Transcript Acid-base titrations

Acid-Base Problems
A complete forward reaction or
equilibrium?
• If the problem involves only strong acids
and/or bases, the reaction goes to completion
(→).
• If the problem involves a weak acid or base,
the reaction involves equilibrium (↔).
If the problem involves only strong acids and/or
bases, the reaction goes to completion (→).
• If only strong acid or base, use one of these
four relationships to solve:
– pH = -log [H+]
pOH = -log[OH-]
– [H+][OH-] = Kw
pH + pOH = 14
• For titration (acid + base), use regular
stoichiometry to find molarity, then use log
formula to find pH if needed. (or use pH to
find molarity depending on given.)
If the problem involves a weak acid or base,
the reaction involves equilibrium (↔).
• Use RICE table.
– Weak acids hydrolyze to produce H+ and a conjugate
base. Ka is used.
– Weak bases hydrolyze to produce OH- and a conjugate
acid. Kb is used.
• For titration (acid + base), two steps are needed:
1. Use regular stoichiometry to find which species are
left after neutralization.
2. Use RICE table and ionization constant (Ka or Kb) to
solve with remaining species.
– Note that Ka is needed if H+ is a product in the
equilibirium and Kb is needed if OH- is a product.
To do all titration problems:
I.
•
•
•
Start with stoichiometry:
Write balanced equation for reaction.
Find # moles H+ (or OH-) you are starting with.
If amount of titrant is given, find how many moles of
that you used.
• Use stoichiometry to figure out what is left over. (If
you have more H+ than OH-, then all OH- will be
converted to water.)
• 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).
Special case
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:
• HA
  H+ + A -
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 
.412 M
-x
.412-x
H+ + CN0
0
+x +x
x
x
Ka = x2
= 6.2 x 10-10
.412
x = [H+] = 1.598 x 10 -5 M
pH = -log[H+] = 4.80
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 LH2O, HCN and CN- are present
Concentrations:
H2O = N/A [HCN] = .01146mol = .221 M
.0519 L
Since acid conc is higher, use Ka.
Equilibrium:
HCN 
.221 M
-x
.221-x
H+ + CN0 .163
+x
+x
x .163 + x
[CN-] = .00844 mol = .163 M
.0519 L
x(.163) = 6.2 x 10-10
.221
x = [H+] = 8.4 x 10 -10 M
pH = -log[H+] = 9.08
Ka =
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?
Stoichiometry:
2HCN + Ba(OH)2  2H2O + Ba(CN)2
Ba(CN)2 is water soluble.
(.010 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
Equivalence point:
.00671 mol H+ + .00671 mol OH- = 0 mol HCN left
(& .00671 mol CN- liberated)
Analyze what is left:
Total volume = =.0164L + .019L =.0354L
H2Oand CN- are present
Concentrations:
H2O = N/A
[CN-] = .00671 mol = .190 M CN.0354 L
Since only a conjugate base is present, use Kb: Kb = 1 x 10 -14 = 1.61 x 10-5
6.2 x 10-10
Equilibrium:
CN+ H2O  HCN + OH.190 M
0
0
-x
+x
+x
.190-x
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 – pOH = 9.93
When a 19.4 mL sample of a 0.382 M aqueous hydrocyanic 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?
Stoichiometry:
2HCN + Ba(OH)2  2H2O + Ba(CN)2
Ba(CN)2 is water soluble.
.0194L (.382 mol/L HCN)
= .0074 mol H+
.0135L|.413 mol Ba(OH)2 | 2 mol OH|
L
|1 mol Ba(OH)2
= .01115 mol OH-
There are more OH- ions, so the solution is basic:
01115 moles OH-- .0074 mol H+ = .00375 mol OH- left and .0074 mol CN- hanging
around.
More OH- than H+:
.00375 mol OH- left and .0074 mol CN- liberated)
Analyze what is left:
Total volume = =.0194L + .0135L =.0329 L
[OH- ] = .00375 mol/ .0329 L = .114 M OH-
OH-, H2O and CN- are present
Concentrations:
H2O = N/A [OH- ] = .00375 mol/ .0329 L = .114 M OH[CN-] = .0074 mol = .225 M CN.0329 L
BUT…CN- is a weak base, and the hydroxide ions it will generate in water will be
very few relative to the amount from the strong base used in titration.
Relative to the OH-, the CN- is negligible,
so pOH = -log [OH-] = .94
pH = 14 - .94 = 13.06