Question

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(15.0 \mathrm{~g}\) of the following compounds in water to make \(655 \mathrm{~mL}\) of solution? (a) scandium(III) iodide (b) sodium carbonate (c) magnesium phosphate (d) potassium oxide

Short answer

Answer: The molarity of each ion in the aqueous solutions are as follows: (a) Scandium(III) iodide - Sc3+: 0.102 M and I-: 0.305 M (b) Sodium carbonate - Na+: 0.433 M and CO3(2-): 0.216 M (c) Magnesium phosphate - Mg2+: 0.262 M and PO4(3-): 0.174 M (d) Potassium oxide - K+: 0.487 M and O(2-): 0.243 M

Step-by-step solution

Identify the formula and ions of each compound

For the given compounds: (a) Scandium(III) iodide - ScI3, ions produced: Sc3+, 3I- (b) Sodium carbonate - Na2CO3, ions produced: 2Na+, CO3(2-) (c) Magnesium phosphate - Mg3(PO4)2, ions produced: 3Mg2+, 2PO4(3-) (d) Potassium oxide - K2O, ions produced: 2K+, O(2-)

Calculate the molar mass of each compound

(a) Scandium(III) iodide - ScI3 Molar mass = (1 * 44.96) + (3 * 126.90) = 225.66 g/mol (b) Sodium carbonate - Na2CO3 Molar mass = (2 * 22.99) + (1 * 12.01) + (3 * 16.00) = 105.99 g/mol (c) Magnesium phosphate - Mg3(PO4)2 Molar mass = (3 * 24.31) + (2 * (1 * 30.97 + 4 * 16.00)) = 262.84 g/mol (d) Potassium oxide - K2O Molar mass = (2 * 39.10) + (1 * 16.00) = 94.20 g/mol

Calculate the moles and ion moles for each compound

Given mass of the compound = 15.0 g (a) Scandium(III) iodide - ScI3 Moles = Mass / Molar mass = 15.0 g / 225.66 g/mol ≈ 0.0665 mol Moles of Sc3+ = 0.0665 mol Moles of I- = 3 * 0.0665 mol = 0.1995 mol (b) Sodium carbonate - Na2CO3 Moles = Mass / Molar mass = 15.0 g / 105.99 g/mol ≈ 0.1416 mol Moles of Na+ = 2 * 0.1416 mol = 0.2832 mol Moles of CO3(2-) = 0.1416 mol (c) Magnesium phosphate - Mg3(PO4)2 Moles = Mass / Molar mass = 15.0 g / 262.84 g/mol ≈ 0.0571 mol Moles of Mg2+ = 3 * 0.0571 mol = 0.1713 mol Moles of PO4(3-) = 2 * 0.0571 mol = 0.1142 mol (d) Potassium oxide - K2O Moles = Mass / Molar mass = 15.0 g / 94.20 g/mol ≈ 0.1593 mol Moles of K+ = 2 * 0.1593 mol = 0.3186 mol Moles of O(2-) = 0.1593 mol

Calculate the molarity of each ion

Molarity = moles of the ion / volume (L) Given volume = 655 mL = 0.655 L (a) Scandium(III) iodide Molarity of Sc3+ = 0.0665 mol / 0.655 L ≈ 0.102 M Molarity of I- = 0.1995 mol / 0.655 L ≈ 0.305 M (b) Sodium carbonate Molarity of Na+ = 0.2832 mol / 0.655 L ≈ 0.433 M Molarity of CO3(2-) = 0.1416 mol / 0.655 L ≈ 0.216 M (c) Magnesium phosphate Molarity of Mg2+ = 0.1713 mol / 0.655 L ≈ 0.262 M Molarity of PO4(3-) = 0.1142 mol / 0.655 L ≈ 0.174 M (d) Potassium oxide Molarity of K+ = 0.3186 mol / 0.655 L ≈ 0.487 M Molarity of O(2-) = 0.1593 mol / 0.655 L ≈ 0.243 M The molarity of each ion in the aqueous solutions are as follows: (a) Sc3+: 0.102 M and I-: 0.305 M (b) Na+: 0.433 M and CO3(2-): 0.216 M (c) Mg2+: 0.262 M and PO4(3-): 0.174 M (d) K+: 0.487 M and O(2-): 0.243 M

Key concepts

Ions

An ion is an atom or group of atoms that carries a charge due to the loss or gain of electrons. Ions are crucial in chemistry as they participate in creating electrical currents and forming chemical compounds. When ionic compounds dissolve in water, they dissociate into individual ions. This dissociation explains many properties of solutions, such as conductivity. To understand ions better:

• An example of cations includes Scandium ion ($Sc^{3+}$), which is positively charged due to losing three electrons.
• An example of anions is Iodide ion ($I^-$), formed by gaining an electron.
• Polyatomic ions like $CO_3^{2-}$ or $PO_4^{3-}$, where two or more atoms hold a charge collectively, are common in many compounds.

Recognizing how these ions contribute to solutions can help students predict reactions and understand balance within solutions.

Molar Mass

The molar mass of a substance is the weight of one mole of that substance, expressed in grams per mole (g/mol). It represents the total mass of all the atoms present in a molecule, and supports stoichiometric calculations in chemistry. For instance:

• Scandium(III) iodide, $ScI_3$, has a molar mass calculated by adding the atomic masses of one scandium atom and three iodine atoms, totaling 225.66 g/mol.
• sodium carbonate (Na$_2$CO$_3$) includes two sodium, one carbon, and three oxygen atoms, resulting in a molar mass of 105.99 g/mol.

Knowing molar masses enables chemists to convert between the mass of a compound and the number of moles, which is essential for laboratory work and theoretical calculations. Understanding these calculations helps to determine the proportion of elements in compounds, ensuring reactions progress correctly.

Chemical Compounds

Chemical compounds consist of two or more different elements bonded together in a specific ratio and structure. The properties of a compound can differ significantly from the properties of its constituent elements. They are often categorized into molecular compounds and ionic compounds. Ionic compounds like $Na_2CO_3$ and $Mg_3(PO_4)_2$ allow us to explore the behavior of ions in water. A compound's formula indicates the type and number of atoms involved. Some key points about compounds:

• Molecular compounds form when atoms share electrons.
• Ionic compounds form from electrostatic attractions between cations and anions.
• The chemical formula gives specific insights into a compound's composition.
• Predicting the formation of compounds can be aided by understanding ions.

Grasping these fundamentals allows students to comprehend how different elements interact, and the fundamental principles of compound formation.

Aqueous Solutions

An aqueous solution arises when a substance is dissolved in water. Water is a universal solvent, meaning it can dissolve many substances, facilitating various chemical reactions in an aqueous medium. Key attributes of aqueous solutions include:

• The solute is the substance dissolved, while the solvent is the liquid, typically water.
• Ionic compounds in aqueous solutions dissociate into ions, allowing them to conduct electricity.
• Stoichiometry in aqueous solutions often involves calculating the molarity (moles of solute per liter) to ensure accurate reaction stoichiometry.

Understanding aqueous solutions is crucial because many chemical reactions, especially in biological and environmental contexts, occur in water. By mastering this concept, students can predict reaction outcomes when compounds dissolve in water.