15.11: Exercises
- Page ID
- 357340
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)15.4: Acids and Bases Defined
- Write equations showing H2PO4− acting as both a Brønsted-Lowry acid and as a base with water.
- Answer
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acid: H2PO4− + H2O ⇌ HPO42− + H3O+; base: H2PO4− + H2O ⇌ H3PO4 + OH−
- Write the equation for the chemical reaction that occurs when caffeine (C8H10N4O2) acts as a Brønsted-Lowry base in water.
- Answer
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C8H10N4O2 + H2O ⇌ HC8H10N4O2+ + OH−
- Citric acid (C6H8O7) is the acid found in citrus fruits. It can lose a maximum of three H+ ions in the presence of a base. Write the chemical equations for citric acid acting stepwise as a Brønsted-Lowry acid in water.
- Answer
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C6H8O7 + H2O ⇌ C6H7O7− + H3O+; C6H7O7− + H2O ⇌ C6H6O72− + H3O+; C6H6O72− + H2O ⇌ C6H5O73− + H3O+
- What is the conjugate acid of each of the following?
- OH−
- H2O
- HCO3−
- NH3
- HSO4−
- H4N2
- Answer
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- H2O
- H3O+
- H2CO3
- NH4+
- H2SO4
- H5N2+
- What is the conjugate base of each of the following?
- H2PO4−
- HS−
- H3O2+
- H4N2
- H2S
- CH3OH
- Answer
-
- HPO42−
- S2−
- H2O2
- H3N2−
- HS−
- CH3O−
- Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- HNO3 + H2O ⇌ H3O+ + NO3−
- CN− + H2O ⇌ HCN + OH−
- H2SO4 + Cl− ⇌ HCl + HSO4−
- HSO4− + OH− ⇌ SO42− + H2O
- Answer
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- BL acid = HNO3; conjugate base = NO3−; BL base = H2O; conjugate acid = H3O+
- BL acid = H2O; conjugate base = OH−; BL base = CN−; conjugate acid = HCN
- BL acid = H2SO4; conjugate base = HSO4−; BL base = Cl−; conjugate acid = HCl
- BL acid = HSO4−; conjugate base = SO42−; BL base = OH−; conjugate acid = H2O
- Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- NO2− + H2O ⇌ HNO2 + OH−
- HBr + H2O ⇌ H3O+ + Br−
- HS− + H2O ⇌ H2S + OH−
- H2PO4− + HCl ⇌ H3PO4 + Cl−
- Answer
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- BL acid = H2O; conjugate base = OH−; BL base = NO2−; conjugate acid = HNO2
- BL acid = HBr; conjugate base = Br−; BL base = H2O; conjugate acid = H3O+
- BL acid = H2O; conjugate base = OH−; BL base = HS−; conjugate acid = H2S
- BL acid = HCl; conjugate base = Cl−; BL base = H2PO4−; conjugate acid = H3PO4
15.5: Strong and Weak Acids and Bases
- Identify each of the following acids or bases as strong or weak:
- HC2H3O2
- Ba(OH)2
- HClO4
- CH3NH2
- Answer
-
- weak acid
- strong base
- strong acid
- weak base
- Identify each of the following acids or bases as strong or weak:
- NaOH
- C17H21NO4 (cocaine)
- H2SO3
- HBr
- Answer
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- strong base
- weak base
- weak acid
- strong acid
15.6: Water as an Acid and as a Base
- Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain.
- Answer
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Inversely proportional. As the concentration of one increases, the concentration of the other will decrease.
- The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10−6 M. What is the concentration of hydroxide ions in the rainwater?
- Answer
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[OH−] = 5.9 × 10−9 M
- The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M. What is the concentration of hydronium in the solution?
- Answer
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[H3O+] = 3.1 × 10−12 M
- Calculate the hydroxide ion concentration for each of the following solutions. Is the solution acidic, basic, or neutral?
- [H3O+] = 3.5 × 10−4 M
- [H3O+] = 8.1 × 10−8 M
- [H3O+] = 7.0 × 10−3 M
- Answer
-
- [OH−] = 2.9 × 10−11 M; acidic
- [OH−] = 1.2 × 10−7 M; basic
- [OH−] = 1.4 × 10−12 M; acidic
- Calculate the hydronium ion concentration for each of the following solutions. Is the solution acidic, basic, or neutral?
- [OH−] = 1.0 × 10−7 M
- [OH−] = 9.3 × 10−11 M
- [OH−] = 2.5 × 10−2 M
- Answer
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- [H3O+] = 1.0 × 10−7 M; neutral
- [H3O+] = 1.1 × 10−4 M; acidic
- [H3O+] = 4.0 × 10−13 M; basic
- Calculate the hydronium and hydroxide ion concentrations of the following solutions:
- 0.200 M HCl
- 0.0143 M NaOH
- 3.0 M HNO3
- 0.0031 M Ca(OH)2
- Answer
-
- [H3O+] = 0.200 M; [OH−] = 5.0 × 10−14 M
- [H3O+] = 6.99 × 10−13 M; [OH−] = 0.0143 M
- [H3O+] = 3.0 M; [OH−] = 3.3 × 10−15 M
- [H3O+] = 1.6 × 10−12 M; [OH−] = 0.0062 M
- Calculate the hydronium and hydroxide ion concentrations of each of the following solutions:
- 0.000259 M HClO4
- 0.21 M NaOH
- 0.000071 M Ba(OH)2
- 2.5 M KOH
- Answer
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- [H3O+] = 0.000259 M; [OH−] = 3.86 × 10−11 M
- [H3O+] = 4.8 × 10−14 M; [OH−] = 0.21 M
- [H3O+] = 7.1 × 10−11 M; [OH−] = 0.00014 M
- [H3O+] = 4.0 ×10−15 M; [OH−] = 2.5 M
15.7: An Introduction to pH
- Identify each of the following solutions as acidic, basic, or neutral.
- pH = 8.2 (baking soda)
- pH = 2.54 (vinegar)
- pH = 9.4 (laundry detergent)
- Answer
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- basic
- acidic
- basic
- Identify each of the following solutions as acidic, basic, or neutral.
- pH = 7.44 (human tears)
- pH = 3.4 (white wine)
- pH = 4.85 (coffee)
- Answer
-
- basic
- acidic
- acidic
- Normal clean rain has a pH of 5.0 to 5.5. Typical acid rain has a pH of 4.0. How much more acidic is acid rain than normal rain with a pH of 5.0?
- Answer
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One pH unit corresponds to a factor of 10. So typical acid rain is 10 times more acidic than normal rain.
- When a sparkling water was first opened, it had a pH of 3.7. Tap water was found to have a pH of 6.7. Which water is more basic? How much more basic is it?
- Answer
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The tap water has a higher pH, so it is more basic. The tap water's pH is 3 pH units greater. This means that the tap water is 103 or 1000 times more basic.
15.8: pH and pOH Calculations
- What is the pOH range for an acidic solution?
- Answer
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The pOH is greater than 7 for an acidic solution (typically up to 14).
- What is the pOH range for a basic solution?
- Answer
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The pOH is less than 7 for a basic solution (typically down to 0).
- Calculate the pH of each solution and classify it as acidic or basic.
- pOH = 5.18
- pOH = 9.26
- pOH = 2.84
- Answer
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- pH = 8.82; basic
- pH = 4.74; acidic
- pH = 11.16; basic
- Calculate the pOH of each solution and classify it as acidic or basic.
- pH = 8.31
- pH = 5.20
- pH = 7.43
- Answer
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- pOH = 5.69; basic
- pOH = 8.80; acidic
- pOH = 6.57; basic
- The concentration of commercial HCl is about 12 M. What is its pH and pOH?
- Answer
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pH = −1.08; pOH = 15.08
- The concentration of concentrated H2SO4 is about 18 M. Assuming only one H+ comes off the H2SO4 molecule, what is its pH and pOH?
- Answer
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pH = −1.26; pOH = 15.26
- Calculate the pH and pOH of each of the following solutions:
- 0.200 M HCl
- 0.0143 M NaOH
- 3.0 M HNO3
- 0.0031 M Ca(OH)2
- Answer
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- pH = 0.699; pOH = 13.301
- pH = 12.155; pOH = 1.845
- pH = −0.48; pOH = 14.48
- pH = 11.79; pOH = 2.21
- Calculate the pH and pOH of each of the following solutions:
- 0.000259 M HClO4
- 0.21 M NaOH
- 0.000071 M Ba(OH)2
- 2.5 M KOH
- Answer
-
- pH = 3.587; pOH = 10.413
- pH = 13.32; pOH = 0.68
- pH = 10.15; pOH = 3.85
- pH = 14.40; pOH = −0.40
- What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
- Answer
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[H3O+] = 3.0 × 10−7 M; [OH−] = 3.3 × 10−8 M
- What are the hydronium and hydroxide ion concentrations in a solution whose pH is 9.20?
- Answer
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[H3O+] = 6.3 × 10−10 M; [OH−] = 1.6 × 10−5 M
- What are the hydronium and hydroxide ion concentrations in a solution whose pOH is 2.18?
- Answer
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[H3O+] = 1.5 × 10−12 M; [OH−] = 6.6 × 10−3 M
15.9: Reactions of Acids and Bases
- Write the balanced chemical equation between Zn metal and HCl(aq). The other product is ZnCl2.
- Answer
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Zn(s) + 2 HCl(aq) ⇌ ZnCl2(aq) + H2(g)
- Write the neutralization reaction in which ZnCl2, also found in Exercise 32, is the salt product.
- Answer
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Zn(OH)2 + 2 HCl ⇌ ZnCl2 + 2 H2O
- Write the balanced chemical equation between aluminum hydroxide and sulfuric acid.
- Answer
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2 Al(OH)3 + 3 H2SO4 ⇌ Al2(SO4)3 + 6 H2O
- Write the balanced chemical equation between phosphoric acid and barium hydroxide.
- Answer
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2 H3PO4 + 3 Ba(OH)2 ⇌ Ba3(PO4)2 + 6 H2O
- Complete and balance the following acid-base reactions:
- HCl solution reacts with solid Ca(OH)2(s)
- a solution of Sr(OH)2 is added to a solution of HNO3
- Answer
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- 2 HCl(aq) + Ca(OH)2(s) ⇌ CaCl2(aq) + 2 H2O(l)
- Sr(OH)2(aq) + 2 HNO3(aq) ⇌ Sr(NO3)2(aq) + 2 H2O(l)
- Complete and balance the following acid-base reactions:
- aqueous H2SO4 reacts with NaOH
- Ba(OH)2 reacts with HF gas
- Answer
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- H2SO4(aq) + 2 NaOH(aq) ⇌ Na2SO4(aq) + 2 H2O(l)
- Ba(OH)2(aq) + 2 HF(g) ⇌ BaF2(aq) + 2 H2O(l)
- Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction.
- Answer
-
2 HCl(aq) + Mg(OH)2(s) ⇌ MgCl2(aq) + 2 H2O(l)
This page was adapted from "Beginning Chemistry (Ball)" by LibreTexts and "Chemistry (OpenStax)" by LibreTexts and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Vicki MacMurdo (Anoka-Ramsey Community College) and Lance S. Lund (Anoka-Ramsey Community College).