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Chapter 4: Problem 33

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in 100.0 \(\mathrm{mL}\) of solution b. 2.5 moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 1.25 \(\mathrm{L}\) of solution c. 5.00 \(\mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in 500.0 \(\mathrm{mL}\) of solution d. 1.00 \(\mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}\) in 250.0 \(\mathrm{mL}\) of solution

Short Answer

Expert verified
In summary, the concentrations of the ions in each solution are as follows: a. Ca²⁺: 1.00 M, NO₃⁻: 2.00 M b. Na⁺: 4.00 M, SO₄²⁻: 2.00 M c. NH₄⁺: 0.187 M, Cl⁻: 0.187 M d. K⁺: 0.0564 M, PO₄³⁻: 0.0188 M

Step by step solution

01

(a) 0.100 mole of Ca(NO3)₂ in 100.0 mL of solution

1. Calculate the moles of ions produced: One mole of Ca(NO3)₂ produces one mole of Ca²⁺ and two moles of NO₃⁻ ions when it dissolves. Therefore, we have 0.100 moles of Ca²⁺ and 0.100 * 2 = 0.200 moles of NO₃⁻ ions. 2. Calculate the molarity of each ion: Since the solution has a volume of 100.0 mL (0.100 L), we can compute the concentration of the ions. Molarity of Ca²⁺ = Moles of Ca²⁺ / Volume of solution in Liters = 0.100 moles / 0.100 L = 1.00 M Molarity of NO₃⁻ = Moles of NO₃⁻ / Volume of solution in Liters = 0.200 moles / 0.100 L = 2.00 M
02

(b) 2.5 moles of Na₂SO₄ in 1.25 L of solution

1. Calculate the moles of ions produced: One mole of Na₂SO₄ produces two moles of Na⁺ and one mole of SO₄²⁻ ions when it dissolves. So, we have 2.5 * 2 = 5.0 moles of Na⁺ and 2.5 moles of SO₄²⁻ ions. 2. Calculate the molarity of each ion: Molarity of Na⁺ = Moles of Na⁺ / Volume of solution in Liters = 5.0 moles / 1.25 L = 4.00 M Molarity of SO₄²⁻ = Moles of SO₄²⁻ / Volume of solution in Liters = 2.5 moles / 1.25 L = 2.00 M
03

(c) 5.00 g of NH₄Cl in 500.0 mL of solution

1. Convert grams to moles: To convert grams to moles, we need to use the molar mass of NH₄Cl, which is approximately 53.49 g/mol. Moles of NH₄Cl = 5.00 g / 53.49 g/mol ≈ 0.0935 moles 2. Calculate the moles of ions produced: One mole of NH₄Cl produces one mole of NH₄⁺ and one mole of Cl⁻ ions when it dissolves. So, we have 0.0935 moles of NH₄⁺ and 0.0935 moles of Cl⁻ ions. 3. Calculate the molarity of each ion: Molarity of NH₄⁺ = Moles of NH₄⁺ / Volume of solution in Liters = 0.0935 moles / 0.500 L = 0.187 M Molarity of Cl⁻ = Moles of Cl⁻ / Volume of solution in Liters = 0.0935 moles / 0.500 L = 0.187 M
04

(d) 1.00 g K₃PO₄ in 250.0 mL of solution

1. Convert grams to moles: To convert grams to moles, we need to use the molar mass of K₃PO₄, which is approximately 212.27 g/mol. Moles of K₃PO₄ = 1.00 g / 212.27 g/mol ≈ 0.00471 moles 2. Calculate the moles of ions produced: One mole of K₃PO₄ produces three moles of K⁺ and one mole of PO₄³⁻ ions when it dissolves. So, we have 0.00471 * 3 ≈ 0.0141 moles of K⁺ and 0.00471 moles of PO₄³⁻ ions. 3. Calculate the molarity of each ion: Molarity of K⁺ = Moles of K⁺ / Volume of solution in Liters = 0.0141 moles / 0.250 L = 0.0564 M Molarity of PO₄³⁻ = Moles of PO₄³⁻ / Volume of solution in Liters = 0.00471 moles / 0.250 L = 0.0188 M

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Most popular questions from this chapter

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

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

Many oxidation–reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. $\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)$ b. $\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)$ c. $\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(i)$ d. $\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$

For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed together. Indicate “none” if no precipitate will form $\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{3} \mathrm{PO}_{4}(a q) \longrightarrow$__________________(s) $\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s) $\mathrm{NaCl}(a q)+\mathrm{KNO}_{3}(a q) \longrightarrow$__________________(s) $\mathrm{KCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow$__________________(s) $\mathrm{FeCl}_{3}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow$__________________(s)

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?
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