To do all titration problems:
I. Start with stoichiometry:
1. Write balanced equation for reaction.
2. Find # moles H+ (or OH-) you are starting with.
3. If amount of titrant is given, find how many moles of that you used.
4. Use stoichiometry to figure out what is left over. (If you have more H+
than OH-, then all OH- will be converted to water.)
5. Subtract the # moles used from the dominant species. The # moles
water will also result in the “liberation” of the same # moles of the
conjugate.
II. Analyze what is left after neutralization. If only base is left, use Kb and write
the ICE table as basic. If only acid is left, use Ka and write ICE table as
hydrolysis of the H+.
III. Do equilibrium problem. (ICE table) Do not forget to include the starting
concentration of the conjugate (A- if you started with HA).
IV. Special cases:
½ way to equivalence point, pH = pKa (This is a short cut!)
At equivalence point, all of the H+ is neutralized.
Note: HA + H2O   H3O+ + A- I will write it in this shortened form:
  H+ + A-
HA
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A 27.1 mL sample of a 0.412 M aqueous hydrocyanic acid solution is titrated with a 0.444 M aqueous barium hydroxide
solution. What is the pH at the start of the titration, before any barium hydroxide has been added?
Nothing has been added, so it is not a titration yet. Solve like any equilibrium
problem:
HCN
H+ + CN-

.412 M
-x
.412-x
0
+x
x
0
+x
x
Ka =
x2
.412
= 6.2 x 10-10
x = [H+] = 1.598 x 10
pH = -log[H+] = 4.80
-5
M
A 41.9 mL sample of a 0.475 M aqueous hydrocyanic acid solution is titrated with a 0.422 M aqueous barium hydroxide
solution. What is the pH after 10.0 mL of base have been added?
Stoichiometry:
2HCN + Ba(OH)2  2H2O + Ba(CN)2
Ba(CN)2 is water soluble.
(.0419 L)(.475 mol/L HCN) = .0199 mol HCN (so .0199 mol H+ is available)
(.010L)(.422 mol Ba(OH)2)(2mol OH-) = .00844 mol OHL
L
More H+ than OH-:
.0199 mol H+ - .00844 mol OH- = .01146 mol HCN left (& .00844 mol CN- liberated)
Analyze what is left:
Total volume = .0419 L + .010 L = .0519 L
H2O, HCN and CN- are present
Concentrations: H2O = N/A
[HCN] = .01146mol = .221 M [CN-] = .00844 mol = .163 M
.0519 L
.0519 L
Since acid conc is higher, use Ka.
Equilibrium:
HCN

.221 M
-x
.221-x
H+ + CN0
+x
x
.163
+x
.163 + x
Ka =
x(.163) = 6.2 x 10-10
.221
x = [H+] = 8.4 x 10 -10 M
pH = -log[H+] = 9.08
When a 29.5 mL sample of a 0.497 M aqueous hydrofluoric acid solution is titrated with a 0.334 M aqueous potassium
hydroxide solution, what is the pH at the midpoint in the titration?
At midpoint, pH = pKa , so pH = -log (7.2 x 10-4) = 3.14
You could prove this by finding that it would take .0439 mol OH- added to reach
equivalence, divide by 2 to get 21.945 mL added at midpoint. Then do ICE to get [H+]
= 7.2 x 10-4M.
What is the pH at the equivalence point in the titration of a 19.0 mL sample of a 0.353 M aqueous hydrocyanic acid
solution with a 0.408 M aqueous potassium hydroxide solution?
(.019 L)(.353 mol/L HCN) = .00671 mol H+ all neutralized by KOH, so:
.00671 mol OH- | 1 L
= .0164 L KOH used at equivalence
| .408 mol KOH
Total vol =.0164 + .019 =.0354L
At equivalence, all H+ and OH- are water,
so only species left is CN- (weak base) = .00671 mol CN-/.0354L = .190 M
Ka for HCN is 6.2 x 10-10 so Kb for CN- = 1x10-14/Ka = 1.61 x 10-5
CN- + H2O
.190 M
-x
.190-x
 HCN + OH0
+x
x
0
+x
x
Kb =
x2
= 1.61 x 10-5
.190
x = [OH-] = 8.5 x 10 -5 M
pOH = -log[OH-] = 4.07
pH = 14-4.07 = 9.93
When a 19.4 mL sample of a 0.382 M aqueous hydrofluoric acid solution is titrated with a 0.413 M aqueous barium
hydroxide solution, what is the pH after 13.5 mL of barium hydroxide have been added?
.0194L (.382 mol/L HF) = .0074 mol H+
.0135L|.413 mol Ba(OH)2 | 2 mol OH= .01115 mol OH- more OH- so basic
|
L
|1 mol Ba(OH)2
.01115 - .0074 = .00375 mol OH- left and .0074 mol F- hanging around.
[OH- ] = .00375 mol/(.0194+.0135)L = .114 M OHRelative to the OH-, the F- is negligible, so pOH = -log [OH-] = .94, and pH = 13.06
An example equation: NH3 + H2O   NH4
+
+ OH-
1. Before adding acid, use ICE and Kb to find OH-. To get from pOH to pH: pH = 14-pOH
2. Before equivalence: NH3 + H+   NH4 + (use stoichiometry-1 mole NH3 to 1 mole of
monoprotic acid)
3. At and after equivalence: all NH3 has been converted to NH4+ so equation switches to
NH4+   NH3 + H+ with Ka = 1x10-14 /Kb for ammonia.
You should have an example of this in your book, but here are some values for you to
work and check if you like:
Titrate 100.0 mL or 0.050 M NH3 with 0.10 M HCl:
1.
2.
3.
4.
5.
Before titration, H+ = 1.1 x 10-11 M and pH = 10.96
After 10 mL HCl added: pH = 9.85
After 25 mL added: pH = 9.25 (midpoint)
After 50 mL added, pH = 5.36 (equivalence)
After 60 mL added, pH = 2.21 (excess H+)
Lastly, strong acid strong base, should be the easiest, right?!?!
This is just simple stoichiometry to find out whether you have more H or OH. However many
moles you have, divide by total volume to get molarity.
Once you know molarity, use these relationships to find what you need:
pH = -log[H+]
pH +pOH = 14
[H+][OH-] = 1 x 10-14
pOH = -log[OH-]
An Example:
20.0 mL of .100 M NaOH is added to 50.0 mL of 0.20 M HNO3.
(Total volume = 20 + 50 = 70 mL)
.020 L (.100 mol/L NaOH) = .002 mole OH.050 L (.200 mol/L HNO3) = .01 mole H+
H+ is majority, so it is still acidic.
.01 mole H+ - .002 mole OH- = .008 mole H+ left
[ H+] = .008 mol H+ / .070L = .114 M H+
pH = -log(.114) = 0.942