Chapter4Worksheet

1、Chang, 7thEdition, Chapter 15, Worksheet #1S. B. Piepho, Fall 2002Acid-Base CalculationsThe Ion-Product Constant for Water, KwWater undergoes ionization to a small extent:+H20(l)H (aq) + OH (aq)The equilibrium constant for the reaction is theion-product constant for water Kw:KwH OH1.01014(1)+This is

2、 a key equation in acid-base chemistry. Note that the product of H and OH is aoconstant at a given temperature (Eq(1) value is for 25C). Thus as the hydrogen ionconcentration of a solution increases, the hydroxide ion concentration decreases (andvice versa).+The pH scale is widely used to report the

3、 molar concentration of hydrogen ion H(aq) inaqueous solution. The pH of a solution is defined aspHlog10H(2)Similarly, pOH and pKware defined aspOHlog10OH (3)pKwlog10(Kw)14.00(4)If you take the log10of both sides of Eq(1), multiply the resulting equation by (-1), anduse the definitions of pH, pOH an

4、d pKwabove, the result is the very useful equationpH + pOH = pKw= 14.00+Equations (2) and (3) above may be solved for H and OH respectively to giveH 10pHOH10pOH(Here we use the well known rule that iflog10yx, theny10 x.) In practice, the pH scaleis only used when H+(aq) is less than 1.0 M.Acidic, ba

5、sic, and neutral solutions can be distinguished as shown below:Type of SolutionpHH+Acidic1.0 10Neutral= 7.00=1.0 10Basic 7.001.0 10777Color of litmuspinkin betweenbluePage 1 of 6Chang, 7thEdition, Chapter 15, Worksheet #1S. B. Piepho, Fall 2002pH and H+ Calculations for Strong Acids and BasesBy defi

6、nition, strong acids and bases are 100% ionized in water solution. Ionization of a+strong acid gives rise to H ions, and ionization of a strong base produces OHions. Theequilibrium constant for a strong acid or strong base is undefined, since the reactionthe ionization is complete. There is no equil

7、ibrium!+In nearly all cases of practical interest the H for a strong acid (or the OH for a strongbase) is determined completely by the stoichiometry of the reaction. Once the OHor pOH+is known for a base, the H or the pH of the base may be calculated using Eq(1)and/or Eq(5).Exercises1.Complete the f

8、ollowing table:+Acidic,pHHpOHbasic, orOH (a)5.4 x 104neutral?7.8 x 10-10(b)(c)10.75(d)5.00Answers:(a)1111pH = 3.27; pOH = 10.73; OH = 1.85 x 10= 1.9 x 10, acidic (since pH 7).(b)pH = 4.89, H+ = 1.3 x 105, pOH = 9.11, acidic (since pH 7). = 5.6 x 10(d)pH = 9.00, H+ = 1.0 x 109, OH = 1.0 x 105, basic

9、(since pH 7).Calculate the pH of a 0.0430 M HNO3solution.Answer:Since HNO3is a strong acid, the nitric acid solution will be 100% ionized. Thus H+=NO3 = 0.0430 M. ThepH = 1.37 (use Eq(2).3. Calculate the pH of a 0.020 M Ba(OH)2(aq) solution.Page 2 of 6Chang, 7thEdition, Chapter 15, Worksheet #1S. B.

10、 Piepho, Fall 2002Answer:Since Ba(OH)2is a strong base it is 100% ionized. Note that ionization gives 2 OHions for each mole ofBa(OH)2. Thus OH = 2 x 0.020 M = 0.040 M. Eq(3) gives pOH = 1.40. Then using Eq(5), pH = 12.60.pH and H+ Calculations for Weak Acids and BasesWeak acids and bases are usuall

11、yess than 5% ionized. The equilibrium constant for aweak acid equilibrium is the acid ionization constant Ka, and for a weak base equilibriumis the base ionization constantKb.A typical monoprotic weak acid equilibrium can be written in two forms, the second ofwhich emphasizes the Br?nsted acid-base

12、nature of the reaction:HA+H(aq) + A (aq)HA + H2O+(8)H3O(aq) + A (aq)+In Eq(9) the Br?nsted acid HA donates a proton H to the Br?nsted base H2O to form H3Oand the conjugate base A. The acid ionization constant (using the second form) isKaH3O A(9)HAA typical weak base equilibrium is+(10)B + H2OBH (aq)

13、 + OH (aq)+In Eq(10) the Br?nsted base B accepts a proton Hfrom the Br?nsted base H2O to form the+conjugate acid BHand OH . The base ionization constant isBH OH(11)KbBExercises4. Calculate (a) the pH and (b) the percent ionization of a 0.250 M HC2H3O2solution.Ka(HC2H3O2) = 1.8 x 10-5. (The formula f

14、or acetic acid may also be written asCH3COOH.) HINT: Begin by filling out the equilibrium table below.Balanced EquationHC2H3O2H+C2H3O2Initial Concentration (M)Change (M)Equilibrium Concentration (M)Page 3 of 6Chang, 7thEdition, Chapter 15, Worksheet #1S. B. Piepho, Fall 2002Answer:H+Balanced Equatio

15、nHC2H3O2+C2H3O2Initial Concentration (M)0.25000Change (M)- xxxEquilibrium Concentration (M)0.250 - xxx(a)KaH C2H3O21.8105x2x2. This approximation is OK if the %HC2H3O20.250- x0.250 x2= 4.5 x 10-6; x = 2.12 x 10-3= H+. pH =ionization is 5%; it is in this case - see answer to (b) below. Thus2.67.(b)%

16、ionizat ion =x100%2.12 1030.85%.100%0.2500.250Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb= 4.4 x 104.HINT: Methylamine is a weak base. First write the equation for the reaction followingthe pattern of Eq(10). Then fill out the equilibrium table below.Balanced EquationCH3NH2CH3NH

17、3+OHInitial Concentration (M)Change (M)Equilibrium Concentration (M)Answer:Since CH3NH2is a weak base, the balanced equation for the reaction is CH+3NH2+ H2OCH3NH3+OH .BalancedEquationCH3NH2CH3NH3+InitialConcentration (M)0.6000Change(M)- xxEquilibrium Concentration (M)0.600 - xxKbBH OHCH3NH3OH x2x24

18、.4 104. Thus x = 1.62 x 10-2=BCH3NH20.600 x0.600OH, and pOH = 1.79. It follows from Eq(5) that pH = 12.21. NOTE: The approximation used is OKsince the % ionization is 2.7% (i.e., less than 5 %).6. The pH of a 0.10 M solution of a weak base is 9.67. What is theKbof the base?Page 4 of 6Chang, 7thEditi

19、on, Chapter 15, Worksheet #1S. B. Piepho, Fall 2002Answer:The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is given below.Balanced EquationBBH+OHInitial Concentration (M)0.1000Change (M)- xxxEquilibrium Concentration (M)0.10 - xxxAt equilibrium, OH+=BH = x.

20、Use the pH to calculate the OH at equilibrium (which is the value ofx).Here pOH = 14.00 pH = 14.009.67 = 4.33. ThusOH 10pOH104.334.68 105x.KbBHOHx2x2(4.68105)22.2108. The approximationis OK sinceB0.10 x0.100.10the % ionization is well under 5%.Relationship between Kafor a Weak Acid and Kbfor its Con

21、jugate BaseThe relationship betweenKafor a weak acid HA and Kbfor its conjugate base A is-14(12)Ka(HA)Kb(A ) = Kw= 1.0 x 10If we define pKa= - log10(Ka) and pKb= - log10(Kb), the logarithmic form of Eq(12) isThe stronger the acid, the larger theKaand the smaller the pKa. Likewise the stronger thebas

22、e, the larger theKband the smaller the pKb. Eqs(12) and (3) show that as theKaincreases(and the pKadecreases), theKbdecreases (and the Kpbincreases). These equations givequantitative support to the statement the stronger“the acid, the weaker the conjugate base.”The justification for Eq(12) follows f

23、rom the equations below. Recall that if Eq(1) + Eq(2)= Eq(3), then K1K2= K3.Eq(1), Weak Acid:HA + H2OEq(2), Conjugate Base:A (aq) + H2O+H3O (aq) + A (aq)HA( aq) + OH (aq)H3O A Ka(HA)HAHAOHKb(A )A _Eq(3) = Eq(1) + Eq(2)+KwH3O OH 2 H20(l)H3O (aq) + OH (aq)Relationship between Kbfor a Weak Base and Kafor its Conjugate AcidAnalogous equations to Eqs(12) and (13) above can be written the relationshipbetweenK for a weak base B andKafor its c