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Chapter 4: Problem 111

The average concentration of bromide ion in seawater is \(65 \mathrm{mg}\) of bromide ion per \(\mathrm{kg}\) of seawater. What is the molarity of the bromide ion if the density of the seawater is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

Short Answer

Expert verified
The molarity of bromide ion in seawater is approximately \(8.34 \times 10^{-4} \, \mathrm{mol/L}\).

Step by step solution

01

Convert the concentration of bromide ion to moles

First, we need to convert the concentration of the bromide ion, given in mg/kg, to moles. To do this, we will use the molar mass of bromide ion (Br鈦�), which is approximately 79.90 g/mol. The given concentration of bromide ion is 65 mg of Br鈦� per kg of seawater. We need to convert this into grams and then to moles using the molar mass of Br鈦�: \(65 \, mg = 0.065 \, g \, \text{of Br鈦粆\). Now, we can convert 0.065 g of Br鈦� to moles of Br鈦� using the molar mass: \(n = \frac{0.065 \, \text{g}}{79.90 \, \text{g/mol}} = 8.1352 \times 10^{-4} \, \text{mol}\). In 1 kg of seawater, there are \(8.1352 \times 10^{-4}\) moles of Br鈦�.
02

Convert the mass of seawater to volume

Next, we will convert the mass of seawater (1 kg) to its volume. We are given the density of seawater, which is 1.025 g/mL. To do this, we will use the following equation: Density = Mass / Volume We will first need to convert 1 kg of seawater to grams (1000 g), then solve for the volume in milliliters: \(1.025 \, \text{g/mL} = \frac{1000 \, \text{g}}{V_{mL}}\) Now, we solve for the volume of seawater in milliliters (V_mL): \(V_{mL} = \frac{1000 \, \text{g}}{1.025 \, \text{g/mL}} = 975.61 \, \text{mL}\). Finally, we convert the volume of seawater to liters (V_L): \(V_{L} = \frac{975.61 \, \text{mL}}{1000 \, \text{mL/L}} = 0.97561 \, \text{L}\).
03

Calculate the molarity of bromide ion

Now that we know the number of moles of Br鈦� and the volume of seawater in liters, we can calculate the molarity (M) using the following equation: Molarity (M) = Number of moles of solute / Volume of solution (L) Using the values from the previous steps, we get: \(M = \frac{8.1352 \times 10^{-4} \, \text{mol}}{0.97561 \, \text{L}} = 8.3367 \times 10^{-4} \, \text{mol/L}\). Therefore, the molarity of bromide ion in seawater is approximately \(8.34 \times 10^{-4} \, \mathrm{mol/L}\).

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Key Concepts

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

Bromide ion concentration
To understand bromide ion concentration, it's important to start with what a "concentration" means. Concentration refers to the amount of a substance present in a certain volume of solution. In our case, it's about the bromide ion in seawater.
Bromide ion concentration is given as 65 mg of Br鈦� per kg of seawater. This might seem straightforward, but it's important to convert this measurement into something more useful for calculations - such as moles per liter, which is called molarity.
Understanding this concentration helps us evaluate how much bromide is present and its impact on the chemistry of seawater. Knowing this, we can work towards solving chemistry problems related to seawater composition with much greater ease.
Seawater density
Seawater density is a critical concept in chemistry, especially when dealing with solutions. Density is defined as mass per unit volume. In this case, seawater has a density of 1.025 grams per milliliter (g/mL).
This density value allows us to translate from how much something weighs to how much space it occupies. In problems involving liquids, converting mass to volume is essential because many chemical calculations, including concentration calculations, are based on volume.
This density conversion is especially useful in chemistry when looking to convert between different units - such as mass (grams) into volume (milliliters). Knowing seawater's density allows us to proceed with these conversions logically and correctly.
Concentration conversion
Converting concentration units is an important skill in chemistry that makes solving various problems more manageable. Initially, the bromide ion concentration was given in milligrams per kilogram (mg/kg). However, for calculating molarity, we need this to be converted into moles per liter (mol/L).
Conversion steps include changing milligrams to grams since molar mass (in grams per mole) is used for this conversion. Afterward, converting to moles involves dividing by the molar mass of bromide ion. Finally, volumetric conversions using the density of seawater result in getting the volume in liters.
These steps might seem tedious at first, but they become intuitive with practice. Accurately converting concentration units is crucial for precise scientific calculations.
Moles calculation
Calculating moles is a central part of chemistry problem-solving. Moles are like a bridge between the mass of a substance and the number of atoms or molecules that substance contains. For the bromide ion concentration in seawater, we calculated the number of moles present using its molar mass.
The molar mass of a bromide ion (Br鈦�) is approximately 79.90 g/mol. Given that we have 65 mg (or 0.065 g), we used the formula below:
  • n = mass (g) / molar mass (g/mol)
where "n" represents the number of moles.
This computation is fundamental in calculating the concentration of elements within a solution and paves the way for arriving at molarity. In chemical analysis, mole calculations allow chemists to quantify substances and maintain accuracy in reaction predictions and formulations.

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Most popular questions from this chapter

Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the HCl(aq) in the stomach and each of the following substances used in various antacids: (a) \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\) (c) \(\mathrm{MgCO}_{3}(s)\) (d) \(\mathrm{NaAl}\left(\mathrm{CO}_{3}\right)(\mathrm{OH})_{2}(s),(\mathrm{e}) \mathrm{CaCO}_{3}(s)\)

A solid sample of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(0.350 \mathrm{~L}\) of \(0.500 \mathrm{M}\) aqueous HBr. The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(88.5 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. What mass of \(\mathrm{Zn}(\mathrm{OH})_{2}\) was added to the HBr solution?

(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly \(550 \mathrm{~mL}\) of solution. (b) How many moles of KBr are present in \(150 \mathrm{~mL}\) of a \(0.275 \mathrm{M}\) solution? (c) How many milliliters of 6. \(1 \mathrm{M}\) HCl solution are needed to obtain \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl} ?\)

A \(3.455-\mathrm{g}\) sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?

An \(8.65-g\) sample of an unknown group 2 A metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color signaling that the equivalence point has been reached after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}, \mathrm{Ba}^{2+} ?\)
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