Skip to main content
Chemistry LibreTexts

10.E: Exercises

  1. Last updated
  2. Save as PDF
  • Page ID
    83120

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    10.1 Arrhenius Definition of Acids and Bases

    Exercises

    1. Give two examples of Arrhenius acids and two of Arrhenius bases.

    2. List the general properties of acids.

    3. List the general properties of bases.

    4. Write a balanced chemical equation for the neutralization of Ba(OH)2(aq) with HNO3(aq).

    5. How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC2H3O2)? (Hint: begin by writing a balanced chemical equation for the process.)

    6. Hydrazoic acid (HN3) can be neutralized by a base. (The azide ion, N3-, is a polyatomic ion with three nitrogen atoms and a -1 charge.)

      A) Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide. B) How many milliliters of 0.0245 M Ca(OH)2 are needed to neutralize 0.564 g of HN3?

    7. Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How is the molarity of 101 mL of Mg(OH)2 solution if it can neutralize 158 mL of 0.106 M HCl(aq)? You will need to write the balanced chemical equation.

    Answers

    1. HCl and HNO3 are acids and NaOH and NH3 are bases. (answers will vary)
    2. sour taste, react with metals, react with bases, and turn litmus red
    3. perhaps hydroxic acid
    4. 2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O(ℓ)
    5. 0.844 mol
    6. A)  2HN3(aq) + Ca(OH)2 → Ca(N3)2 + 2H2O;  B) 268 mL
    7. (0.106 mol HCl / L HCl soln) (0.158 L HCl soln) (1 mol Mg(OH)2 / 2 mol HCl) (1 / 0.101 L Mg(OH)2 soln) = 0.0829 mol Mg(OH)2 / L Mg(OH)2 soln = 0.0829 M Mg(OH)2

    10.2 Brønsted-Lowry Definition of Acids and Bases

    Exercises

    1. Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base.

      HCl(aq) + NH3(aq) → NH4+(aq) + Cl−(aq)

    2. Explain why a Brønsted-Lowry acid can be called a proton donor.

    3. Write the chemical equation of the reaction of ammonia in water and label the Brønsted-Lowry acid and base.

    4. Demonstrate that the dissolution of HNO3 in water is actually a Brønsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Brønsted-Lowry acid and base.

    Answers

    1. HCl: Brønsted-Lowry acid; NH3: Brønsted-Lowry base
    2. A Brønsted-Lowry acid gives away an H+ ion—nominally, a proton—in an acid-base reaction.
    3. NH3 + H2O → NH4+ + OH−; NH3: Brønsted-Lowry base; H2O: Brønsted-Lowry acid
    4. HNO3 + H2O → H3O+ + NO3−; HNO3: Brønsted-Lowry acid; H2O: Brønsted-Lowry base

    10.3 Water: Both an Acid and a Base

    Exercises

    1. Is H2O(ℓ) acting as an acid or a base?

      H2O(ℓ) + NH4+(aq) → H3O+(aq) + NH3(aq)

    2. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C2H3O2− solutions, the following reaction can occur:

      C2H3O2−(aq) + H2O(ℓ) → HC2H3O2(aq) + OH−(aq)

      Is H2O acting as an acid or a base in this reaction?

    Answers

    1. base
    2. acid

     

    10.4 The Strengths of Acids and Bases

    Concept Review Exercises

    1. Explain the difference between a strong acid or base and a weak acid or base.
    2. Explain what is occurring when a chemical reaction reaches equilibrium.
    3. Define pH.

    Answers

    1. A strong acid or base is 100% ionized in aqueous solution; a weak acid or base is less than 100% ionized.

    2. The overall reaction progress stops because the reverse process balances out the forward process.

    3. pH is a measure of the hydrogen ion concentration.

     

    Exercises

    1. Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: Consult Table 10.4.1)

      (A) HF;  (B) HC2H3O2; (C) HCl; (D) HClO4

    2. Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: Consult Table 10.4.1)

      (A) NH3; (B) NaOH; (C) Mg(OH)2; (D) Cu(OH)2

    3. Write the chemical equation for the equilibrium process for HC2H3O2.

    4. Write the chemical equation for the equilibrium process for NH3.

    5. Consider two different solutions, 0.15 M HCl and 0.15 M HF. (A) Which would have lower pH? (B) Which would have lower freezing point?

    Answers

    1. (A) weak; (B) weak; (C) strong; (D) strong
    2. (A) weak; (B) strong; (C) strong; (D) weak
    3. HC2H3O2(aq) ⇆ H+(aq) + C2H3O2−(aq)
    4. NH3(aq) + H2O ⇆ NH4+(aq) + OH−(aq)
    5. (A) The HCl solution will have lower pH ( because larger concentration of H+). (Both are the same concentration, but HF is a weak acid so does not ionize completely so there will be a lower [H+] and thus higher pH for HF.) (B) The HCl will have lower freezing point because there is a larger concentration of dissolved particles (since it completely ionizes).

    10.5: Buffers

    Concept Review Exercise

    1. Explain how a buffer prevents large changes in pH.

    Answer

    1. A buffer has components that react with both strong acids and strong bases to resist sudden changes in pH.

    Exercises

    1. Describe a buffer. What two related chemical components are required to make a buffer?

    2. Which solute combinations can make a buffer? Assume all are aqueous solutions.

      A) HCl and NaCl; B) HNO2 and NaNO2; C) NH4NO3 and HNO3; D) NH4NO3 and NH3

    3. For each combination in Exercise 2 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.

    4. The complete phosphate buffer system is based on four substances: H3PO4, H2PO4−, HPO42−, and PO43−. What different buffer solutions can be made from these substances?

    Answers

    1. A buffer resists sudden changes in pH. It has a weak acid or base and a salt of that weak acid or base.
    2. A) not a buffer; B) buffer; C) not a buffer; D) buffer
    3. #2b: strong acid: H+ + NO2− → HNO2; strong base: OH− + HNO2 → H2O + NO2−; #2d: strong acid: H+ + NH3 → NH4+; strong base: OH− + NH4+ → H2O + NH3
    4. Buffers can be made by combining H3PO4 and H2PO4−, H2PO4− and HPO42−, and HPO42− and PO43−.

     

    10.E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    1. Back to top
    • Was this article helpful?

    Recommended articles

    1. Article type
      Section or Page
      License
      CC BY-NC-SA
      License Version
      4.0
      Show Page TOC
      no on page
    2. Tags
      This page has no tags.