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Chapter 4: Problem 69

(a) Which will have the highest concentration of sodium ions: \(0.25 \mathrm{M} \mathrm{NaCl}, 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) or \(0.075 \mathrm{MNa}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of sodium ion: \(20.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaHCO}_{3}\) or \(15.0 \mathrm{~mL}\) of \(0.04 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S} ?\)

Short Answer

Expert verified
(a) Na鈧侰O鈧 has the highest concentration of sodium ions (0.30 M). (b) NaHCO鈧 contains more moles of sodium ions (0.003 mol).

Step by step solution

01

Determine Sodium Ion Concentration from Each Compound

For each compound, calculate the concentration of sodium ions based on its dissociation in water. - **NaCl:** Dissociates into Na鈦 and Cl鈦. Thus, 1 mol of NaCl yields 1 mol of Na鈦. Concentration of Na鈦 = 0.25 M - **Na鈧侰O鈧:** Dissociates into 2 Na鈦 and CO鈧兟测伝. Thus, 1 mol of Na鈧侰O鈧 yields 2 mol of Na鈦. Concentration of Na鈦 = 2 脳 0.15 M = 0.30 M - **Na鈧働O鈧:** Dissociates into 3 Na鈦 and PO鈧劼斥伝. Thus, 1 mol of Na鈧働O鈧 yields 3 mol of Na鈦. Concentration of Na鈦 = 3 脳 0.075 M = 0.225 M.
02

Identify the Highest Concentration of Sodium Ions

Compare the concentrations calculated in Step 1: - NaCl: 0.25 M Na鈦 - Na鈧侰O鈧: 0.30 M Na鈦 - Na鈧働O鈧: 0.225 M Na鈦 The highest concentration of sodium ions is 0.30 M from Na鈧侰O鈧.
03

Calculate Moles of Sodium Ions in Each Solution

Determine the moles of sodium ions for each given solution: - **NaHCO鈧:** Dissociates into Na鈦 and HCO鈧冣伝. Thus, 1 mol of NaHCO鈧 yields 1 mol of Na鈦. Moles of Na鈦 in 20.0 mL of 0.15 M NaHCO鈧 = 0.020 L 脳 0.15 mol/L = 0.003 mol - **Na鈧係:** Dissociates into 2 Na鈦 and S虏鈦. Thus, 1 mol of Na鈧係 yields 2 mol of Na鈦. Moles of Na鈦 in 15.0 mL of 0.04 M Na鈧係 = 0.015 L 脳 0.04 mol/L 脳 2 = 0.0012 mol
04

Compare Moles of Sodium Ions

Compare the moles calculated in Step 3: - NaHCO鈧: 0.003 mol of Na鈦 - Na鈧係: 0.0012 mol of Na鈦 The solutions made from NaHCO鈧 contain more moles of sodium ions.

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Key Concepts

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

Sodium Ion Calculation
Determining the concentration of sodium ions in a solution is a key task in chemistry. Many compounds, when dissolved in water, release sodium ions (Na鈦), but the number of ions released depends on the chemical formula of the compound. For instance, NaCl will dissociate completely into one Na鈦 ion for every molecule, while Na鈧侰O鈧 will release two Na鈦 ions. To find the concentration of Na鈦, you need to multiply the molarity (M) of the compound by the number of sodium ions it can release.
  • Consider the compound's formula. For example, in Na鈧働O鈧, each formula unit releases three Na鈦 ions when dissociated.
  • Use the formula:
    Concentration of Na鈦 = Molarity of compound 脳 Number of Na鈦 ions released.
This is a simple method to calculate the amount of sodium ions in a solution, which is crucial for many chemical applications.
Chemical Dissociation
When a chemical compound dissolves in water, it often separates into its component ions through a process called dissociation. Understanding this process is vital for calculating ion concentrations.
  • For NaCl, the formula shows it's made up of Na鈦 and Cl鈦 ions. When NaCl is dissolved, it separates into one Na鈦 and one Cl鈦 ion.
  • Compounds like Na鈧侰O鈧 dissociate differently, splitting into two Na鈦 ions and one CO鈧兟测伝 ion.
  • A more complex example is Na鈧働O鈧, which dissociates to release three Na鈦 ions and one PO鈧劼斥伝 ion.
Understanding the dissociation pattern of a compound allows you to properly calculate how many of each ion are present in a solution. Chemical dissociation is essential for predicting the outcomes of reactions and the behavior of solutions under different conditions.
Molarity
Molarity is a concentration unit in chemistry that is defined as the number of moles of solute per liter of solution. It is denoted by the symbol "M". Molarity is central to understanding solution concentrations and is one of the first steps in calculating ion concentrations.
  • To calculate molarity, divide the moles of solute by the volume of the solution in liters: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
  • This unit helps chemists compare the concentration of solutions and determine reaction ratios and stoichiometry.
When you know the molarity of a compound, and how it dissociates, you can calculate individual ion concentrations, as shown in sodium ion calculations.
Moles Calculation
The concept of 'the mole' is crucial for converting between atoms, molecules, or ions and measurable quantities. A mole is Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\). When calculating moles, you're converting measurable mass or volume into a countable number of particles.
  • To calculate moles from volume and molarity, use the formula: \[ \text{Moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
  • This calculation allows for determining the exact number of sodium ions or any other species in a given volume of solution.
  • For example, calculating the moles of Na鈦 ions in a 20 mL solution of 0.15 M NaHCO鈧 involves converting mL to L (0.020 L) and using the equation to find 0.003 moles of Na鈦.
Mastering mole calculations enables accurate chemical analysis and is foundational for more advanced chemistry topics.

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Most popular questions from this chapter

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) HF, (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (benzoic acid), (c) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (benzene), (d) \(\mathrm{CoCl}_{3}\), (e) \(\mathrm{AgNO}_{3}\).

Three solutions are mixed together to form a single solution; in the final solution, there are \(0.2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\), \(0.1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S}\), and \(0.1 \mathrm{~mol} \mathrm{CaCl}_{2}\) present. What solid(s) will precipitate?

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-}\), (b) \(\mathrm{NO}_{3}^{-}\) (c) \(\mathrm{NH}_{4}^{+}\), (d) \(\mathrm{S}^{2-}\), (e) \(\mathrm{SO}_{4}^{2-}\).

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other platinum. You have three solutions at your disposal: \(1 \mathrm{M}\) sodium nitrate, \(1 M\) nitric acid, and \(1 M\) nickel nitrate. How could you use these solutions to determine the identities of each metal powder?

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a \(42.3 \mathrm{~mL}\) aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?
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