Chapter 9 Acids And Bases - Websites.rcc.edu
Chapter 9–1Chapter 9 Acids and BasesSolutions to In-Chapter Problems9.1 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a net positiveor negative charge. Use Example 9.1 to help determine which of the compounds are Brønsted–Lowryacids.c. H2PO4–: contains a H atom,a Brønsted–Lowry acidd. Cl–: no H atoma. HI: contains a H atom,a Brønsted–Lowry acidb. SO42–: no H atom9.2 A Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a netnegative charge. Use Example 9.2 to help determine which of the compounds are Brønsted–Lowrybases.a. Al(OH)3: lone pairs on OH,a Brønsted–Lowry baseb. Br–: lone pairs on Br,a Brønsted–Lowry basec. NH4 : no lone pair of electronsd. CN–: lone pairs on C and N,a Brønsted–Lowry base9.3 In each equation, the Brønsted–Lowry acid is the species that loses a proton and the Brønsted–Lowrybase is the species that gains a proton. Use Example 9.3 to help determine which reactant is an acidand which is a base.a.HCl(g) NH3(g)acid acidOH–(aq)baseNH4 (aq) baseb. CH3COOH(l)c.Cl–(aq)H2O(l)CH3COO–(aq) H3O (aq)base HSO4–(aq)H2O(l) SO42–(aq)acid9.4 Use Example 9.4 to draw the conjugate acid of each species. Conjugate acid–base pairs differ by thepresence of a proton. To draw a conjugate acid from a base, add a proton, H . Then add 1 to thecharge of the base to give the charge on the conjugate acid.a. H2 O: add one H to make H3 O .b. I–: add one H to make HI.c. HCO3–: add one H to make H2 CO3.9.5 Use Example 9.5 to draw the conjugate base of each species. Conjugate acid–base pairs differ by thepresence of a proton. To draw a conjugate base from an acid, remove a proton, H . Then, add –1 tothe charge of the acid to give the charge on the conjugate base.a. H2S: remove one H to make HS– .b. HCN: remove one H to make CN– .c. HSO4–: remove one H to make SO42–.
Acids and Bases 9–29.6 Use Sample Problem 9.6 to label the acid and base in each reaction. The Brønsted–Lowry acid loses aproton to form its conjugate base. The Brønsted–Lowry base gains a proton to form its conjugateacid.acidbasea.H2O(l)b.CH3COOH(l) basebasec.Br (aq)I–(aq)HI(g)acid–conjugate base NH3(g)acid HNO3(aq)conjugate acidH3O (aq) conjugate baseCH3COO–(aq)conjugate acidHBr(aq)conjugate acid NH4 (aq)conjugate base NO3–(aq)9.7 Draw the conjugate acid and base of ammonia as in Examples 9.4 and 9.5.a. NH3: add one H to make NH4 .b. NH3: remove one H to make NH2–.9.8 The stronger the acid, the more readily it dissociates to form its conjugate base. In molecular art, thestrongest acid has the most A– and H3 O ions, and the fewest molecules of undissociated HA.D (smallest amount of A– and H3 O ) F E (largest amount of A– and H3 O )9.9 Use Table 9.1 to determine which acid is stronger. The stronger the acid, the weaker the conjugatebase.a. H2SO4 is the stronger acid; H3PO4 has the stronger conjugate base.b. HCl is the stronger acid; HF has the stronger conjugate base.c. H2CO3 is the stronger acid; NH4 has the stronger conjugate base.d. HF is the stronger acid; HCN has the stronger conjugate base.9.10Draw the conjugate acid of each species as in Example 9.4. Then compare the acids.a. NO2–: The conjugate acid is HNO2.NO3–: The conjugate acid is HNO3 .b. Since NO2– is the stronger base, it has a weaker conjugate acid. Therefore, HNO3 is thestronger acid.9.11To determine if the reactants or products are favored at equilibrium: Identify the acid in the reactants and the conjugate acid in the products. Determine the relative strength of the acid and the conjugate acid. Equilibrium favors the formation of the weaker acid.a.HF(g) OH–(aq)acidb. NH4 (aq)F–(aq) H2O(l)conjugate acidweaker acidProducts are favored. acidweaker acidReactants are favored.Cl–(aq)NH3(g) HCl(aq)conjugate acid
Chapter 9–3c.HCO3–(aq) H3O (aq)acid9.12H2CO3(aq) H2O(l)conjugate acidweaker acidProducts are favored.Compare the acid and conjugate acid to determine if the reactants or products are favored.a. C3H6O3(aq) H2O(l)C3H5O3–(aq) acidweaker acidReactants are favored.b. C3H6O3(aq)conjugate acid HCO3–(aq)C3H5O3–(aq) H2CO3(aq)conjugate acidweaker acidProducts are favored.acid9.13H3O (aq)Use Table 9.2 to find the Ka for each acid as in Example 9.7. The acid with the larger Ka is thestronger acid.a. increasing acid strength: HPO42–, H2PO4–, H3PO4b. increasing acid strength: HCN, CH3 COOH, HF9.14The stronger acid has the larger Ka. The weaker acid has the stronger conjugate base.a. H3PO4 is a stronger acid than CH3COOH.b. Remove one H to draw the conjugate base: H2 PO4– CH3COO– (the stronger base is formedfrom the weaker acid).9.15To determine the direction of equilibrium, identify the acid in the reactants and the conjugate acidin the products as in Sample Problem 9.10. Then compare their Ka values. Equilibrium favorsthe formation of the acid with the smaller Ka value.reactants favoredHCO3–(aq) acidKa 5.6 x 10–11smaller Kaweaker acid9.16NH3(aq)CO32–(aq) NH4 (aq)conjugate acidKa 5.6 x 10–10larger Kastronger acidCompare the acids by comparing their Ka values from Table 9.2.a. H2CO3 has the larger Ka.b. H2CO3 is stronger.c. HCN has the stronger conjugate base.d. H2CO3 has the weaker conjugate base.e. When H2CO3 is dissolved in water, the equilibrium lies further to the right.
Acids and Bases 9–49.17Use the equation [OH–] Kw/[H3 O ] to calculate the hydroxide ion concentration as in SampleProblem 9.11. When [OH–] [H3 O ], the solution is basic. When [OH–] [H3 O ], the solution isacidic.a. [OH–]b. [OH–]c. [OH–]d. [OH–]9.18[H3O ]Kw [H3O ]Kw [H3O ]Kw [H3O ]1.0 x 10–14 10–31.0 x 10–14 10–111.0 x 10–14 2.8 x 10–101.0 x 10–14 5.6 x 10–4 10–11 Macidic 10–3 Mbasic3.6 x 10–5 M basic1.8 x 10–11 M acidicUse the equation [H3O ] Kw/[OH–] to calculate the hydronium ion concentration as in SampleProblem 9.11.a.b.c.d.9.19Kw [H3O ] [H3O ] [H3O ] [H3O ] Kw[OH–]Kw[OH–]Kw[OH–]Kw[OH–]1.0 x 10–14 10–61.0 x 10–14 10–91.0 x 10–14 5.2 x 10–111.0 x 10–14 7.3 x 10–4 10–8 Mbasic 10–5 Macidic 1.9 x 10–4 Macidic 1.4 x 10–11 MbasicSince NaOH is a strong base that completely dissociates to form Na and OH–, the concentrationof NaOH gives the concentration of OH– ions. The [OH –] can then be used to calculate [H3 O ]from the expression for Kw. Similarly, since HCl is a strong acid and completely dissociates, the[H3 O ] can then be used to calculate [OH–] from the expression for Kw. See Example 9.8. a. [H3O ]b. [OH–] Kw[OH–] 1 x 10–1410–3concentration of OH–Kw1 x 10–14 [H3O ]10–3concentration of H3O 10–11 Mconcentration of H3O 10–11 Mconcentration of OH–
Chapter 9–5c.[OH–] Kw1 x 10–14 [H3O ] 1.56.7 x 10–15 Mconcentration of OH–concentration of H3O d.[H3O ] Kw1 x 10–14 [OH–] –13.0 x 103.3 x 10–14 Mconcentration of H3O concentration of OH–9.20When the coefficient of a number written in scientific notation is one, the pH equals the value ofx in 10–x.a. 1 10–6 M: pH 6b. 1 10–12 M: pH 129.21c. 0.000 01 M 10–5: pH 5d. 0.000 000 000 01 M 10–11: pH 11The pH equals the value of x in 10–x, and this gives the H3 O concentration. An acidic solutionhas a pH 7. A basic solution has a pH 7. A neutral solution has a pH 7.a. pH 13, [H3 O ] 1 10–13 M: basicb. pH 7, [H3O ] 1 10–7 M: neutral9.22Use a calculator to determine the antilogarithm of (– pH); [H3 O ] antilog(–pH).a. [H3O ] antilog(–pH)[H3O ]b. [H3O ] antilog(–pH)[H3O ]9.23 antilog(–10.2) 6 x 10–11 M antilog(–7.8) 2 x 10–8 Mc. [H3O ] antilog(–pH)[H3O ] antilog(–4.3) 5 x 10–5 MUse a calculator to determine the logarithm of a number that contains a coefficient other than onein scientific notation; pH –log [H3 O ].a. pH –log [H3O ] b. pH –log [H3O ] 9.24c. pH 3, [H3 O ] 1 10–3 M: acidicc. pH –log [H3O ] –log(1.8 x 10–6)–(–5.74) 5.74–log(9.21 x 10–12)d. pH –log [H3O ] –(–11.036) 11.036–log(8.8 x 10–5)–(–4.06) 4.06–log(7.62 x 10–11)–(–10.118) 10.118Label the organs based on the definitions listed in Answer 9.21.saliva—acidic to slightly basicpancreas—basicsmall intestines—basicurine—acidic to basicblood—basicstomach—acidiclarge intestines—acidic to neutral
Acids and Bases 9–69.25Write the balanced equations as in Section 9.7A.a.HNO3(aq) NaOH(aq)H2O(l) NaNO3(aq)waterb.H2SO4(aq) KOH(aq)H2O(l)salt K2SO4(aq)saltwaterH2SO4(aq) 2 KOH(aq)2 H2O(l)Place a 2 to balance H and O.Place a 2 to balance K.9.26A net ionic equation contains only the species involved in a reaction.H (aq)9.27 K2SO4(aq) OH–(aq)H2O(l) for Problem 9.25a, bThe acid and base react to form a salt and carbonic acid (H2CO3), which decomposes to CO2 andH2 O as in Sample Problem 9.17.H2SO4(aq) CaCO3(s)CaSO4(aq) H2O(l) CO2(g)from H2CO39.28The acid and base react to form a salt and carbonic acid (H2CO3), which decomposes to CO2 andH2 O as in Sample Problem 9.17.a.HNO3(aq) NaHCO3(aq)NaNO3(aq) H2O(l) CO2(g)from H2CO3b. 2 HNO3(aq) MgCO3(s)Mg(NO3)2(aq) H2O(l) CO2(g)from H2CO3
Chapter 9–79.29Follow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine whattypes of acid and base (strong or weak) are used to form the salt. When the ions in the salt comefrom a strong acid and strong base, the solution is neutral. When the ions come from acids andbases of different strength, the ion derived from the stronger reactant determines the acidity.a.c.KII–K from KOHstrong basefrom HIstrong acidK2CO3K from HCO3–weak acidNH4Id.I–from NH3weak basefrom HIstrong acidacidic solutionA basic solution has a pH 7.a.c.LiClLi from LiOHstrong baseCl–from HClstrong acidK2CO3K from KOHstrong baseCO32–from HCO3–weak acidbasic solutionpH 7NH4BrNH4 from NH3weak baseneutral solutionb.from Ba(OH)2strong baseBr–from HBrstrong acidacidic solutiond.MgCO3Mg2 from Mg(OH)2strong baseCO32–from HCO3–weak acidbasic solutionpH 7Cl–from HClstrong acidneutral solutionf.NH4 BaCl2Ba2 neutral solutionbasic solution9.30from HNO3strong acidfrom Ca(OH)2strong baseCO32–from KOHstrong basee.NO3–Ca2 neutral solutionb.Ca(NO3)2Na3PO4Na from NaOHstrong basePO43–from HPO42–weak acidbasic solution
Acids and Bases 9–89.31Calculate the molarity of the solution using a titration as in Example 9.11.25.5 mL NaOH molarity9.321L 0.0061 mol HCl L15 mL solution 0.0061 mol NaOH0.0061 mol HClx1000 mL 1L0.41 M HClAnswerCalculate the number of milliliters of NaOH solution needed.5.0 mL H2SO40.030 mol H2SO40.060 mol NaOH9.331 mol HCl1 mol NaOHmol0.24 mol NaOHx1000 mLx0.0061 mol NaOHM1Lx1Lx1000 mLxx6.0 mol H2SO4x2 mol NaOH1 mol H2SO41L 1L2.0 mol NaOH 0.030 mol H2SO40.060 mol NaOHx1000 mL1L 30. mL NaOH solutionA buffer is a solution whose pH changes very little when acid or base is added. Most buffersare solutions composed of approximately equal amounts of a weak acid and the salt of itsconjugate base.a. A solution containing HBr and NaBr is not a buffer, because it contains a strong acid, HBr.b. A solution containing HF and KF is a buffer since HF is a weak acid and F– is its conjugatebase.c. A solution containing CH3 COOH alone is not a buffer since it contains a weak acid only.9.34a. HCO3– and CO32– are both needed because the H3O concentration depends on two terms—Ka,which is a constant, and the ratio of the concentrations of the weak acid and its conjugate base.If these concentrations do not change much, the concentration of H3 O and therefore the pH donot change much.b. When a small amount of acid is added, [HCO3–] increases and [CO32–] decreases.c. When a small amount of base is added, [HCO3–] decreases and [CO32–] increases.
Chapter 9–99.35Calculate the pH of the buffer as in Example 9.12. The pH is the same if equal concentrations ofthe weak acid and conjugate base are present.a.[H3O ] Ka[H2PO4–]x[HPO42–][H3O ]pHb. [H3O ]–log [H3O ] pH Kac. 2–[HPO4 ][H3O ]–log [H3O ] pH Ka9.36 [HPO42–]–log [H3O ] pH [0.10 M] (6.2 x 10–8) x 6.2 x 10–8 M[1.0 M][1.0 M]7.21[H3O ]pH6.2 x 10–8 M[0.10 M]–log(6.2 x 10–8)[H2PO4–]x x7.21[H3O ]pH(6.2 x 10–8)–log(6.2 x 10–8)[H2PO4–]x (6.2 x 10–8) 6.2 x 10–8 Mx[0.50 M][0.50 M]–log(6.2 x 10–8)7.21Calculate the pH of the buffer as in Example 9.12.[H3O ] Kax[CH3COOH][CH3COO–][H3O ]pH –log [H3O ] pH (1.8 x 10–5) x 2.4 x 10–5 M–log(2.4 x 10–5)4.62[0.20 M][0.15 M]
Acids and Bases 9–10Solutions to End-of-Chapter Problems9.37 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a netpositive or negative charge. Use Example 9.1 to help determine which of the compounds areBrønsted–Lowry acids.a. HBr: contains a H atom,a Brønsted–Lowry acidb. Br2: no H atomc. AlCl3: no H atom9.38A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a netpositive or negative charge. Use Example 9.1 to help determine which of the compounds areBrønsted–Lowry acids.a. H2 O: contains a H atom,a Brønsted–Lowry acidb. I–: no H atomc. HOCl: contains a H atom,a Brønsted–Lowry acid9.39d. FeBr3: no H atome. CH3CH2COOH: contains a H atom,a Brønsted–Lowry acidf. CO2: no H atomA Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a netnegative charge. Use Example 9.2 to help determine which of the compounds are Brønsted–Lowry bases.a. OH–: lone pairs on OH,a Brønsted–Lowry baseb. Ca2 : no lone pair of electronsc. C2H6: no lone pair of electrons9.40d. HCOOH: contains a H atom,a Brønsted–Lowry acide. NO2–: no H atomf. HNO2: contains a H atom,a Brønsted–Lowry acidd. PO43–: lone pairs on O,a Brønsted–Lowry basee. OCl–: lone pairs on O and Cl,a Brønsted–Lowry basef. MgCO3: lone pairs on O,a Brønsted–Lowry baseA Brønsted–Lowry base must contain a lone pair of electrons, but it may be neutral or have a netnegative charge. Use Example 9.2 to help determine which of the compounds are Brønsted–Lowry bases.a. Cl–: lone pairs on Cl,a Brønsted–Lowry baseb. BH3: no lone pair of electronsc. H2 O: lone pairs on O,a Brønsted–Lowry based. Na : no lone pair of electronse. Ca(OH)2: lone pairs on O,a Brønsted–Lowry basef. HCOO–: lone pairs on O,a Brønsted–Lowry base
Chapter 9–119.41Draw the conjugate acid of each species as in Example 9.4.a. HS–: add one H to make H2S.b. CO32–: add one H to make HCO3–.Hd.c. NO2–: add one H to make HNO2.9.42HCNHHHadd one H to makeHHHCNHH HDraw the conjugate acid of each species as in Example 9.4.a. Br–: add one H to make HBr2–Ob.HOPOOc.HHOCCadd one H to makeHHHCOHadd one H to makeH9.43add one H to makePOHOHHOCCOHHHd.O–O–OHHHCO HHDraw the conjugate base of each species as in Example 9.5.a. HNO2: remove one H to make NO2–.b. NH4 : remove one H to make NH3.9.44c. H2 O2: remove one H to make HO2–.Draw the conjugate base of each species as in Example 9.5.a. H3 O : remove one H to make H2 Oc. H2Se: remove one H to make HSe–9.45c. HSO4–: remove one H to make SO42–The acid loses a proton (gray sphere) to form its conjugate base, while the base gains a proton toform its conjugate acid.a. acid baseconjugate base b.baseconjugate acidacidconjugate acidconjugate base
Acids and Bases 9–129.46The acid loses a proton to form its conjugate base, while the base gains a proton to form itsconjugate acid.a. acidb.basebaseconjugate baseb.I–(aq)conjugate baseNH3(g) baseNH4 (aq)conjugate acidHCOOH(l) acidHCOO–(aq)conjugate baseO (aq)conjugate acidH2SO4(aq)conjugate acidHI(g)acidH2O(l) basec.H3HSO4–(aq) baseH2O(l)–OH te acidacidSO42—(aq)conjugate baseHPO42–(aq)acidPO43–(aq)conjugate base–OH(aq)baseH2O(l)conjugate acidNH3(aq)baseNH4 (aq)conjugate acidHF(aq)acidF–(aq)conjugate baseDraw the conjugate acid and base of HCO3–.b. conjugate base: CO32–a. conjugate acid: H2 CO39.50Draw the conjugate acid and base of H2PO4–.b. conjugate base: HPO42–a. conjugate acid: H3 PO49.51conjugate baseLabel the conjugate acid–base pairs in each reaction as in Answer 9.6.a.9.49conjugate acidLabel the conjugate acid–base pairs in each reaction as in Answer 9.6.a.9.48conjugate acid acid9.47conjugate baseDraw the acid–base reaction.HNO3(aq) H2O (l)H3O (aq) NO3–(aq)
Chapter 9–139.52Draw the acid–base reaction.H3O (aq) HCOO–(aq)HCOOH(aq) H2O(l)9.53A represents HCl because it shows a fully dissociated acid. B represents HF, only partiallydissociated.9.54A represents HCN since HCN is a weak acid and will only partially dissociate in water.9.55a. B represents a strong acid because HZ is completely dissociated, forming H3O and Z–.b. A represents a weak acid because most of the acid HZ remains, and few ions (H3 O and Z–) areformed.9.56a. H2Zb. H2Oc. H2SO4 is stronger than HSO4–.Use Tables 9.1 and 9.2 to determine the stronger acid as in Example 9.7. The acid with the largerKa is the stronger acid.c. HF is stronger than H2 O.The weaker acid has the stronger conjugate base.a. H2 O9.60H3O Use Tables 9.1 and 9.2 to determine the stronger acid as in Example 9.7. The acid with the largerKa is the stronger acid.a. HCN is stronger than HPO42–.b. HSO4– is stronger than NH4 .9.59H3O HZ–a. CH3COOH is stronger than H2 O.b. H3PO4 is stronger than HCO3–.9.582Z2–H2O H2Z9.572b. HCO3–c. HSO4–The weaker acid has the stronger conjugate base.a. HPO42–b. NH4 c. H2 O9.61a. An acid that dissociates to a greater extent in water is a stronger acid: A is the stronger acid.b. An acid with a smaller Ka is weaker (A is weaker): B is the stronger acid.c. An acid with a stronger conjugate base is a weaker acid (A is weaker): B is the stronger acid.9.62a. An acid that dissociates to a greater extent in water is the stronger acid: B is the stronger acid.b. An acid with a larger Ka is stronger (A is stronger): A is the stronger acid.c. An acid with a stronger conjugate base is a weaker acid (B is weaker): A is the stronger acid.
Acids and Bases 9–149.63The acid with the larger Ka is the stronger acid. The stronger acid has a weaker conjugate base.a.SO42–HSO4–stronger acid conjugate baseKa 1.2 x 10–2larger Kastronger acidb.9.64HPO42–Ka 6.2 x 10–8conjugate basestronger baseCH3COOHCH3COO–CH3CH2COOHCH3CH2COO–stronger acidKa 1.8 x 10–5larger Kastronger acidconjugate baseKa 1.3 x 10–5conjugate basestronger baseThe acid with the larger Ka is the stronger acid. The stronger acid has a weaker conjugate base.a.H2PO4–H3PO4stronger acid conjugate baseKa 7.5 x 10–3larger Kastronger acidb.HCOO–HCOOHstronger acidKa 1.8 x 10–4larger Kastronger acid9.65H2PO4–conjugate baseHCOOHHCOO–Ka 1.8 x 10–4conjugate basestronger baseC6H5COOHKa 6.5 x 10–5C6H5COO–conjugate basestronger baseThe stronger acid has more dissociated ions. The stronger acid has the larger Ka.a. B has more dissociated ions (A– and H3 O ) and is therefore the stronger acid.b. B has the larger Ka since it is the stronger acid.9.66The stronger acid has the weaker conjugate base. Since A is the weaker acid it will have thestronger conjugate base.9.67To determine if the reactants or products are favored at equilibrium: Identify the acid in the reactants and the conjugate acid in the products. Determine the relative strength of the acid and the conjugate acid. Equilibrium favors the formation of the weaker acid.a.H3PO4(aq) CN–(aq)H2PO4–(aq)acidb.Br–(aq) HCN(aq)conjugate acidweaker acidProducts are favored. HSO4–(aq)acidweaker acidReactants are favored.SO42–(aq) HBr(aq)conjugate acid
Chapter 9–15c. CH3COO–(aq) H2CO3(aq)acidweaker acidReactants are favored.9.68 HF(aq)conjugate acidNH4 (aq)NH3(aq)acidb.F–(aq) conjugate acidweaker acidProducts are favored.Br–(aq)c. HBr(aq)acidweaker acidReactants are favored.conjugate acidHCN(aq)HCO3–(aq) –OH(aq) H2O(l)H2CO3(aq) –CN(aq)conjugate acidacidweaker acidReactants are favored.Use the equation [OH–] Kw/[H3 O ] to calculate the hydroxide ion concentration as in Answer9.17. When [OH–] [H3 O ], the solution is basic. When [OH–] [H3O ], the solution is acidic.a. [OH–]b. [OH–]c. [OH–]d. [OH–]9.70HCO3–(aq)To determine if the reactants or products are favored at equilibrium: Identify the acid in the reactants and the conjugate acid in the products. Determine the relative strength of the acid and the conjugate acid. Equilibrium favors the formation of the weaker acid.a.9.69 CH3COOH(aq) Kw[H3O ]Kw[H3O ]Kw [H3O ]Kw [H3O ] 1.0 x 10–1410–81.0 x 10–1410–101.0 x 10–143.0 x 10–41.0 x 10–142.5 x 10–11 10–6 Mbasic 10–4 Mbasic 3.3 x 10–11 Macidic 4.0 x 10–4 MbasicUse the equation [–OH] Kw/[H3 O ] to calculate the hydroxide ion concentration as in Answer9.17. When [– OH] [H3 O ], the solution is basic. When [–OH] [H3O ], the solution is acidic.a. [–OH] Kw [H3O ] 1.0 x 10–1410–1 10–13 Macidic
Acids and Bases 9–16b. [–OH] Kw 1.0 x 10–1410–13[H3O ]c. [–OH] Kw [H3O ]d. [–OH]Kw 2.6 3.8 x 10–8 Macidic8.3 x 10–3 Mbasic 1.2 x10–12Use the equation [H3O ] Kw/[OH–] to calculate the hydronium ion concentration as in SampleProblem 9.11.a.b.c.d.9.7210–1 Mbasicx10–71.0 x 10–14 [H3O ]9.711.0 x 10–14 [H3O ] [H3O ] [H3O ] [H3O ] Kw[OH–]Kw[OH–]Kw[OH–]Kw[OH–] 1.0 x 10–1410–2 1.0 x 10–144.0 x 10–8 1.0 x 10–146.2 x 10–7 1.0 x 10–148.5 x 10–13 10–12 Mbasic 2.5 x 10–7 Macidic 1.6 x 10–8 Mbasic 1.2 x 10–2 MacidicUse the equation [H3O ] Kw/[–OH] to calculate the hydroxide ion concentration as in SampleProblem 9.11. When [H3 O ] [– OH], the solution is acidic. When [H3 O ] [–OH], the solution isbasic.a.b.c.d.[H3O ][H3O ][H3O ][H3O ] Kw[–OH]Kw– 10–12 [ OH] Kw[–OH]Kw[–OH]1.0 x 10–141.0 x 10–145.0 x 10–2 Macidic 2.0 x 10–5 M10–101.0 x 10–146.0 x 10–4 1.0 x 10–148.9 x 10–11acidic 1.7 x 10–11 Mbasic 1.1 x 10–4 Macidic
Chapter 9–179.73Use a calculator to determine the logarithm of a number that contains a coefficient other than onein scientific notation; pH –log [H3 O ] as in Answer 9.23.a. pH –log [H3O ] –log(10–12) –(–12)b. pH –log [H3O ] 9.74 12–log(2.5 x 10–7)d. pH –log [H3O ] –(–6.60) 6.60–log(1.6 x 10–8)–(–7.80) 7.80–log(1.2 x 10–2)–(–1.92) 1.92Use a calculator to determine the logarithm of a number that contains a coefficient other than onein scientific notation; pH –log [H3 O ] as in Answer 9.25.a. pH –log [H3O ] –log(10–2) c. pH –log [H3O ] 2 –log(2.0 x 10–5)d. pH –log [H3O ] –(–2)b. pH –log [H3O ] 9.75c. pH –log [H3O ] –(–4.70) 4.70–log(1.7 x 10–11)–(–10.77) 10.77–log(1.1 x 10–4)–(–3.96) 3.96[H3O ]5.3 10–35.0 10–74 10–51.5 10–5[OH–]1.9 10–122.0 10–82.5 10–106.8 acidicacidic[H3O ]6.3 10–44.0 10–62.5 10–51.5 10–11[OH–]1.6 10–112.5 10–94.0 10–106.8 basicbasic9.769.77Use a calculator to determine the antilogarithm of (–pH); [H3O ] antilog(–pH).a. [H3O ] antilog(–pH)[H3O ]b. [H3O ] antilog(–pH)[H3O ] antilog(–12) 1 x 10–12 M antilog(–1) 1 x 10–1 Mc. [H3O ] antilog(–pH)[H3O ]d. [H3O ] antilog(–pH)[H3O ] antilog(–1.80) 1.6 x 10–2 M antilog(–8.90) 1.3 x 10–9 M
Acids and Bases 9–189.78Use a calculator to determine the antilogarithm of (–pH); [H3O ] antilog(–pH).a. [H3O ] antilog(–pH)[H3O ]b. [H3O ] antilog(–pH)[H3O ]9.79 antilog(–pH)[H3O ][OH–]Kw 1 x 10–4 M antilog(–8) 1 x10–8 antilog(–pH)[H3O ]d. [H3O ] antilog(–pH)[H3O ]M[H3O ] antilog(–5.90) 1.3 x 10–6 M1.0 x 10–14 antilog(–pH)[H3O ][–OH]Kw antilog(–2.60) 2.5 x 10–3 M antilog(–11.30) 5.0 x 10–12 M1.3 x 10–6 7.7 x 10–9 M[H3O ] antilog(–8.2) 6 x 10–9 M1.0 x 10–146 x 10–9 2 x 10–6 MUse the equations in Answers 9.77 and 9.69 to calculate the concentrations of H3O and OH– inthe sample.[H3O ] antilog(–pH)[H3O ][OH–]9.82 c. [H3O ]Use the equations in Problems 9.77 and 9.69 to calculate the concentrations of H3O and – OH inthe sample.[H3O ]9.81antilog(–4)Use the equations in Answers 9.77 and 9.69 to calculate the concentrations of H3O and OH– inthe sample.[H3O ]9.80 Kw [H3O ] antilog(–4.10) 7.9 x 10–5 M1.0 x 10–14 7.9 x 10–5 1.3 x 10–10 MUse the equations in Problems 9.77 and 9.69 to calculate the concentrations of H3O and – OH inthe sample.[H3O ] antilog(–pH)[H3O ][–OH] Kw[H3O ] antilog(–3.15) 7.1 x 10–4 M1.0 x 10–147.1 x 10–4 1.4 x 10–11 M
Chapter 9–199.83Use a calculator to determine the logarithm of a number that contains a coefficient other than onein scientific notation; pH –log [H3 O ].a. pH –log [H3O ] Kwb. [H3O ] 9.84[OH–] –log(2.5 x 10–3)–(–2.60)1.0 x 10–141.5 x 10–2 2.60 6.7 x 10–13 MpH –log [H3O ] –log(6.7 x 10–13)–(–12.17) Use a calculator to determine the logarithm of a number that contains a coefficient other than onein scientific notation; pH –log [H3 O ].a. pH –log [H3O ] Kwb. [H3O ] [–OH] –log(0.015)–(–1.82)1.0 x 10–140.0025 1.82 4.0 x 10–12 MpH –log [H3O ] –log(4.0 x 10–12)–(–11.40) 9.85The pH of 0.10 M HCl is lower than the pH of 0.1 M CH3COOH solution (1.0 vs. 2.9) becauseHCl is fully dissociated to H3 O and Cl–, whereas CH3 COOH is not fully dissociated.9.86The pH of 0.0050 M CH3COOH solution is higher than the pH of 0.0050 M HCl (3.5 vs. 2.3)because HCl is fully dissociated to H3O and Cl–, whereas CH3COOH is not fully dissociated.9.87Write the balanced equations as in Section 9.7.a. HBr(aq)b. KOH(aq)2 HNO3(aq) Ca(OH)2(aq)c. HCl(aq)d. H2SO4(aq)9.8812.17 NaHCO3(aq) Mg(OH)2(aq)KBr(aq) H2O(l)2 H2O(l) Ca(NO3)2(aq)NaCl(aq) H2O(l) CO2(g)2 H2O(l) MgSO4(aq)Write the balanced equations as in Section 9.7.a. HNO3(aq) LiOH(aq)LiNO3(aq) H2O(l)b.H2SO4(aq) 2 NaOH(aq)2 H2O(l) Na2SO4(aq)c.K2CO3(aq) 2 HCl(aq)H2O(l) CO2(g) 2 KCl(aq)11.40
Acids and Bases 9–20d.9.89HI(aq) NaHCO3(aq)H2O(l) CO2(g) NaI(aq)Write the balanced equation.Ca(NO3)2(aq) CO2(g) H2O(l)2 HNO3(aq) CaCO3(s)9.90Write the balanced equation.Al(OH)3(s) 3 HCl(aq)9.91AlCl3(aq) 3 H2O(l)Follow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine whattypes of acid and base (strong or weak) are used to form the salt. When the ions in the salt comefrom a strong acid and strong base, the solution is neutral. When the ions come from acids andbases of different strength, the ion derived from the stronger reactant determines the acidity.a.c.NaII–Na from NaOHstrong basefrom HIstrong acidNH4 Li from LiOHstrong baseF–from HFweak acidbasic solution9.92from Mg(OH)2strong baseK from KOHstrong baseHCO3–from H2CO3weak acidbasic solutionBr–from HBrstrong acidneutral solutionf.KHCO3MgBr2Mg2 acidic solutiond.LiFe.NO3–from HNO3strong acidfrom NH3weak baseneutral solutionb.NH4NO3NaH2PO4Na from NaOHstrong baseH2PO4–from H3PO4weak acidbasic solutionFollow Example 9.10 to determine the acidity when a salt is dissolved in water. Determine whattypes of acid and base (strong or weak) are used to form the salt. When the ions in the salt comefrom a strong acid and strong base, the solution is neutral. When the ions come from acids andbases of different strength, the ion derived from the stronger reactant determines the acidity.
Chapter 9–21a.c.NaBrNa K Br–neutral solutionb.from NaOHstrong based.Cs CN–from HCNweak acidfrom Ca(OH)2strong basefrom HBrstrong acidneutral solutionf.K3PO4F–from CsOHstrong baseBr–K from KOHstrong basefrom HFweak acidPO43–from H3PO4weak acidbasic solutionbasic solutionCalculate the molarity of the solution using a titration as in Example 9.11.35.5 mL NaOHM 1Lx1000 mL25 mL solution 0.0036 mol NaOH0.0036 mol HCl0.0036 mol HCl Lmolarity1L 1 mol NaOHmol0.10 mol NaOHx1 mol HClx0.0036 mol NaOH9.94from CH3COOHweak acidCsFbasic solution9.93Ca2 basic solutionNaCNNa CaBr2CH3COO–from KOHstrong basefrom NaOHfrom HBrstrong base strong acide.KCH3COOx1000 mL 1L0.14 M HClAnswerCalculate the molarity of the solution using a titration as in Example 9.11.17.2 mL NaOH0.0026 mol NaOHMmolarity 1LxmolL1000 mL1 mol HClx1 mol NaOH 0.15 mol NaOHx1L 0.0026 mol HCl5.00 mL solution 0.0026 mol NaOH0.0026 mol HClx1000 mL1L 0.52 M HClAnswer
Acids and Bases 9–229.95Calculate the molarity of the solution using a titration as in Example 9.11.15.5 mL NaOH molarity9.96 L 0.0031 mol NaOH0.0031 mol CH3COOH0.0031 mol CH3COOH1000 mLx25 mL solutionM molarity1Lx0.0033 mol NaOH 1L0.12 M CH3COOHAnswer1 mol H2SO4x0.18 mol NaOHx1000 mL2 mol NaOHmol 1L0.0033 mol NaOH0.0017 mol H2SO4 0.0017 mol H2SO4 L1000 mLx25.0 mL solution1L 0.068 M H2SO4AnswerCalculate the number of milliliters of solution needed.10.0 mL CH3COOH0.025 mol NaOH1Lxx1000 mLx0.025 mol CH3COOH9.98 1LCalculate the molarity of the solution using a titration as in Example 9.11.18.5 mL NaOH9.971 mol CH3COOH1 mol NaOHmol0.20 mol NaOHx1000 mLx0.0031 mol NaOHM1Lx1 mol NaOH2.5 mol CH3COOH1L 0.025 mol CH3COOH0.025 mol NaOH 1 mol CH3COOH1Lx1000 mLx1.0 mol NaOH1L 25 mL NaOH solution 0.028 mol H2SO4Calculate the number of milliliters of solution needed.8.0 mL H2SO4x1000 mLx0.028 mol H2SO40.056 mol NaOH1Lx2 mol NaOH3.5 mol H2SO41L0.056 mol NaOH 1 mol H2SO4x1L2.0 mol NaOHx1000 mL1L 28 mL NaOH solution
Chapter 9–239.999.100B is a buffer because it contains equal amounts of the acid and its conjugate base. A is not abuffer since it contains only acid.A is a buffer because it contains equal amounts of the acid and its conjugate base. B is not abuffer since it contains only conjugate base.9.101A buffer is most effective at minimizing pH changes when the concentrations of the weak acidand its conjugate base are equal because when acid or base is added to a buffer, a proton donorand a proton acceptor are available to react with either of them. Since the ratio of theconcentration of the acid and conjugate base is one to begin with, the ratio stays close to one afteracid or base is added.9.102A buffer is a solution whose pH changes very little when acid or base is added. A buffer can alsobe prepared from a weak base and its conjugate acid because a proton donor and a protonacceptor are available to react with any added acid or base.9.103Yes, a buffer can be prepared from equal amounts of NaCN and HCN. HCN is a weak acid andCN– is its conjugate base, so in equal amounts they form a buffer.9.104No, a buffer cannot be prepared from equal amounts of HNO3 and KNO3. A buffer is preparedfrom a weak acid (or weak base) and its conjugate base (or conjugate acid). HNO3 is a strong acidand completely dissociates in solution. KNO3 is a soluble salt and also completely dissociates insolutio
Chapter 9–1 Chapter 9 Acids and Bases Solutions to In-Chapter Problems 9.1 A Brønsted–Lowry acid must contain a hydrogen atom, but it may be neutral or contain a net positive or negative charge. Use Example 9.1 to help determine which of the compounds are Brønsted–Lowry acids. a. HI: contains a H atom, a Brønsted–Lowry acid c. H 2PO 4
2. Describe the common properties of acids and bases 3. Identify acids and bases using indicators, pH papers 4. Name some common lab acids and bases, acids at and bases at home 5. Describe reactions of acids with metals, bases and carbonates 6. Describe the application of acids, bases and p
_7. Which statement describes an alternate theory of acids and bases? (1) Acids and bases are both H acceptors. (2) Acids and bases are both H donors. (3) Acids are H acceptors, and bases are H donors. (4) Acids are H donors, and bases are H acceptors. _8. Which substance is the
Properties of Acids and Bases Return to the Table of contents Slide 5 / 208 What is an Acid? Acids release hydrogen ions into solutions Acids neutralize bases in a neutralization reaction. Acids corrode active metals. Acids turn blue litmus to red. Acids taste sour. Properties of Acids Slide 6 / 208 Properties
Lecture Notes for Chapter 16: Acids and Bases I. Acids and Bases a. There are several ways to define acids and bases. Perhaps the easiest way to start is to list some of the properties of acids and bases. b. The table below summarizes some properties that will be helpful
Acids, Bases, and pH The hydrohalic acids (HX (aq)), where X represents a halogen) include HF, HCl, HBr, and HI. Only HF is a weak acid. The rest are strong acids. What factors account for this difference? 8.1 Explaining the Properties of Acids and Bases 8.2 The Equilibrium of Weak Acids and Bases 8.3
properties of acids and bases. 5. Describe the colors that form in acidic and basic solutions with litmus paper and phenolphthalein. 6. Explain the difference between strong acids or bases and weak acids or bases. 7. Memorize the strong acids and bases. 8. Define the terms polyprotic and amphiprotic. 9. Perform calculations using the following .
Unit 12 Acids and Bases- Funsheets Part A: Name and write the formula for the following acids and bases. 1) Carbonic acid _ . Part E: Using your knowledge of acids and bases, answer the following questions. 1) Fill in the following Venn diagram about properties of acids and bases. You must fill in at least 4 facts in each.
Unit 11 Objectives: Acids & Bases Acid-Base Nomenclature Content Objectives: I can name ionic compounds containing acids, and bases, using (IUPAC) nomenclature rules. I can write the chemical formulas of acids and bases. Criteria for Success: I can identify an acid as a binary acid or an oxyacid. I can name common binary acids, oxyacids, and bases given their chemical formula.
