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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

What volume of \(0.2500 \mathrm{M}\) cobalt(III) sulfate is required to react completely with (a) \(25.00 \mathrm{~mL}\) of \(0.0315 \mathrm{M}\) calcium hydroxide? (b) \(5.00 \mathrm{~g}\) of sodium carbonate? (c) \(12.50 \mathrm{~mL}\) of \(0.1249 M\) potassium phosphate?

Calcium in blood or urine can be determined by precipitation as call cium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\). The precipitate is dissolved in strong acid and titrated with potassium permanganate. The products of the reaction are carbon dioxide and manganese(II) ion. A 24-hour urine sample is collected from an adult patient, reduced to a small volume, and titrated with \(26.2 \mathrm{~mL}\) of \(0.0946 \mathrm{M} \mathrm{KMnO}_{4}\). How many grams of calcium oxalate are in the sample? Normal range for \(\mathrm{Ca}^{2+}\) output for an adult is 100 to \(300 \mathrm{mg}\) per 24 hour. Is the sample within the normal range?

Analysis of an unknown acid shows that \(24.55 \mathrm{~mL}\) of \(0.128 \mathrm{M} \mathrm{NaOH}\) are required to react completely with \(0.566 \mathrm{~g}\) of the acid. The equation for. the reaction is $$ \mathrm{HB}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{B}^{-}(a q) $$ What is the molar mass of the acid?

Write a net ionic equation for any precipitation reaction that occurs when \(0.1 \mathrm{M}\) solutions of the following are mixed. (a) zinc nitrate and nickel(II) chloride (b) potassium phosphate and calcium nitrate (c) sodium hydroxide and zinc nitrate (d) iron(III) nitrate and barium hydroxide

The standard set by OSHA for the maximum amount of ammonia permitted in the workplace is \(5.00 \times 10^{-3} \%\) by mass. To determine a factory's compliance, \(10.00 \mathrm{~L}\) of air \((d=1.19 \mathrm{~g} / \mathrm{L})\) is bubbled into \(100.0 \mathrm{~mL}\) of \(0.02500 \mathrm{M} \mathrm{HCl}\) at the same temperature and pressure. Ammonia in the air bubbled in reacts with \(\mathrm{H}^{+}\) as follows: $$ \mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{NH}_{4}^{+}(a q) $$ The unreacted hydrogen ions required \(57.00 \mathrm{~mL}\) of \(0.03500 \mathrm{M} \mathrm{NaOH}\) for complete neutralization. Is the factory compliant with the OSHA standards for ammonia in the workplace?
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