the given exercise, silver nitrate ( AgNO_3 ) acts as the titrant, and it is carefully added to a potassium bromide ( KBr ) solution.
The endpoint in this method is determined using the Mohr titration, where a visible color change marks the completion of the reaction. However, the process involves performing a blank titration to account for any impurities or reactions that do not involve the analyte. This step is crucial to ensure accuracy, and the volume used in the blank titration must be subtracted from the total to find the net volume of silver nitrate needed to react with the potassium bromide.
Thus, titration provides a reliable way to measure an unknown concentration by referencing the stoichiometry of the reaction involved in the method. By carefully measuring the volume of the titrant and using the known concentration, you can deduce the number of moles of each reactant participating in the chemical reaction.
Mole Calculations
Mole calculations play a central role in understanding chemical quantities in reactions. They allow us to convert between mass, moles, and number of molecules using the concept of the mole, which is a standard unit in chemistry. In the exercise, mole calculations are crucial to determine how much KBr is in the sample.
Initially, the volume of AgNO_3 used in the titration is converted into moles using its molarity: \( \text{moles} = M \times V \), where \( M \) stands for molarity, and \( V \) is the volume in liters. The reaction between AgNO_3 and KBr involves a 1:1 mole ratio, meaning one mole of AgNO_3 reacts with one mole of KBr.
By calculating the moles of AgNO_3 used, you directly determine the moles of KBr, as they are equal at the endpoint of this titration. This involves multiplication with the molar mass of KBr to find the mass of KBr in the original sample. Understanding this stoichiometric relationship helps deduce the quantity of substance involved in the chemical process.
Weight Percentage
Weight percentage (% w/w) is a common way to express concentrations in chemical mixtures and solutions. It quantifies how much part of a substance makes up the whole in terms of weight. In simpler terms, it tells you the proportion of a particular component within a mixture based on weight.
In this example, the weight percentage of KBr is calculated using the formula: \( \% \text{w/w} = \left( \frac{\text{mass of KBr}}{\text{mass of sample}} \right) \times 100 \% \). This percentage indicates how much of the sample's weight is due to KBr, providing insight into its concentration.
Using the calculated mass of KBr in the sample, and the total mass of the original sample, the weight percentage is calculated to be 43.8\%. This helps chemists and engineers to determine the purity of materials, evaluate their properties, or assess their suitability for specific applications. Expressing concentration in weight percentage is particularly useful when the total weight of the sample or batch is more important than the volume.

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Most popular questions from this chapter

The concentration of \(o\) -phthalic acid in an organic solvent, such as \(n\) butanol, is determined by an acid-base titration using aqueous \(\mathrm{NaOH}\) as the titrant. As the titrant is added, the \(o\) -phthalic acid extracts into the aqueous solution where it reacts with the titrant. The titrant is added slowly to allow sufficient time for the extraction to take place. (a) What type of error do you expect if the titration is carried out too quickly? (b) Propose an alternative acid-base titrimetric method that allows for a more rapid determination of the concentration of \(o\) -phthalic acid in \(n\) -butanol.

The amount of calcium in physiological fluids is determined by a complexometric titration with EDTA. In one such analysis a \(0.100-\mathrm{mL}\) sample of a blood serum is made basic by adding 2 drops of \(\mathrm{NaOH}\) and titrated with \(0.00119 \mathrm{M}\) EDTA, requiring \(0.268 \mathrm{~mL}\) to reach the end point. Report the concentration of calcium in the sample as milligrams Ca per \(100 \mathrm{~mL}\).

The following data for the titration of a monoprotic weak acid with a strong base were collected using an automatic titrator. Prepare normal, first derivative, second derivative, and Gran plot titration curves for this data, and locate the equivalence point for each. $$ \begin{array}{cccc} \text { Volume of } \mathrm{NaOH}(\mathrm{ml}) & \mathrm{pH} & \text { Volume of } \mathrm{NaOH}(\mathrm{mL}) & \mathrm{pH} \\ \hline 0.25 & 3.0 & 49.95 & 7.8 \\ 0.86 & 3.2 & 49.97 & 8.0 \\ 1.63 & 3.4 & 49.98 & 8.2 \\ 2.72 & 3.6 & 49.99 & 8.4 \\ 4.29 & 3.8 & 50.00 & 8.7 \\ 6.54 & 4.0 & 50.01 & 9.1 \\ 9.67 & 4.2 & 50.02 & 9.4 \\ 13.79 & 4.4 & 50.04 & 9.6 \\ 18.83 & 4.6 & 50.06 & 9.8 \\ 24.47 & 4.8 & 50.10 & 10.0 \\ 30.15 & 5.0 & 50.16 & 10.2 \\ 35.33 & 5.2 & 50.25 & 10.4 \\ 39.62 & 5.4 & 50.40 & 10.6 \\ 42.91 & 5.6 & 50.63 & 10.8 \\ 45.28 & 5.8 & 51.01 & 11.0 \\ 46.91 & 6.0 & 51.61 & 11.2 \\ 48.01 & 6.2 & 52.58 & 11.4 \\ 48.72 & 6.4 & 54.15 & 11.6 \\ 49.19 & 6.6 & 56.73 & 11.8 \\ 49.48 & 6.8 & 61.11 & 12.0 \\ 49.67 & 7.0 & 68.83 & 12.2 \\ 49.79 & 7.2 & 83.54 & 12.4 \\ 49.87 & 7.4 & 116.14 & 12.6 \\ 49.92 & 7.6 & & \end{array} $$

Before the introduction of EDTA most complexation titrations used \(\mathrm{Ag}^{+}\) or \(\mathrm{CN}^{-}\) as the titrant. The analysis for \(\mathrm{Cd}^{2+},\) for example, was accomplished indirectly by adding an excess of \(\mathrm{KCN}\) to form \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\), and back titrating the excess \(\mathrm{CN}^{-}\) with \(\mathrm{Ag}^{+},\) forming \(\mathrm{Ag}(\mathrm{CN})_{2}^{-} .\) In one such analysis a \(0.3000-\mathrm{g}\) sample of an ore is dissolved and treated with \(20.00 \mathrm{~mL}\) of \(0.5000 \mathrm{M} \mathrm{KCN}\). The excess \(\mathrm{CN}^{-}\) requires \(13.98 \mathrm{~mL}\) of \(0.1518 \mathrm{M} \mathrm{AgNO}_{3}\) to reach the end point. Determine the \(\% \mathrm{w} / \mathrm{w}\) Cd in the ore.

The amount of uranium in an ore is determined by an indirect redox titration. The analysis is accomplished by dissolving the ore in sulfuric acid and reducing \(\mathrm{UO}_{2}^{2+}\) to \(\mathrm{U}^{4+}\) with a Walden reductor. The solution is treated with an excess of \(\mathrm{Fe}^{3+}\), forming \(\mathrm{Fe}^{2+}\) and \(\mathrm{U}^{6+}\). The \(\mathrm{Fe}^{2+}\) is titrated with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) In a typical analysis a 0.315 -g sample of ore is passed through the Walden reductor and treated with \(50.00 \mathrm{~mL}\) of \(0.0125 \mathrm{M} \mathrm{Fe}^{3+}\). Back titrating with 0.00987 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) requires \(10.52 \mathrm{~mL}\). What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{U}\) in the sample?

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