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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, Na鈧働O鈧�: \[6.0 \mathrm{~M}\] Na鈦� and \[2.0 \mathrm{~M}\] PO鈧劼斥伝. b. Barium nitrate, Ba(NO鈧�)鈧�: \[0.500 \mathrm{~M}\] Ba虏鈦� and \[1.00 \mathrm{~M}\] NO鈧冣伝. c. Potassium chloride, KCl: \[0.0268 \mathrm{~M}\] K鈦� and \[0.0268 \mathrm{~M}\] Cl鈦�. d. Ammonium sulfate, (NH鈧�)鈧係O鈧�: \[1.33 \mathrm{~M}\] NH鈧勨伜 and \[0.667 \mathrm{~M}\] SO鈧劼测伝.

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: Na鈧働O鈧� 鈫� 3Na鈦� + PO鈧劼斥伝 Concentration of Na鈦� = [(0.0200 mol * 3) / 0.0100 L] = 6.0 M Concentration of PO鈧劼斥伝 = [(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(NO鈧�)鈧� 鈫� Ba虏鈦� + 2NO鈧冣伝 Concentration of Ba虏鈦� = [(0.300 mol * 1) / 0.600 L] = 0.500 M Concentration of NO鈧冣伝 = [(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 鈫� K鈦� + Cl鈦� Concentration of K鈦� = [(0.0134 mol * 1) / 0.500 L] = 0.0268 M Concentration of Cl鈦� = [(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 (NH鈧�)鈧係O鈧� in the solution: Molar mass of (NH鈧�)鈧係O鈧� = (2 * (14.01 + (4 * 1.01))) + (32.07 + (4 * 16.00)) = 132.14 g/mol moles of (NH鈧�)鈧係O鈧� = 132 g / 132.14 g/mol = 1.00 mol Now, calculate the concentrations of ions: (NH鈧�)鈧係O鈧� 鈫� 2NH鈧勨伜 + SO鈧劼测伝 Concentration of NH鈧勨伜 = [(1.00 mol * 2) / 1.50 L] = 1.33 M Concentration of SO鈧劼测伝 = [(1.00 mol * 1) / 1.50 L] = 0.667 M

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion Concentration Calculation
When dealing with solutions of strong electrolytes, understanding how to calculate the concentration of ions is essential. Strong electrolytes dissociate completely in water, splitting into their constituent ions. To find the concentration of ions, you need to follow these steps:
  • Identify the formula for the electrolyte and determine how it dissociates into ions. For instance, sodium phosphate (\(\text{Na}_3\text{PO}_4\)) dissociates into three sodium ions (\(\text{Na}^+\)) and one phosphate ion (\(\text{PO}_4^{3-}\)).
  • Convert the given volume of the solution to liters if necessary, as this will be needed to calculate molarity.
  • Calculate the concentration of each ion by using the formula:\[\text{Concentration of ion} = \frac{\text{moles of ion}}{\text{volume of solution in L}}\]
  • Multiply the moles of the compound by the number of each type of ion it produces upon dissociation.
Understanding this process enables you to accurately quantify the amount of each ion present in a chemical solution.
Molarity
Molarity is a key concept when dealing with solutions and is defined as the number of moles of solute per liter of solution. This is an important metric because it tells us how concentrated a solution is. When calculating molarity, the following steps are involved:
  • First, you must convert the volume of the solution to liters if it isn't already.
  • Next, determine the number of moles of the solute. If given in grams, the moles can be calculated using the molar mass of the solute. For example, if you have 132 g of ammonium sulfate (\((NH_4)_2SO_4\)), you use its molar mass (132.14 g/mol) to find 1.00 mole of the compound.
  • Finally, use the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
This calculation tells us the strength of a solution, which is crucial in chemistry when reactions depend on the concentration of solutions.
Chemical Solutions
Chemical solutions consist of a solute, which is the substance being dissolved, and a solvent, which is the medium that dissolves the solute. In most cases, the solvent is water, especially when dealing with aqueous solutions. Here are some basics:
  • The complexity of chemical solutions arises from the varied nature of solutes and their interactions with the solvents.
  • Understanding the properties of chemical solutions, such as conductivity and pH, can vary widely depending on the concentration and type of solute used.
  • In the context of strong electrolytes, as the solute dissolves, it fully dissociates into its constituent ions.
The dissolution and dissociation process impacts how chemical solutions are used in practical settings, such as in titrations or other chemical reactions, showcasing the versatile and complex nature of these fluids.
Chemical Equations
Chemical equations are representations of chemical reactions, where the reactants transform into products. For strong electrolytes, these equations represent the dissociation into ions. Here's how they work:
  • The chemical formula of a compound shows the ratio of each element that composes the compound. For instance, potassium chloride (\(\text{KCl}\)) shows a 1:1 ratio of potassium to chloride ions.
  • A balanced chemical equation respects the conservation of mass, meaning the number of each type of atom on the reactants side is equal to the number on the products side.
  • In a dissociation equation, emphasis is put on displaying the ions separately, such as \(\text{Ba(NO}_3)_2 \rightarrow \text{Ba}^{2+} + 2\text{NO}_3^{-}\), indicating that one barium nitrate turns into one barium ion and two nitrate ions.
Understanding chemical equations allows one to predict and balance chemical reactions effectively, especially when calculating the outcome of reactions and the resulting concentrations of products.

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Most popular questions from this chapter

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2}\), or \(200.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NaCl} ?\)

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{FeSO}_{4}(a q)+\mathrm{KCl}(a q)\) b. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) c. \(\mathrm{CaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) d. \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Hydrochloric acid \((75.0 \mathrm{~mL}\) of \(0.250 M\) ) is added to \(225.0 \mathrm{~mL}\) of \(0.0550 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

What volume of \(0.0200 M\) calcium hydroxide is required to neutralize \(35.00 \mathrm{~mL}\) of \(0.0500 M\) nitric acid?

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} .\) A \(0.250-\mathrm{g}\) sample of citric acid dissolved in \(25.0 \mathrm{~mL}\) of water requires \(37.2 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{NaOH}\) for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?
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