A 33.0 mL sample of a 0.406 M aqueous hypochlorous acid solution is titrated with a 0.387 M aqueous barium
hydroxide solution. What is the pH after 6.54 mL of base have been added?
pH =
5.737
7.24
Feedback:
Please note that while the results of the intermediate calculations below are displayed rounded to 3 significant digits,
the actual calculations are done internally without rounding.
When the strong base, Ba(OH)2, is added, the weak acid reacts to neutralize the OH-.The balanced equation is:
2 HClO + Ba(OH)2
Ba(ClO)2 + 2 H2O
The initial number of moles of the weak acid HClO is given by 0.406 M
Similarly, the number of moles of Ba(OH)2 added is given by 0.387 M
3.30E-2 L = 1.34E-2 mol
6.54E-3 L = 2.53E-3 mol
Note that the amount of base added is not enough to neutralize all of the acid, so the final solution will contain both
the remaining weak acid and its conjugate base. This is a buffer solution. In order to calculate the pH, calculate the
moles of ClO- formed and the moles of HClO remaining.
The number of moles of HClO used is
2 mol HClO
2.53E-3 mol Ba(OH)2
1 mol Ba(OH)2
= 5.06E-3 mol HClO used
The amount of HClO that remains is 1.34E-2 mol - 5.06E-3 mol = 8.34E-3 mol HClO remaining.
The number of moles of ClO- formed is
1 mol Ba(ClO)2
2.53E-3 mol Ba(OH)2
1 mol Ba(OH)2
2 mol ClO-
= 5.06E-3 mol ClO- formed
1 mol Ba(ClO)2
The final solution thus contains 8.34E-3 mol HClO and 5.06E-3 mol ClO-. The pH of this buffer solution can
conveniently be calculated using the Henderson-Hasselbalch equation:
[ClO-]
pH = pKa + log
[HClO]
The tabulated value of Ka for HClO is 3.5E-8.
pKa = - log(3.5E-8) = 7.46
Since the volumes cancel out in the ratio of molarities, substitute moles for M. Then:
5.06E-3
pH = 7.46 + log
= 7.24
8.34E-3
What is the pH at the equivalence point in the titration of a 25.0 mL sample of a0.332 M
aqueous hypochlorous acid solution with a 0.371 M aqueouspotassium
hydroxide solution?
pH =
7.46
10.35
Feedback:
Please note that while the results of the intermediate calculations below are displayed
rounded to 3 significant digits, the actual calculations are done internally without rounding.
When the strong base, KOH, is added, it reacts to neutralize the weak acid. The balanced
equation is:
HClO + KOH
KClO + H2O
At the equivalence point, the number of moles of OH- added is equal to the number of
moles of weak acid initially present and all of the HClO has been converted to ClO-.
HClO + OH-
ClO- + H2O
The initial number of moles of the weak acid HClO is given by 0.332M
= 8.30E-3 mol
2.50E-2 L
The number of moles of KOH required to neutralize the acid is
1 mol KOH
= 8.30E-3 mol KOH
8.30E-3 mol HClO
1 mol HClO
The volume of KOH required is thus 8.30E-3 mol / 0.371 M = 2.24E-2 L
... and the total volume at the equivalence point is 2.50E-2 L + 2.24E-2 L = 4.74E-2 L
The number of moles of ClO- that is formed is
1 mol ClO8.30E-3 mol HClO
1 mol HClO
= 8.30E-3 mol ClO-
... and the concentration of ClO- at the equivalence point is 8.30E-3 mol ClO- /4.74E-2 L
= 0.175 M
The tabulated value of Ka for HClO is 3.50E-8.
The value of Kb for the weak base ClO- can be calculated from the tabulated value of Ka for
its conjugate acid HClO.
1.0e-14
= 2.86E-7
Kb =
3.50E-8
Solve for the [OH-] in a 0.175 M solution of ClO- using the ICE method:
ClO- + H2O
OH- + HClO
Initial
0.175
0
0
Change
-x
+x
+x
Equilibrium
0.175 - x
x
x
Kb = 2.86E-7 = [OH ][HClO]
=
x2
[ClO-]
0.175 - x
Assume that x is small relative to 0.175. This is acceptable if [ClO-]o > 100·Kb :0.175 >
100·2.86E-7
Then:
2.86E-7=
x2
(0.175)
x = [(2.86E-7)(0.175)]1/2 = 2.24E-4 M = [OH-]
[H3O+] =
1.0e-14
= 4.47E-11 M
2.24E-4
pH = -log [H3O+] = 10.35
A 25.8 mL sample of 0.345 M methylamine, CH3NH2, is titrated with 0.367 Mhydrochloric
acid.
After adding 10.8 mL of hydrochloric acid, the pH is
9.6
10.72
Use the Tables link on the toolbar for any equilibrium constants that are required.
Feedback:
Please note that while the results of the intermediate calculations below are displayed
rounded to 3 significant digits, the actual calculations are done internally without rounding.
The reaction that occurs is: CH3NH2(aq) + HCl(aq)
The net ionic equation is: CH3NH2(aq) + H3O+(aq)
[CH3NH3]Cl(aq)
CH3NH3+(aq) + H2O(l)
mol CH3NH2 = 0.345 M
2.58E-2 L = 8.90E-3 mol
mol H3O+ = 0.367 M
1.08E-2 L = 3.96E-3 mol
There is excess base, so all of the added H3O+ reacts. The resulting solution
contains both CH3NH2 and CH3NH3+ and is a buffer solution.
Initial(mol)
Change(mol)
After rxn(mol)
CH3NH2 +
H3O+
8.90E-3
3.96E-3
-3.96E-3
-3.96E-3
4.94E-3
0
CH3NH3+ + H2O
0
+3.96E-3
3.96E-3
Volume total = 2.58E-2 L + 1.08E-2 L = 3.66E-2 L
4.94E-3 mol
Molarity of CH3NH2 =
= 0.135 M
3.66E-2 L
Molarity of CH3NH3+=
3.96E-3 mol
= 0.108 M
3.66E-2 L
CH3NH3+ + OH0.108 + x
x
CH3NH2 + H2O
0.135 - x
At equilibrium (M)
(0.108 + x)(x)
≈
Kb = 4.2E-4 =
0.135 - x
(0.108)(x)
x = 5.23E-4 = [OH-]
0.135
The approximation is acceptable: [CH3NH2]o > 100·Kb
(0.135 > 100·4.20E-4)
pOH = - log(5.23E-4) = 3.28
pH = 14 - 3.28 = 10.72
Alternatively, the Henderson-Hasselbalch Equation may be used:
Ka(CH3NH3+)=
Kw
1.0e-14
Kb(CH3NH2)
=
4.2E-4
[CH3NH2]
pH = pKa + log
[CH3NH3+]
(0.135)
pH = -log(2.38E-11) + log
.108
= 10.72
= 2.38E-11