Chem2000
Spring 2004
Assignment 4
Friday Feb 6,
2004
1) A sample of 3.7400 g of propionic acid (CH3CH2COOH) was dissolved water to make
100.00 ml of solution. This solution was titrated with 0.525 M NaOH. (Ka = 1.3*10-5)
a) Determine the equivalence volume.
Let CH3CH2COOH = HA
Fw(HA) = 74.060 g/mol nHA = 3.74 g/74.060 g/mol = 0.05050 mol = 50.50 mmol.
[HA] = 50.50 mmol/100.00 ml = 0.5050 M
nOH = [OH-]*V = 50.50 mmol = (0.525 M)*Ve +. Ve = 96.19 ml.
b) Determine the pH of the solution before any tritrant is added.
Weak Acid Problem
I
∆
E
HA
0.05050
-x
0.5050 –x
+ H2O U
H3O+
0
+x
+x
+
A0
+x
+x
Ka = 1.3 *10-5 = x2/(0.05050 –x) ª x2/0.05050 for x<< 0.05050
6.56*10 –7 = x2
x = 8.1*10-4 M = [H3O+]
pH = 3.10
c) Determine the pH of this solution when 50.00 ml of NaOH solution is added.
Buffer Problem
nHA= nHAo – nOH- = (50.50 mmol) – (50.00 ml)*(0.525 M) = 24.45 mmol.
[HA] = (24.45 mmol)/(100.00 + 50.00 ml) = 0.1630 M
nA-= nOH- = (50.00 ml)*(0.525 M) = 26.25 mmol
[A-] = (26.25 mmol)/(150.00 ml) = 0.1750 M
Henderson-Hasselbach Equation:
pH = pKa – log[HA]/[A-] =-log(1.3*10-5) – log(0.1630/0.1750) = 4.89 – 0.031 = 4.92
d) Determine the pH of this solution at the equivalence point.
Weak Base Problem
At the equivalence point there is only A- present nA- = nOH- = nHAo = 50.50 mmol
[A-] = (50.50 mmol)/(100.00 ml + 96.19 ml) = 0.2574 M
I
∆
E
A+
0.2574
-x
0.2574 –x
H2O
U
HA
0
+x
+x
+
OH0
+x
+x
Kb = Kw/Ka = 1.01*10-14/1.3*10-5 = 7.8*10-10 = x2/(0.2574 – x) ª x2/0.2574
for x<<0.2574
x = 1.4 * 10-5 M=[OH-] pOH = 4.85
pH = pKw – pOH = 14.00 –4.85 = 9.15
e) Determine the pH of this solution once 100 ml of solution is added.
Determine excess OH- and use it to compute pH.
nOH- (excess) = nOH- - nHAo = (100.00 ml)*(0.525 M) – 50.50 mmol = 2.00 mmol
[OH-] = (2.00 mmol)/(100.00 ml + 100.00 ml) = 0.0100 M
pOH = 2.00
pH = 12.00
2) A sample of 1.1000 g of methylamine (CH3NH2) was dissolved water to make 100.00 ml of
solution. This solution was titrated with 0.4210 M HCl. (Kb = 4.4*10-4)
a) Determine the equivalence volume.
Let CH3NH2 = B
Fw(B) = 31.058 g/mol nB = 1.1000 g/31.058 g/mol = 0.035418 mol = 35.418 mmol.
[B] = 35.418 mmol/100.00 ml = 0.35418 M
nH3O = [H3O+]*V = 35.418 mmol = (0.4210 M)*Ve Ve = 84.13 ml.
b) Determine the pH of the solution before any tritrant is added.
Weak Base Problem
I
∆
E
B
0.35418
-x
0.35418 –x
+ H2O U
OH0
+x
+x
+
HB+
0
+x
+x
Ka = 4.4 *10-4 = x2/(0.35418 –x) ª x2/0.35418 for x<< 0.35418
1.6*10 –4 = x2
pOH = 1.90
x=
1.25*10-2 M = [OH-]
pH= pKw – pOH = 14.00 – 1.90 = 12.10
c) Determine the pH of this solution when 50 ml of HCl solution is added.
Buffer Problem
nB= nBo – nH3O = (35.418 mmol) – (50.00 ml)*(0.4210 M) = 14.368 mmol.
[B] = (14.368 mmol)/(100.00 + 50.00 ml) = 0.0958 M
nHB+ = nH3O+ = (50.00 ml)*(0.4210 M) = 21.05 mmol
[HB+] = (26.25 mmol)/(150.00 ml) = 0.1403 M
Ka = Kw/Kb = 1.01 * 10 –14/4.4*10-4 = 2.3*10-11
pKa = 10.64
Henderson-Hasselbach Equation:
pH = pKa – log[HB+]/[B] =10.64 – log(0.1430/0.0958) = 10.47
d) Determine the pH of this solution at the equivalence point.
Weak Acid Problem
At the equivalence point there is only HB+ present nHB+ = nHAo = 35.418 mmol
[HB+] = (35.418 mmol)/(100.00 ml + 84.13 ml) = 0.1924 M
I
∆
E
HB+ +
0.1924
-x
0.1924 –x
H2O
U
H3O+ +
0
+x
+x
B
0
+x
+x
Ka = Kw/Kb = 1.01*10-14/4.4*10-4 = 2.3*10-11 = x2/(0.1924 – x) ª x2/0.1924
for x<<0.1924
x = 2.1 * 10-6 M = [H3O+] pH = 5.68
e) Determine the pH of this solution once 100 ml of solution is added
Determine excess H3O+ and use it to compute pH.
nH3O+ (excess) = nH3O+ - nBo = (100.00 ml)*(0.4210 M) – 35.418 mmol = 6.68 mmol
[H3O+] = (6.68 mmol)/(100.00 ml + 100.00 ml) = 0.0334 M
pH = 1.476
3) Calculate the equilibrium concentration of H2CO3, HCO3-, CO32-, and H3O+ in a 100 ml
solution which was initially prepared as 0.045 mol/L in H2CO3. (Ka1 = 4.3* 10-7, Ka2 =
4.8*10-11). If the pH is adjusted to 10 determine the concentrations of H2CO3, HCO3-, CO32-.
Since Ka1 >>Ka2 we can treat them as two independent equilibria where the first
equilibrium establishes the initial conditions for the second.
I
∆
E
H2CO3 (aq)
0.045
−x
0.045 –x
+
H2O(l) U
HCO3-(aq) +
0
+x
+x
H3O+(aq)
0
+x
+x
Ka1 = 4.3*10-7 = x2/(0.045 –x) ª x2/(0.045 ) for x << 0.045
x = 1.4*10-4 = [H3O+] =[ HCO3-]
I
∆
E
HCO3- (aq)
0.00014
−x
0.00014 – x
+
H2O(l) U
CO32-(aq) +
0
+x
+x
Ka2 = 4.8*10-11 = x(0.00014 + x)/(0.00014 –x) ª x
H3O+(aq)
0.00014
+x
+x
for x << 0.00014
[CO32-] = 4.8*10-11 M
[H3O+] = 0.00014 + 4.8*10-11ª 0.00014 M (pH = 3.85 <pKa1 => H2CO3 dominates)
[HCO3-] = 0.00014 - 4.8*10-11ª 0.00014 M
[H2CO3] = 0.045 – 0.00014 ª 0.045 M
pH has been changed to 10.0.
pH = 10 => [H3O+] = 1.0 * 10 –10
Ka1 = [HCO3-][H3O+]/[H2CO3] => [H3O+]/Ka1 = [H2CO3]/ [HCO3-]
= 1.0 * 10 –10/4.3*10-7
= 2.33*10-5
Ka2= [CO32-][H3O+]/[HCO3-] => Ka2/[H3O+] = [CO32-]/[HCO3-]
= 4.8*10-11/1.0*10-10
= 0.48
fraction of [CO32-] = [CO32-]/{[H2CO3] + [HCO3-] +[CO32-]}
= {[CO32-]/[HCO3-]}/{[H2CO3]/ [HCO3-] + [HCO3-] /[HCO3-]+[CO32-]/[HCO3-]}
= 0.48 /(2.33*10-5 + 1.00 +0.48) = 0.48/1.48 = 0.324
fraction of [HCO3-]= [HCO3-]/{[H2CO3] + [HCO3-] +[CO32-]}
= {[HCO3-]/[HCO3-]}/{[H2CO3]/ [HCO3-] + [HCO3-] /[HCO3-]+[CO32-]/[HCO3-]}
= 1.00/1.48 = 0.676
fraction of [H2CO3]= [HCO3-]/{[H2CO3] + [HCO3-] +[CO32-]}
= {[H2CO3]/[HCO3-]}/{[H2CO3]/ [HCO3-] + [HCO3-] /[HCO3-]+[CO32-]/[HCO3-]}
= 2.33*10-5/1.48 = 1.57*10 -5
[H2CO3]o = [H2CO3] + [HCO3-] +[CO32-] = 0.045 M
[CO32-] = 0.324*0.045 = 0.0146 M
[HCO3-] = 0.676 *0.045 = 0.0304 M
[H2CO3] = 1.57*10-5 *0.045 = 7.07*10-7 M
(pH = 10 ª pKa2 =10.31 => [HCO3-] ª [CO32-] )
4) For each of the following reactions identify the Lewis acid and the Lewis base.
Lewis Acid
Lewis Base
a)
SiF4 + 2 F U SiF6
b)
4 NH3 + Zn2+ U Zn(NH3)42+
c)
2Cl- + HgCl2 U HgCl42-
d)
CO2 + H2O U H2CO3
O=C=O ↔ -O-C+=O this resonance form can accept a l.p of electrons from H2O, hence
CO2 can be regarded as a Lewis acid and H2O as a Lewis base.
5) Classify each of the following as a Lewis acid or Lewis base:
Lewis Acid
Lewis Base
a)
b)
c)
d)
e)
f)
g)
h)
CNH+
H2O
Fe3+
OHCO2
P(CH3)3
B(CH3)3