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Chapter 15: Problem 17

Determine the molarity of each of the ions present in the following aqueous salt solutions: (assume \(100 \%\) ionization) (a) \(1.25 M \mathrm{CuBr}_{2}\) (b) \(0.75 \mathrm{M} \mathrm{NaHCO}_{3}\) (c) \(3.50 M \mathrm{~K}_{3} \mathrm{AsO}_{4}\) (d) \(0.65 M\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) Cu^{2+} 1.25 M, Br^- 2.50 M. (b) Na^+ 0.75 M, HCO_3^- 0.75 M. (c) K^+ 10.50 M, AsO_4^{3-} 3.50 M. (d) NH_4^+ 1.30 M, SO_4^{2-} 0.65 M.

Step by step solution

01

Determine Ionization

Identify the ions produced and their coefficients when each salt dissociates completely in water.
02

Calculate Molarity of Each Ion (a)

For the salt \( \text{CuBr}_2 \), dissociates into \( \text{Cu}^{2+} + 2 \text{Br}^- \). Given the molarity of \( \text{CuBr}_2 \) is 1.25 M, then the molarity of \( \text{Cu}^{2+} \) is 1.25 M and the molarity of \( \text{Br}^- \) is \( 2 \times 1.25 \text{ M} = 2.50 \text{ M} \).
03

Calculate Molarity of Each Ion (b)

For the salt \( \text{NaHCO}_3 \), dissociates into \( \text{Na}^+ + \text{HCO}_3^- \). Given the molarity of \( \text{NaHCO}_3 \) is 0.75 M, the molarity of \( \text{Na}^+ \) and \( \text{HCO}_3^- \) are both 0.75 M.
04

Calculate Molarity of Each Ion (c)

For the salt \( \text{K}_3 \text{AsO}_4 \), dissociates into \( 3 \text{K}^+ + \text{AsO}_4^{3-} \). Given the molarity of \( \text{K}_3 \text{AsO}_4 \) is 3.50 M, then the molarity of \( \text{AsO}_4^{3-} \) is 3.50 M and the molarity of \( \text{K}^+ \) is \( 3 \times 3.50 \text{ M} = 10.50 \text{ M} \).
05

Calculate Molarity of Each Ion (d)

For the salt \( \text{(NH}_4)_2 \text{SO}_4 \), dissociates into \( 2 \text{NH}_4^+ + \text{SO}_4^{2-} \). Given the molarity of \( \text{(NH}_4)_2 \text{SO}_4 \) is 0.65 M, the molarity of \( \text{SO}_4^{2-} \) is 0.65 M and the molarity of \( \text{NH}_4^+ \) is \( 2 \times 0.65 \text{ M} = 1.30 \text{ M} \).

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Key Concepts

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

Ionization in Aqueous Solutions
When salts dissolve in water, they break apart into their constituent ions, a process known as ionization. This means that each salt dissociates into positive and negative ions.
For example:
  • CuBr_2 dissociates into Cu2+ and 2 Br-
  • NaHCO_3 dissociates into Na+ and HCO_3-
  • K_3AsO_4 dissociates into 3 K+ and AsO_43-
  • (NH_4)_2SO_4 dissociates into 2 NH_4+ and SO_42-

Understanding this process helps in calculating the concentration of each ion in the solution. If a compound is said to be 100% ionized, it completely separates into these ions in water.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. In the context of ionization:
  • The stoichiometric coefficients (the numbers in front of each ion) help determine the concentration of ions produced from a given amount of salt.
  • For instance, in CuBr_2, 1 unit of CuBr_2 produces 1 unit of Cu2+ and 2 units of Br-.
  • Similarly, 1 unit of NaHCO_3 produces 1 unit of Na+ and 1 unit of HCO_3-.
  • This stoichiometric relationship allows us to calculate the concentration of individual ions when we know the concentration of the original salt solution.
Chemical Dissociation
Chemical dissociation is the process where molecules split into smaller particles such as atoms, ions, or radicals, usually in a reversible manner. When salts dissociate in water:
  • The ions are surrounded by water molecules, stabilizing them in solution
  • For example, K_3AsO_4 dissociates into 3 K+ and AsO_43-, meaning each unit of K_3AsO_4 will yield three potassium ions and one arsenate ion in solution.
    The concentration of these ions is then calculated based on the initial molarity of the salt solution, taking into account the number of ions each salt yields upon dissociation.
Salt Solutions
Salt solutions are mixtures where salts are dissolved in water, resulting in ionization. The properties of these solutions depend on:
  • The type of salt
  • The concentration of the salt
  • The extent of dissociation (often assumed to be 100% for simplicity).
    For example:
    • In a 1.25 M CuBr_2 solution,
    • The concentration of Cu2+ ions would be 1.25 M and the concentration of Br- ions would be 2.5 M
    • Similarly, in a 0.75 M NaHCO_3 solution, both Na+ and HCO_3- would have concentrations of 0.75 M.
      Thus, being able to determine the concentration of ions in these salt solutions helps predict the behavior of the substances in various chemical reactions and applications.

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Most popular questions from this chapter

A \(10.0-\mathrm{mL}\) sample of \(\mathrm{HNO}_{3}\) was diluted to a volume of \(100.00 \mathrm{~mL}\). Then \(25 \mathrm{~mL}\) of that diluted solution were needed to neutralize \(50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{KOH}\). What was the concentration of the original nitric acid?

Calculate the \(\mathrm{pH}\) of solutions having the following \(\left[\mathrm{H}^{+}\right]\): (a) \(0.0020 \mathrm{M}\) (b) \(7.0 \times 10^{-8} M\) (c) \(3.0 \mathrm{M}\)

Calculate the \(\left[\mathrm{H}^{+}\right]\)for (a) black coffee, with a \(\mathrm{pH}\) of \(5.0\) (b) black tea with a pH of \(6.8\) (c) green tea with a pH of \(7.9\) (d) laundry detergent with a pH of \(9.7\)

For each of the given pairs, determine which solution is more acidic. All are water solutions. Explain your answer. (a) \(1 \mathrm{M} \mathrm{HCl}\) or \(2 \mathrm{M} \mathrm{HCl}\) ? (b) \(1 M \mathrm{HNO}_{3}\) or \(1 M \mathrm{H}_{2} \mathrm{SO}_{4}\) ?

If \(380 \mathrm{~mL}\) of \(0.35 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) are added to 500 . \(\mathrm{mL}\) of \(0.65\) \(M \mathrm{HCl}\), will the mixture be acidic or basic? Find the \(\mathrm{pH}\) of the resulting solution.
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