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Chapter 4: Problem 3

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

Expert verified
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) Co³⁺: 0.0269 M, Cl⁻: 0.0807 M (b) Ni³⁺: 0.0219 M, SO₄²⁻: 0.0329 M (c) Na⁺: 0.0313 M, MnO₄⁻: 0.0313 M (d) Fe²⁺: 0.0206 M, Br⁻: 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: Co³⁺ and a chloride ion with a -1 charge: Cl⁻. To balance the charges, we need 3 chloride ions for each cobalt ion. Thus, the formula for cobalt(III) chloride is CoCl₃.
02

(a) Calculate the molar mass of CoCl₃

The molar mass of CoCl₃ 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 CoCl₃ is: Molar mass = \(58.93 + (3 \times 35.45) = 58.93 + 106.35 = 165.28 \mathrm{~g/mol}\)
03

(a) Find the moles of CoCl₃

Given the mass of CoCl₃ as 20.00 g, we can find the moles of CoCl₃ 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 CoCl₃ dissolved in 4.50 L of the solution. This means: Molarity of Co³⁺ = moles of Co³⁺ / volume of solution = \(0.121 / 4.50 = 0.0269 \mathrm{~M}\) Since there are 3 moles of Cl⁻ for every mole of Co³⁺, the moles of Cl⁻ = \(3 \times 0.121 = 0.363 \mathrm{~mol}\). Molarity of Cl⁻ = moles of Cl⁻ / 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: Ni³⁺ and a sulfate ion with a -2 charge: SO₄²⁻. To balance the charges, we need 2 nickel ions and 3 sulfate ions. Thus, the formula for nickel(III) sulfate is Ni₂(SO₄)₃.
06

(b) Calculate the molar mass of Ni₂(SO₄)₃

The molar mass of Ni₂(SO₄)₃ 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 (SO₄) is 96.06 g/mol. So, the molar mass of Ni₂(SO₄)₃ 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 Ni₂(SO₄)₃

Given the mass of Ni₂(SO₄)₃ as 20.00 g, we can find the moles of Ni₂(SO₄)₃ 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 Ni₂(SO₄)₃ dissolved in 4.50 L of the solution: Molarity of Ni³⁺ = (2 × moles of Ni³⁺) / volume of solution = \((2 \times 0.0493) / 4.50 = 0.0219 \mathrm{~M}\) Since there are 3 moles of SO₄²⁻ for every mole of Ni₂(SO₄)₃, the moles of SO₄²⁻ = \(3 \times 0.0493 = 0.1479 \mathrm{~mol}\). Molarity of SO₄²⁻ = moles of SO₄²⁻ / 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⁺ and a permanganate ion with a -1 charge: MnO₄⁻. The charges are already balanced, so the formula for sodium permanganate is NaMnO₄.
10

(c) Calculate the molar mass of NaMnO₄

The molar mass of NaMnO₄ 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 NaMnO₄ 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 NaMnO₄

Given the mass of NaMnO₄ as 20.00 g, we can find the moles of NaMnO₄ 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 NaMnO₄ dissolved in 4.50 L of the solution. Molarity of Na⁺ = moles of Na⁺ / volume of solution = \(0.141 / 4.50 = 0.0313 \mathrm{~M}\) Molarity of MnO₄⁻ = moles of MnO₄⁻ / 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: Fe²⁺ and a bromine ion with a -1 charge: Br⁻. To balance the charges, we need 2 bromine ions for each iron ion. Thus, the formula for iron(II) bromide is FeBr₂.
14

(d) Calculate the molar mass of FeBr₂

The molar mass of FeBr₂ 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 FeBr₂ is: Molar mass = \(55.85 + (2 \times 79.90) = 55.85 + 159.80 = 215.65 \mathrm{~g/mol}\)
15

(d) Find the moles of FeBr₂

Given the mass of FeBr₂ as 20.00 g, we can find the moles of FeBr₂ 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 FeBr₂ dissolved in 4.50 L of the solution. Molarity of Fe²⁺ = moles of Fe²⁺ / volume of solution = \(0.0927 / 4.50 = 0.0206 \mathrm{~M}\) Since there are 2 moles of Br⁻ for every mole of Fe²⁺, the moles of Br⁻ = \(2 \times 0.0927 = 0.1854 \mathrm{~mol}\). Molarity of Br⁻ = moles of Br⁻ / volume of solution = \(0.1854 / 4.50 = 0.0412 \mathrm{~M}\)

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Most popular questions from this chapter

Mssign oxidation numbers to each element in (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NaMnO}_{4}\) (c) \(\mathrm{SnO}_{2}\) (d) NOF (e) \(\mathrm{NaO}_{2}\)

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