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The mass percentage of chloride ion in a 25.00-mL sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took \(42.58 \mathrm{~mL}\) of \(0.2997 \mathrm{M}\) silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in the seawater if its density is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

Short Answer

Expert verified
First, we find the moles of silver nitrate used in the titration: Moles of AgNO鈧 = \(0.2997 \frac{mol}{L} \times 0.04258 L = 0.01276 mol\) Since the mole ratio between AgNO鈧 and Cl鈦 is 1:1, the moles of Cl鈦 in the seawater sample is also 0.01276 mol. Next, we calculate the mass of chloride ions: Mass of Cl鈦 = Moles of Cl鈦 脳 Molar mass of Cl鈦 = \(0.01276 mol \times 35.45 \frac{g}{mol} = 0.4524 g\) Now, we determine the mass of the seawater sample: Mass of seawater sample = \(1.025 \frac{g}{mL} \times 25.00 mL = 25.625 g\) Finally, we compute the mass percentage of chloride ion in the seawater: Mass Percentage of chloride ion = \(\frac{0.4524 g}{25.625 g} \times 100 = 1.76 \%\) Thus, the mass percentage of chloride ion in the seawater is approximately 1.76%.

Step by step solution

01

Calculate the moles of silver nitrate

In this step, we will calculate the moles of silver nitrate (AgNO鈧) used in the titration using the volume and molarity given. The equation to find moles is: Moles = Molarity 脳 Volume Here, the volume of AgNO鈧 is 42.58 mL and its molarity is 0.2997 M. Remember to convert the volume from mL to L. Moles of AgNO鈧 = \(0.2997 \frac{mol}{L} \times 0.04258 L\)
02

Calculate moles of chloride ions

In this step, we will use stoichiometry to find the moles of chloride ions (Cl鈦) in the seawater sample. The balanced equation for the reaction between silver nitrate and chloride ions is: AgNO鈧 + Cl鈦 鈫 AgCl + NO鈧冣伝 From the balanced equation, we can see that the mole ratio between AgNO鈧 and Cl鈦 is 1:1. Therefore, the moles of Cl鈦 in the seawater sample is equal to the moles of AgNO鈧 used in the titration. Moles of Cl鈦 = Moles of AgNO鈧
03

Calculate the mass of chloride ions

Now, we will convert the moles of chloride ions into mass using the molar mass of Cl鈦 (35.45 g/mol): Mass of Cl鈦 = Moles of Cl鈦 脳 Molar mass of Cl鈦
04

Calculate the mass of the seawater sample

Next, we need to calculate the mass of the seawater sample. We are given the volume of the sample (25.00 mL) and its density (1.025 g/mL). We can calculate the mass using the equation: Mass = Density 脳 Volume Mass of seawater sample = \(1.025 \frac{g}{mL} \times 25.00 mL\)
05

Calculate the mass percentage of chloride ion

Finally, we will calculate the mass percentage of chloride ion in the seawater using the equation: Mass Percentage of chloride ion = \(\frac{Mass~of~Cl^-}{Mass~of~seawater~sample} \times 100\) Now that we have a step-by-step plan, let's calculate the values and find out the mass percentage of chloride ion in the seawater.

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Key Concepts

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

Titration
Titration is an analytical technique used to determine the concentration of an unknown solution. It involves the gradual addition of a solution of known concentration (the titrant) to the unknown solution until the reaction reaches a specific endpoint, which is often indicated by a color change. In the exercise, the mass percentage of chloride ion in seawater is determined through titration using a silver nitrate solution.

In this analytical procedure, a precipitate forms when the silver ions from the silver nitrate solution react with the chloride ions in seawater to form an insoluble compound, silver chloride (AgCl). The point at which stoichiometrically equivalent amounts of reactants have reacted is known as the equivalence point. In the given exercise, the volume of silver nitrate required to reach this point is carefully measured, as it is directly related to the quantity of chloride ions present in the seawater.

To improve understanding, it's important to remember that indicator compounds can be used to signal the equivalence point in some titrations visually. However, in this particular exercise involving the precipitation of AgCl, the formation of an insoluble compound serves as the endpoint signal, negating the need for an additional indicator.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, the mass of reactants and products remains constant.

In the context of the exercise, stoichiometry is key when determining the moles of chloride ions, Cl鈦, present in the sample. The balanced chemical equation AgNO鈧 + Cl鈦 鈫 AgCl + NO鈧冣伝 shows a 1:1 mole ratio between silver nitrate (AgNO鈧) and chloride ions (Cl鈦). This means that for every mole of silver nitrate reacting, there will be one mole of chloride ions reacting as well.

By knowing the number of moles of silver nitrate used in the titration, one can directly find the number of moles of chloride ions present. To truly understand and apply stoichiometry, students should practice with different types of reactions and equations to become comfortable with mole ratios and the conversion between moles and mass using molar masses.
Molarity
Molarity is a measure of the concentration of a solution, defined as the number of moles of solute divided by the volume of solution in liters. It is expressed in units of moles per liter (mol/L).

In the given exercise, molarity plays a crucial role as it allows us to convert the volume of the silver nitrate solution used in the titration into moles. The equation relating moles, molarity, and volume is: Moles = Molarity 脳 Volume. Here, the molarity of the silver nitrate solution is given as 0.2997 M, which signifies that there are 0.2997 moles of silver nitrate in every liter of solution.

When calculating the moles of silver nitrate, it's essential to convert the volume from milliliters (mL) to liters (L) since molarity is expressed per liter. By understanding molarity, students can relate the amount of substance used in a reaction to the volume of solution, making it a fundamental concept in various chemistry disciplines, including titrations.

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Most popular questions from this chapter

The arsenic in a 1.22-g sample of a pesticide was converted to \(\mathrm{AsO}_{4}{ }^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of \(\mathrm{As}\) in \(\mathrm{AsO}_{4}{ }^{3-}\) ? (b) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

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