CHEM 1001 Problem Set #2b - Chemical Equilibria: Acids and Bases

16.2 The conjugate base has one less proton than the conjugate acid. (The conjugate acid-base pairs are highlighted in bold text.)
a) HIO3(aq) + H2O(l) equilibrium arrow IO3-(aq) + H3O+(aq)
b) C6H5COOH(aq) + H2O(l) equilibrium arrow C6H5COO-(aq) + H3O+(aq)
c) HPO42-(aq) + H2O(l) equilibrium arrow PO43-(aq) + H3O+(aq)
d) C2H5NH3+(aq) + H2O(l) equilibrium arrow C2H5NH2(aq) + H3O+(aq)

16.9 Kw = [H3O+][OH-] = 1.00 X 10-14
a) [H3O+] = 0.00165 M, [OH-] = Kw/[H3O+] = 6.06X10-12 M
b) [OH-] = 0.0087 M, [H3O+] = Kw/[OH-] = 1.1X10-12 M
c) [OH-] = 2(0.00213 M) = 0.00426 M, [H3O+] = Kw/[OH-] = 2.35X10-12 M
d) [H3O+] = 5.8X10-4 M, [OH-] = Kw/[H3O+] = 1.7X10-11 M

16.10 a) pH = -log(0.0045) = 2.35
b) pH = -log(6.14X10-4) = 3.212
c) pOH = -log(0.00683) = 2.166, pH = 14-pOH = 11.834
d) pOH = -log{2(4.8X10-3)} = 2.02, pH = 14-pOH = 11.98

16.24 [HC3H5O2]i = 0.275 mol/0.625 L = 0.440 M
HC3H5O2     + H2O      equilibrium arrow   C3H5O2-     +    H3O+
initial 0.440 M        -        -
equil. 0.440 - 0.00239
= 0.438 M		 0.00239 M 0.00239 M
Ka = [HC3H5O2][H3O+]/[C3H5O2-] = 1.30X10-5

16.25 CH2FCOOH  + H2O  equilibrium arrow CH2FCOO- + H3O+
initial    0.318 M           -    -
equil.    0.318 - x           x    x
[H3O+] = x = 10-1.56 = 2.75X10-2 M
Ka = x2/(0.318 - x) = (2.75X10-2)2/(0.318 - 2.75X10-2) = 2.6X10-3

16.27 HC7H5O2  + H2O      equilibrium arrow   C7H5O2-     +    H3O+
initial Ma        -        -
equil. Ma - x        x        x
[H3O+] = x = 10-2.85 = 1.41X10-3 M
Ka = x2/(Ma - x)
Ma = (x2/Ka) + x = {(1.41X10-3)2/(6.3X10-5)} + 1.41X10-3 = 3.31X10-2 M
Also,
Ma = moles of benzoic acid/volume of solution
or
Ma = {(mass of benzoic acid)/(MW of benzoic acid)}/volume of solution
therefore
mass of benzoic acid = (Ma)(MW of benzoic acid)(volume of solution)
= (3.31X10-2 M)(122.123 g mol-1)(350.0X10-3 L)
= 1.4 g

16.56 See Table 16.3 on page 710 of textbook for Ka and Kb values
a) Ka(C5H5NH+) = Kw/Kb(C5H5N) = 6.7x10-6
b) Kb(CHO2-) = Kw/Ka(HCHO2) = 5.6x10-11
c) Kb(C6H5O-) = Kw/Ka(HOC6H5) = 1.0x10-4

16.57 a) neutral (no hydrolysis)
b) alkaline (F- hydrolyzes) : F- + H2O equilibrium arrow HF + OH-
c) neutral (no hydrolysis)
d) alkaline (OCl- hydrolyzes) : OCl- + H2O equilibrium arrow HOCl + OH-

16.61 C6H7O2-  + H2O  equilibrium arrow HC6H7O2  + OH-
initial  0.37 M         -    -
equil.  0.37 - x         x    x
The equilibrium constant associated with this equilibrium is Kb
Kb = Kw/Ka = 1.00X10-14/(10-4.77) = 5.9X10-10
Also,
Kb = x2/(0.37 - x)
5.9X10-10 = x2/(0.37 - x)
assume x << 0.37
5.9X10-10 = x2/0.37, x = 1.5X10-5 M = [OH-] (assumption is valid)
pOH = -log(1.5X10-5) = 4.83, pH = 14 - pOH = 9.17

16.112 moles of OH- = (15.40X10-3 L)(0.394 M) = 6.07X10-3 mol
moles of H3O+ = (24.80X10-3 L)(0.248 M) = 6.15X10-3 mol
moles of H3O+ > moles of OH-
excess moles of H3O+ = (6.15 - 6.07)X10-3 mol = 0.08 X10-3 mol
[H3O+] = 0.08 X10-3 mol/{15.40 + 24.80)X10-3 L = 2X10-3 M
pH = 2.7

17.2 NH3 + H+ equilibrium arrow NH4+, Kb = 1.8X10-5, pKb = 4.74
a) since [NH3] and [NH4+] are similar, this is a buffer and it is valid to use the Henderson-Hasselbalch equation
pOH = pKb + log{[NH4+]/[NH3]} = 4.74 + log{0.102/0.164} = 4.53
[OH-] = 10-4.53 = 2.9X10-5 M
b) [NH4+] = 0.102 M
c) [Cl-] = 0.102 M
d) [H3O+] = Kw/[OH-] = 3.4X10-10 M

17.6 a) [OH-] = 2 X 0.0062 M = 0.012 M
b) pOH = pKb + log{[NH4+]/[NH3]}
= -log(1.8x10-5) + log{(2 X 0.315)/(0.486)}
= 4.857
[OH-] = 10-4.857 = 1.4X10-5 M
c) NH4+        +    OH-      equilibrium arrow     NH3    + H2O
initial 0.264 M 0.196 M        -
equil. 0.264 - 0.196
= 0.068 M		       -   0.196 M
pOH = pKb + log{[NH4+]/[NH3]}
= -log(1.8x10-5) + log{0.068/0.196}
= 4.285
[OH-] = 10-4.285 = 5.2X10-5 M

17.7 pH = pKa + log{[CHO2-]/[HCHO2]}
4.06 = -log(1.8X10-4) + log[CHO2-] - log(0.366)
log[CHO2-] = -0.1212
[CHO2-] = 0.76 M

17.8 pOH = pKb + log{[NH4+]/[NH3]} = 14 - pH
-log(1.8x10-5) + log(0.732) - log[NH3] = 14 - 9.12
log[NH3] = -0.2708
[NH3] = 0.54 M

17.9 a) pH = pKa + log{[C7H5O2-]/[HC7H5O2]}
= -log(6.3x10-5) + log{0.033/0.012}
= 4.64
b) pOH = pKb + log{[NH4+]/[NH3]}
= -log(1.8x10-5) + log{0.153/0.408}
= 4.32
pH = 14 - pOH = 9.68

17.11 a) 0.100 M NaCl is not a buffer: Both members of a conjugate acid-base pair are not present.
b) 0.100 M NaCl - 0.100 M NH4Cl is not a buffer: Although the weak acid NH4+ is present, its conjugate base, NH3 is not.
c) 0.100 M CH3NH2 - 0.150 M CH3NH3+Cl- is a buffer: Both members of a conjugate acid-base pair are present.
d) 0.100 M HCl - 0.050 M NaNO2 is not a buffer: All of the NO2- will be consumed by the HCl giving a mixture of 0.050 M HNO2 (a weak acid) and 0.050 M HCl (a strong acid).
e) 0.100 M HCl - 0.200 M NaC2H3O2 is a buffer: All of the HCl will react with the C2H3O2- to form HC2H3O2 providing a solution of 0.100 M C2H3O2- (a weak base) and 0.100 M HC2H3O2 (its conjugate acid).

17.16 moles of benzoic acid = 2.00 g/122.123 g mol-1 = 1.638X10-2 mol
moles of benzoate = 2.00 g/144.105 g mol-1 = 1.388X10-2 mol
a) pH = pKa + log {moles of benzoate/moles of benzoic acid}
= -log(6.3X10-5) + log{1.388X10-2/1.638X10-2}
= 4.13
b) to lower the pH to 4.00 (i.e. make the solution more acidic), benzoic acid must be added
4.00 = -log(6.3X10-5) + log{1.388X10-2/moles of benzoic acid}
log(moles of benzoic acid) = -1.657
moles of benzoic acid = 10-1.657 = 2.203X10-2 mol
Now,
moles of benzoic acid = initial moles + added moles
2.203X10-2 mol = 1.638X10-2 mol + added moles
therefore,
added moles = 5.65X10-3 mol
mass of benzoic acid to add = (5.65X10-3 mol)(122.123 g mol-1) = 0.69 g

17.98 a) HCHO2 + OH- arrow CHO2- + H2O
CHO2- + H3O+ arrow HCHO2 + H2O
b) C6H5NH3+ + OH- arrow C6H5NH2 + H2O
C6H5NH2 + H3O+ arrow C6H5NH3+ + H2O
c) H2PO4- + OH- arrow HPO42- + H2O
HPO42- + H3O+ arrow H2PO4- + H2O