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What is the molarity of each ion present in aqueous solutions prepared by dissolving \(20.00 \mathrm{~g}\) of the following compounds in water to make 4.50 L of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide

Short Answer

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Question: Calculate the molarity of each ion present in the solutions after dissolving the following compounds in 4.50 L of water: (a) 20.00 g of cobalt(III) chloride, (b) 20.00 g of nickel(III) sulfate, (c) 20.00 g of sodium permanganate, and (d) 20.00 g of iron(II) bromide. Answer: The molarity of each ion present in the solutions is as follows: (a) Co鲁鈦: 0.0269 M, Cl鈦: 0.0807 M (b) Ni鲁鈦: 0.0219 M, SO鈧劼测伝: 0.0329 M (c) Na鈦: 0.0313 M, MnO鈧勨伝: 0.0313 M (d) Fe虏鈦: 0.0206 M, Br鈦: 0.0412 M

Step by step solution

01

(a) Determine the formula of cobalt(III) chloride

Cobalt(III) chloride has a cobalt ion with a +3 charge: Co鲁鈦 and a chloride ion with a -1 charge: Cl鈦. To balance the charges, we need 3 chloride ions for each cobalt ion. Thus, the formula for cobalt(III) chloride is CoCl鈧.
02

(a) Calculate the molar mass of CoCl鈧

The molar mass of CoCl鈧 can be calculated by adding the individual molar masses of cobalt and chloride. The molar mass of cobalt (Co) is 58.93 g/mol, and the molar mass of chloride (Cl) is 35.45 g/mol. Hence, the molar mass of CoCl鈧 is: Molar mass = \(58.93 + (3 \times 35.45) = 58.93 + 106.35 = 165.28 \mathrm{~g/mol}\)
03

(a) Find the moles of CoCl鈧

Given the mass of CoCl鈧 as 20.00 g, we can find the moles of CoCl鈧 using the following formula: Moles = Mass / Molar mass Moles = \(20.00 / 165.28 = 0.121 \mathrm{~mol}\)
04

(a) Calculate the molarity of each ion present

We have a 0.121 mol of CoCl鈧 dissolved in 4.50 L of the solution. This means: Molarity of Co鲁鈦 = moles of Co鲁鈦 / volume of solution = \(0.121 / 4.50 = 0.0269 \mathrm{~M}\) Since there are 3 moles of Cl鈦 for every mole of Co鲁鈦, the moles of Cl鈦 = \(3 \times 0.121 = 0.363 \mathrm{~mol}\). Molarity of Cl鈦 = moles of Cl鈦 / volume of solution = \(0.363 / 4.50 = 0.0807 \mathrm{~M}\)
05

(b) Determine the formula of nickel(III) sulfate

Nickel(III) sulfate has a nickel ion with a +3 charge: Ni鲁鈦 and a sulfate ion with a -2 charge: SO鈧劼测伝. To balance the charges, we need 2 nickel ions and 3 sulfate ions. Thus, the formula for nickel(III) sulfate is Ni鈧(SO鈧)鈧.
06

(b) Calculate the molar mass of Ni鈧(SO鈧)鈧

The molar mass of Ni鈧(SO鈧)鈧 can be calculated by adding the individual molar masses of nickel and sulfate. The molar mass of nickel (Ni) is 58.69 g/mol, and the molar mass of sulfate (SO鈧) is 96.06 g/mol. So, the molar mass of Ni鈧(SO鈧)鈧 is: Molar mass = \((2 \times 58.69) + (3 \times 96.06) = 117.38 + 288.18 = 405.56 \mathrm{~g/mol}\)
07

(b) Find the moles of Ni鈧(SO鈧)鈧

Given the mass of Ni鈧(SO鈧)鈧 as 20.00 g, we can find the moles of Ni鈧(SO鈧)鈧 using the following formula: Moles = Mass / Molar mass Moles = \(20.00 / 405.56 = 0.0493 \mathrm{~mol}\)
08

(b) Calculate the molarity of each ion present

We have 0.0493 mol of Ni鈧(SO鈧)鈧 dissolved in 4.50 L of the solution: Molarity of Ni鲁鈦 = (2 脳 moles of Ni鲁鈦) / volume of solution = \((2 \times 0.0493) / 4.50 = 0.0219 \mathrm{~M}\) Since there are 3 moles of SO鈧劼测伝 for every mole of Ni鈧(SO鈧)鈧, the moles of SO鈧劼测伝 = \(3 \times 0.0493 = 0.1479 \mathrm{~mol}\). Molarity of SO鈧劼测伝 = moles of SO鈧劼测伝 / volume of solution = \(0.1479 / 4.50 = 0.0329 \mathrm{~M}\)
09

(c) Determine the formula of sodium permanganate

Sodium permanganate has a sodium ion with a +1 charge: Na鈦 and a permanganate ion with a -1 charge: MnO鈧勨伝. The charges are already balanced, so the formula for sodium permanganate is NaMnO鈧.
10

(c) Calculate the molar mass of NaMnO鈧

The molar mass of NaMnO鈧 can be calculated by adding the individual molar masses of sodium, manganese, and oxygen. The molar mass of sodium (Na) is 22.99 g/mol, the molar mass of manganese (Mn) is 54.94 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. The molar mass of NaMnO鈧 is: Molar mass = \(22.99 + 54.94 + (4 \times 16.00) = 22.99 + 54.94 + 64.00 = 141.93 \mathrm{~g/mol}\)
11

(c) Find the moles of NaMnO鈧

Given the mass of NaMnO鈧 as 20.00 g, we can find the moles of NaMnO鈧 using the following formula: Moles = Mass / Molar mass Moles = \(20.00 / 141.93 = 0.141 \mathrm{~mol}\)
12

(c) Calculate the molarity of each ion present

We have 0.141 mol of NaMnO鈧 dissolved in 4.50 L of the solution. Molarity of Na鈦 = moles of Na鈦 / volume of solution = \(0.141 / 4.50 = 0.0313 \mathrm{~M}\) Molarity of MnO鈧勨伝 = moles of MnO鈧勨伝 / volume of solution = \(0.141 / 4.50 = 0.0313 \mathrm{~M}\)
13

(d) Determine the formula of iron(II) bromide

Iron(II) bromide has an iron ion with a +2 charge: Fe虏鈦 and a bromine ion with a -1 charge: Br鈦. To balance the charges, we need 2 bromine ions for each iron ion. Thus, the formula for iron(II) bromide is FeBr鈧.
14

(d) Calculate the molar mass of FeBr鈧

The molar mass of FeBr鈧 can be calculated by adding the individual molar masses of iron and bromine. The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of bromine (Br) is 79.90 g/mol. The molar mass of FeBr鈧 is: Molar mass = \(55.85 + (2 \times 79.90) = 55.85 + 159.80 = 215.65 \mathrm{~g/mol}\)
15

(d) Find the moles of FeBr鈧

Given the mass of FeBr鈧 as 20.00 g, we can find the moles of FeBr鈧 using the following formula: Moles = Mass / Molar mass Moles = \(20.00 / 215.65 = 0.0927 \mathrm{~mol}\)
16

(d) Calculate the molarity of each ion present

We have 0.0927 mol of FeBr鈧 dissolved in 4.50 L of the solution. Molarity of Fe虏鈦 = moles of Fe虏鈦 / volume of solution = \(0.0927 / 4.50 = 0.0206 \mathrm{~M}\) Since there are 2 moles of Br鈦 for every mole of Fe虏鈦, the moles of Br鈦 = \(2 \times 0.0927 = 0.1854 \mathrm{~mol}\). Molarity of Br鈦 = moles of Br鈦 / volume of solution = \(0.1854 / 4.50 = 0.0412 \mathrm{~M}\)

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Key Concepts

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

Ionic Compounds
Ionic compounds are formed when elements combine due to opposite electrical charges. A typical ionic bond involves a metal that donates one or more electrons to a non-metal, resulting in positively charged metal cations and negatively charged non-metal anions. For instance, in cobalt(III) chloride, cobalt ions with a +3 charge pair with three chloride ions, each having a -1 charge, to ensure electrical neutrality in the chemical formula CoCl鈧.
Nickel(III) sulfate highlights a slightly different composition with two nickel ions of +3 charge each combining with three sulfate ions, creating the compound Ni鈧(SO鈧)鈧 where the overall charge balance is maintained.
Likewise, sodium permanganate, being NaMnO鈧, pairs the +1 sodium ion with a -1 permanganate ion, exhibiting the simplicity of a one-to-one charge balance. Lastly, in iron(II) bromide (FeBr鈧), an iron ion with a +2 charge bonds with two bromide ions of -1 charge each, keeping the compound electrically neutral.
Understanding these charge interactions is crucial when calculating the molarity of ions in solutions.
Molar Mass Calculation
Calculating molar mass is a fundamental skill in chemistry when working with ionic compounds. The molar mass signifies the mass of a given substance (in grams) as one mole, which is equivalent to the substance's formula weight expressed in atomic mass units. Let's break this process down:
You derive the molar mass by summing the atomic masses of all atoms in a molecule. For example, cobalt(III) chloride's molar mass combines the atomic mass of one cobalt atom (58.93 g/mol) and three chloride atoms (3 x 35.45 g/mol), culminating in 165.28 g/mol.
For nickel(III) sulfate, compute with the molar mass of two nickel atoms (58.69 g/mol each) and three sulfate groups (96.06 g/mol each), leading to a total of 405.56 g/mol. This sum provides the mass of one mole of the compound. Such calculations are essential for determining how many molecules are present in any given mass of the compound, facilitating the subsequent calculations of mole-based concentrations and reactions in aqueous solutions.
Aqueous Solutions
Aqueous solutions are mixtures wherein water acts as the solvent. In chemistry, many reactions and compounds are studied within these solutions due to water's exceptional capacity to dissolve numerous substances. Upon dissolving ionic compounds in water, they dissociate into their respective ions.
When the compound cobalt(III) chloride, for instance, dissolves, it forms distinct cobalt and chloride ions evenly dispersed throughout the liquid. The molarity of each ion in such a solution depends on both the number of moles of the dissolved compound and the volume of the solution at hand.
Molarity is calculated using the formula \( M = \frac{n}{V} \), where \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters. It's in the aqueous framework that we measure ion concentrations to predict reactions and behaviors in different chemical environments. Recognizing the unique dynamics of aqueous solutions allows chemists to tailor their approaches when working with a range of chemical reactions.

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Most popular questions from this chapter

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

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