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Chapter 4: Problem 4

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

Expert verified
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

01

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-)
02

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
03

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
04

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

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Most popular questions from this chapter

Write the formulas of the following compounds and decide which are soluble in water. (a) sodium sulfate (b) iron(III) nitrate (c) silver chloride (d) chromium(III) hydroxide

What is the volume of \(1.222 \mathrm{M}\) sodium hydroxide required to react with (a) \(32.5 \mathrm{~mL}\) of \(0.569 \mathrm{M}\) sulfurous acid? (One mole of sulfurous acid reacts with two moles of hydroxide ion.) (b) \(5.00 \mathrm{~g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) ? (One mole of oxalic acid reacts with two moles of hydroxide ion.) (c) \(15.0 \mathrm{~g}\) of concentrated acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) that is \(88 \%\) by mass pure?

When solutions of aluminum sulfate and sodium hydroxide are mixed, a white gelatinous precipitate forms. (a) Write a balanced net ionic equation for the reaction. (b) What is the mass of the precipitate when \(2.76 \mathrm{~g}\) of aluminum sulfate in \(125 \mathrm{~mL}\) of solution is combined with \(85.0 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) ? (c) What is the molarity of the ion in excess? (Ignore spectator ions and assume that volumes are additive.)

Hydrogen gas is bubbled into a solution of barium hydroxide that has sulfur in it. The unbalanced equation for the reaction that takes place is $$ \mathrm{H}_{2}(g)+\mathrm{S}(s)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{S}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O} $$ (a) Balance the equation. (b) What volume of \(0.349 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is required to react completely with \(3.00 \mathrm{~g}\) of sulfur?

A reagent bottle is labeled \(0.450 \mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{3}\). (a) How many moles of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) are present in \(45.6 \mathrm{~mL}\) of this solution? (b) How many milliliters of this solution are required to furnish \(0.800 \mathrm{~mol}\) of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (c) Assuming no volume change, how many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) do you need to add to \(2.00 \mathrm{~L}\) of this solution to obtain a \(1.000 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (d) If \(50.0 \mathrm{~mL}\) of this solution is added to enough water to make \(125 \mathrm{~mL}\) of solution, what is the molarity of the diluted solution?
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