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Chapter 4: Problem 30

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.0200 \mathrm{~mol}\) of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300 \mathrm{~mol}\) of barium nitrate in \(600.0 \mathrm{~mL}\) of solution c. \(1.00 \mathrm{~g}\) of potassium chloride in \(0.500 \mathrm{~L}\) of solution d. \(132 \mathrm{~g}\) of ammonium sulfate in \(1.50 \mathrm{~L}\) of solution

Short Answer

Expert verified
The concentrations of the ions present in each solution are: a. Sodium phosphate, Na₃PO₄: \[6.0 \mathrm{~M}\] Na⁺ and \[2.0 \mathrm{~M}\] PO₄³⁻. b. Barium nitrate, Ba(NO₃)₂: \[0.500 \mathrm{~M}\] Ba²⁺ and \[1.00 \mathrm{~M}\] NO₃⁻. c. Potassium chloride, KCl: \[0.0268 \mathrm{~M}\] K⁺ and \[0.0268 \mathrm{~M}\] Cl⁻. d. Ammonium sulfate, (NH₄)₂SO₄: \[1.33 \mathrm{~M}\] NH₄⁺ and \[0.667 \mathrm{~M}\] SO₄²⁻.

Step by step solution

01

Solution a: Calculate the concentrations of ions in sodium phosphate solution

First, convert 10.0 mL volume to L: 10.0 mL * (1 L / 1000 mL) = 0.0100 L Now, calculate the concentrations of ions: Na₃PO₄ → 3Na⁺ + PO₄³⁻ Concentration of Na⁺ = [(0.0200 mol * 3) / 0.0100 L] = 6.0 M Concentration of PO₄³⁻ = [(0.0200 mol * 1) / 0.0100 L] = 2.0 M
02

Solution b: Calculate the concentrations of ions in barium nitrate solution

First, convert 600.0 mL volume to L: 600.0 mL * (1 L / 1000 mL) = 0.600 L Now, calculate the concentrations of ions: Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻ Concentration of Ba²⁺ = [(0.300 mol * 1) / 0.600 L] = 0.500 M Concentration of NO₃⁻ = [(0.300 mol * 2) / 0.600 L] = 1.00 M
03

Solution c: Calculate the concentrations of ions in potassium chloride solution

First, calculate the moles of KCl in the solution: Molar mass of KCl = 39.10 (K) + 35.45 (Cl) = 74.55 g/mol moles of KCl = 1.00 g / 74.55 g/mol = 0.0134 mol Now, calculate the concentrations of ions: KCl → K⁺ + Cl⁻ Concentration of K⁺ = [(0.0134 mol * 1) / 0.500 L] = 0.0268 M Concentration of Cl⁻ = [(0.0134 mol * 1) / 0.500 L] = 0.0268 M
04

Solution d: Calculate the concentrations of ions in ammonium sulfate solution

First, calculate the moles of (NH₄)₂SO₄ in the solution: Molar mass of (NH₄)₂SO₄ = (2 * (14.01 + (4 * 1.01))) + (32.07 + (4 * 16.00)) = 132.14 g/mol moles of (NH₄)₂SO₄ = 132 g / 132.14 g/mol = 1.00 mol Now, calculate the concentrations of ions: (NH₄)₂SO₄ → 2NH₄⁺ + SO₄²⁻ Concentration of NH₄⁺ = [(1.00 mol * 2) / 1.50 L] = 1.33 M Concentration of SO₄²⁻ = [(1.00 mol * 1) / 1.50 L] = 0.667 M

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Most popular questions from this chapter

A 230.-mL sample of a \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is \(1.10 \mathrm{M}\). What volume of water evaporated from the \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution?

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 M\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The VO \(^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

What volume of each of the following bases will react completely with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) ? a. \(0.100 \mathrm{M} \mathrm{NaOH}\) b. \(0.0500 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) c. \(0.250 \mathrm{M} \mathrm{KOH}\)
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