Acid Base Solutions - Answers

 

1.         a)  Buret is a long tube marked for volumes, usually in increments of 0.1 mL

            b)  Titrant is the solution in the buret.

            c)  Indicator is a weak acid or base that changes colour, depending on the pH of the solution.

            d)  End-point is the pH at which the indicator changes colour 

pH = -logKa(indicator)

e)  The stopcork is the valve at the bottom of a buret.

f)  A standard solution is a solution of known concentration.

 

2.         Given:  24.00 mL of 0.100 mol/L NaOH (aq) neutralizes 20.00 mL of HCl (aq)

            Required:  concentration of HCl (aq)

 

            Solution:                                   HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

                                    Mole ratio        1 mol          1 mol

           

            nNaOH = MNaOH X VNaOH  = 0.100 mol/L X 0.0240 L = 0.002400 mol

 

            nHCl = nNaOH = 0.002400 mol

 

            Therefore: MHCl = nHCl = 0.002400 mol = 0.120 mol

                                           VHCl      0.02000 L                   L

 

3.         Solution:                       HC2H3O2 (aq) + NaOH (aq) → H2O (l) + NaC2H3O2 (aq)

                a)         mole ratio           1 mol                1 mol           1 mol        1 mol

 

            nNaOH = 0.500 mol/L X 0.01340 L = 0.00670 mol

 

            MHC2H3O2 = 0.00670 mol = 0.670 mol

                                  0.0100 L                    L

 

            b)         0.670 mol HC2H3O2 = 0.670 mol X 1 mL X 60.0 g

                                  1 L                       1000 mL      1.0 g      1 mol

 

                        = 40.2 g HC2H3O2 = 4.02 g HC2H3O2

                                      1000 g vinegar         100 g vinegar

 

                        Therefore, vinegar is 4.02 % by weight acetic acid.

 

4.         Solution:                       HC3H5O3 + NaOH → H2O + NaC3H5O3

                                mole ratio         1 mol           1 mol        

 

            nNaOH = 0.160 mol/L X 0.01835 L = 0.002936 mol

           

            nHC3H5O2 = nNaOH = 2.94 X 10-3 mol

5.         Solution:                                   HC9H7O4 + KOH → KC9H7O4 + H2O

                                    Mole ratio        1 mol           1 mol

 

            nKOH = 0.0300 mol/L X 0.02940 L = 0.0008820 mol

 

            nHC9H7O4 in 0.250 g of pain reliever = 8.820 X 10-4 mol

 

Therefore, the mass of pure HC9H7O4 in 0.250 g of pain reliever =

8.820 X 10-4 mol X 180.0 g = 1.588 X 10-1 g

                                 1 mol

 

% of aspirin in pain reliever = 1.588 X 10-1 g X 100

                                                    0.250 g

                                              = 63.5 %

 

Self-ionization of Water and pH Answers

 

1.         Kw = [H+][OH-]                       in neutral solution [H-]=[OH-]

            2.4 X 10-14 = [H+]2

            [H+] = 1.55 X 10-7 = [OH-]

            pH = -log1.55 X 10-7 = 6.81

            pOH = -log1.55 X 10-7 = 6.81

 

            pH + pOH = 13.62                  pKw = -logKw = -log2.4 X 10-14 = 13.62

 

            Water at 37 C is neutral when pH = 6.81

 

2.         [H3O+] = 0.01 mol/L                Therefore, pH = -log0.010 = 2.0

           

3.         [H3O+] = 0.0050 mol/L            Therefore pH - -log0.0050 = 2.3

 

4.         6.0 g NaOH X 1 mol = 0.150 mol

                1.00 L          40.0 g     1.00 L

            [OH-] = 0.150 mol/L                pOH = -log0.150 = 0.82

           

            pH + pOH = 14.00      pH = 13.18 at 25 C

 

5.         0.837 g Ba(OH)2 = 8.37 g  X  1 mol  = 0.04886 mol

                 100 mL              1000 mL  171.3 g

 

            Therefore 0.04886 mol Ba(OH)2 / L

            But,      Ba(OH)2 → Ba2+ + 2 OH-

 

            Therefore 2 X 0.04886 mol OH/L        [OH-] = 9.77 X 10-2 mol/L

            pOH = -log9.77 X 10-2 = 1.01 pH = 12.99

 

6.         [H3O+] = 1.9 X 10-5 mol/L                   pH = -log1.9 X 10-5 = 4.7

 

7.         pH = -log1.4 X 10-5 = 4.9

 

 

 

Acid and Base Ionization Constants - Answers

 

1.         Ka = [H3O+][C6H5CO2-]

                        [C6H5CO2H]

 

2.         C6H5CO2- (aq) + H2O (l) ↔ C6H5CO2H (aq) + OH- (aq)

            Kb = [C6H5CO2H][OH-]

                        [C6H5CO2-]

 

3.         The lower the pKa, the stronger the acid.  The stronger the acid, the weaker its conjugate base and vice versa.  HF is a stronger acid than HCN, therefore the conjugate base of HF is weaker than the conjugate base of HCN.  Therefore CN is the stronger base.

 

4.                                 HF (aq) + H2O (l) ↔ H3O+ (aq) + F- (aq)                 Ka = 3.5 X 10-4

            Initial                0.15 mol/L                              0              0

                Equilibrium       0.15-x                                      x              x

 

                0.15_____   = 429                   Ka = [H3O+][F-]

            3.5 X 10-4         1                                 [HF]

 

            3.5 X 10-4 = __x2__

                                   0.15

            x = 7.2 X 10-3

                pH = -log7.2 X 10-3 = 2.1

 

5.                                 HC2H3O2 (aq) + H2O (l) ↔ H3O+ (aq) + C2H3O2- (aq)

            Initial                1.0                                           0                      0

            Equilibrium       1.0-x                                        x                      x

           

            Ka = 1.8 X 10-5

 

            Ka = [H3O+][C2H3O2-]

                        [HC2H3O2]

            1.8 X 10-5 = __x2__

                                    1.0

            x = 4.2 X 10-3

            pH = -log4.2 X 10-3 = 2.4

 

 

Hydrolysis of Ions - Answers

 

1.         Basic

2.         Basic.  H2C2O4 is a weak acid, therefore its conjugate base hydrolyzes producing a basic solution.

 

3.         a)  Acidic (NH4)2, HC2H3O3

            b)  Basic KF, KCN

            c)  Neutral NaI, CsNO3, KBr

 

                                    CN- (aq) + H2O (l) ↔ HCN (aq) + OH-       

            Initial (mol/L)    0.20                                         0          0

            Equilibrium       0.20-x                                      x          x

 

            Kb = Kw = 1.0 X 10-14 = 2.0 X 10-5

                      Ka     4.9 X 10-10

           

            Kb = [HCN-][OH-]

                        [CN-]

            2.0 X 10-5 = __x2_

                                    0.20

            x = 2.0 X 10-3 = [OH-]

            pOH = 2.7       pH = 11.3

 

 

                                    NO2- (aq) + H2O (l) ↔ HNO2 (aq) + OH- (aq)

 

            Initial (mol/L)    0.10                             0                      0

            Equilibrium       0.10-x                          x                      x

                       

            Kb = [HNO2][OH-]

                        [NO2-]

            2.2 X 10-11 = __x2__

                                      0.10

            x = 1.5 X 10-6 = [OH-]

            pOH = 5.8

            but pH + pOH = 14

            pH = 8.2         

 

4.                                 CO32- (aq) + H2O (l) ↔ HCO3- (aq) + OH- (aq)     

            Initial (mol/L)    x                                  0                      0

            Equilibrium       x-5 X 10-3                    5 X 10-3           5 X 10-3

           

            Kb = Kw = 1.0 X 10-14 = 2.1 X 10-4

                    Ka      4.7 X 10-11

            pH = 11.70   pOH = 2.30                    [OH-] = 5.0 X 10-3

            Kb = [HCO3-][OH-]

                        [CO32-]

            2.1 X 10-4 = (5.0 X 10-3)2

                                    x

            x = 0.12 mol/L

            but 1 mol of CO32- per 1 mol Na2CO3∙10H2O.  Therefore mass of Na2CO3∙10H2O

            = 0.12 mol X 286.0 g = 34 g

                   1 L           1 mol       1 L

 

6.         pH = 10.00      pOH = 4.00     [OH-] = 10-4.00 = 1.00 X 10-4 mol/L

 

                                    SO3 2- (aq) + H2O (l) ↔ HSO3- (aq) + OH- (aq)

            Initial                x                                  0                      0

            Equilibrium       x-1 X 10-4                    1 X 10-4           1 X 10-4

 

            Kb = Kw = 1.00 X 10-14 = 1.5 X 10-7

                     Ka      6.6 X 10-8

                Kb = [HSO3-][OH-]

                        [SO32-]

            1.5 X 10-7 = (1.00 X 10-4)2

                                    x

            x = 6.7 X 10-2

            6.7 X 10-2 mol/L of SO32-

 

            3.33 X 10-2 mol X 252.1 g = 8.4 g/0.5 L

                        0.5 L            1 mol

 

7.         a)                     HSO4- (aq) + H2O (l) ↔ H3O+ (aq) + SO42- (aq)

                        Ka = [H3O+][SO42-] = 1.0 X 10-2

                                    [HSO4-]

 

            b)                     HSO4- (aq) + H2O (l) ↔ H3O+ (aq) + SO42- (aq)

                            Initial    0.010                           0                      0

                        Eqm.    0.010-x                        x                      x

 

                        1.0 X 10-2 = ___x2__

                                              0.010-x 

                        1.0 X 10-4 1.0 X 10-2 x = x2

                        x2 + 1.0 X 10-2 x 1.0 X 10-4 = 0

                        x = 1.0 X 10-2 +/- √1.0 X 10-4 + 1.0 X 10-4

                                                            2

                          = -1.0 X 10-2 +/- 1.4 X 10-2

                                                2

                          = -1.2 X 10-2 or 0.2 X 10-2

                        x = 0.2 X 10-2

                        [H3O+] = 2.0 X 10-3 mol/L

            c)         Using simplification (0.010 x) = 0.010

                        1.0 X 10-2 = __x2__

                                              0.010

                        x = 1.0 X 10-2

                        [H3O+] = 1.0 X 10-2

 

                d)         % ionization = 2.0 X 10-3   X 100  = 20 %

                                                   0.010

 

            e)         % ionization = 1.0 X 10-2   X 100 = 100%

                                                   0.010

 

 



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