Solutions for Acid-Base Titrations Exercises
1.
If the acid or base is too weak or too dilute, there is very little change in pH at the
equivalence point. This makes it difficult to have a satisfactory end point.
2.
.
C6H5CH2NH2(aq) + H3O+(aq)
3.
K
º C6H5CH2NH3+(aq) + H2O(l)
The equation for the titration of a weak base, B, with a strong acid is
B(aq) + H3O+(l)
6 HB+(aq) + H2O(l)
This produces a solution of the conjugate acid, HB+, at the equivalence point so
the solution is acidic (pH<7).
4.
The titration reaction is HCO2H(aq) + OH-(l) 6 HCO2-(aq) + H2O(l). Since the
analyte and titrant concentrations are equal, it will take 50.0 mL of base to reach
the equivalence point. Note that for formic acid Ka = 1.80 x 10-4 (pKa = 3.74).
20.0 mL base
init mol HCO2H = (0.050 mol HCO2H/L)(0.0500 L) = 0.0025 mol
mol OH- added = (0.0200 L)(0.050 mol OH-/L) = 0.0010 mol
HCO2H(aq) + OH-(l)
6 HCO2-(aq) + H2O(l)
init mol
0.0025
0.0010
0
mol after rxn
0.0015
0
0.0010
A buffer is formed with:
[HCO2H] = (0.0015 mol)/(0.0700 L) = 0.0214 M
[HCO2-] = (0.0010 mol)/(0.0700 L) = 0.0143 M
Use Henderson-Hasselbalch,
Solutions for Acid-Base Titrations Exercises
50.0 mL base
At the equivalence point all of the formic acid has been converted to its conjugate
base so you have a solution of a weak base.
[HCO2-] = (0.0025 mol)/(0.100 L) = 0.0250 M
HCO2-(aq) + H2O(l)
init
0.0250 M
equil 0.0250 - x ~ 0.0250
º HCO2H(aq) + OH-(aq)
0
~10-7 M
x
~10-7 + x ~ x
where x is the increase in [HCO2H] and is assumed to be << 0.025 M and
>> 10-7 M.
pOH = -log (1.18 x 10-6) = 5.93
Y pH = 8.07
60.0 mL base
At this point 10.0 mL extra base has beeen added and you have a solution of a
weak base and strong base. The OH- contribution from the weak base is not
significant so the pH depends on the excess OH- present.
excess mol OH- added = (0.0100 L)(0.050 mol OH-/L) = 0.00050 mol
[OH-] = (0.00050 mol)/(0.110 L) = 0.00455 M
pOH = -log (0.00455) = 2.34
5.
Y pH = 11.66
This occurs when the titrant volume is one-half that needed to reach the
equivalence point. Here, half the base (B) has been converted to its conjugate
acid (HB+) and [B] = [HB+] and pH = pKa. At this point a small addition of acid
will have the least effect on the [B]/[HB+] ratio and hence on the pH.
Solutions for Acid-Base Titrations Exercises
6.
When the titrant volume is one-half that needed to reach the equivalence point.
half the weak acid (HA) has been converted to its conjugate base (A-), buffer is
formed and [HA] = [A-]. The Henderson-Hasselbalch equation looks like this
At this point the principal electrolyte in solution is NaA. Half the starting moles of
A- are now in a volume 50% larger so [NaA] = 0.0333 M. Thus, μ = 0.0333 M.
γHA ~ 1 and
Substituting and solving for pKa gives
7.
When NaCN is titrated with HClO4, the following reaction occurs.
CN-(aq) + H3O+(aq)
6 HCN(aq) + H2O(l)
init mol CN- = (0.100 mol CN-/L)(0.0500 L) = 0.00500 mol
a)
4.20 mL of 0.438 M HClO4 (mol H+ added = (0.00420 L)(0.438 mol H+/L) =
0.00184 mol)
CN-(aq) + H3O+(aq)
init mol
0.00500
0.00184
mol after rxn
0.00316
0
A buffer is formed with:
6 HCN(aq) + H2O(l)
0
0.00184
[HCN] = (0.00184 mol)/(0.05420 L) = 0.0339 M
[CN-] = (0.00316 mol)/(0.05420 L) = 0.0583 M
Solutions for Acid-Base Titrations Exercises
HCN(aq) + H2O(l)
init
0.0339 M
equil ~0.0339 M
º H3O+(aq) + CN-(aq)
~10-7 M
0.0583 M
x
~0.0583 M
b) 11.82 mL of 0.438 M HClO4 (mol H+ added = 0.00518 -
c)
at equivalence point, 11.42 mL of HClO4 have been added and there is an
aqueous solution of HCN and NaClO4
[HCN] = (0.00500 mol)/(0.06142 L) = 0.0814 M
HCN(aq) + H2O(l)
init
more than needed
to reach
equivalence point;
assume pH is
determined by the
excess strong acid)
0.0814 M
equil 0.0814 - x
º H3O+(aq) + CN-(aq)
~10-7 M
0
~x
x
Y x = 7.10 x 10-6 M = [H3O+] Y pH = 5.15
Solutions for Acid-Base Titrations Exercises
8.
The ratio [HIn]/[In-] changes from 10:1 when pH = pKHIn - 1 to 1:10 when pH =
pKHIn + 1. This change is generally sufficient to cause a color change.
9.
Cresol purple at:
pH = 1.0 - red
pH = 2.0 - orange (mix of red & yellow)
pH = 3.0 - yellow
10.
No. The titration produces a solution of the conjugate base at the equivalence
point. This solution will have a pH > 7 which is well outside the color transition
range for bromocresol green.
11.
a)
The titration reaction is F- + H3O+ 6 HF + H2O. Since the
concentration of HClO4 is twice that of the NaF, you need one-half the
volume (V) of the analyte solution. Thus, the formal concentration of HF
at the equivalence point is given by:
At the equivalence point,
HF(aq) + H2O(l)
init
0.0200 M
equil 0.0200 - x
º H3O+(aq) + F-(aq)
~10-7 M
0
~x
x
Y x = 3.688 x 10-3 M (x << 0.02 assumption is not valid! - use successive
approximations)
Solutions for Acid-Base Titrations Exercises
x2 = 3.331 x 10-3
x3 = 3.367 x 10-3
x4 = 3.363 x 10-3 M = [H3O+] Y pH = 2.47
b)
The equivalence point pH is quite acidic so there will not be a large
change in pH around the equivalence point. Note that the pH of the buffer
plateau before the equivalence point is ~ 3.2 and the pH of the titrant is
~1.2. A sharp change in indicator color will not be seen in this titration, but
you could use a color match for the end point.
12.
Sodium carbonate (Na2CO3) is commonly used to standardize HCl solutions.
Potassium hydrogen phthalate (KHC8H4O4) is commonly used to standardize
NaOH solutions (see Prob. 13 below).
13.
Potassium hydrogen phthalate (KHC8H4O4) 204.2 g/mol
HC8H4O4- + OH-
6 H2O + C8H4O42-
mol OH- = mol KHC8H4O4 = (0.05 mol/L)(0.030 L) = 0.0015 mol
mass KHC8H4O4 = (0.0015 mol)(204.2 g/mol) = 0.31 g