Transcript slides(ppt)

Acid Base
Stoichiometry and Titrations
By Marisa Carlson and Lizzie Sokol
Stoichiometry

In an acid, HxA, or base, B(OH)x, x
determines the number of equivalence
points in the titration.
What is a Titration?

A procedure for quantitative analysis of a
substance by an essentially complete
reaction in solution with a measured
quantity of a reagent of known
concentration.
Finding Equivalence Points

n1V1C1 = n2V2C2
 n is the number of moles of H+ or OH- in the acid
and base to be titrated and/or to titrate
 C is the concentration of the acid and base
 V is the volume, one quantity is given, the other
one is unknown and determines how much titrant
is needed to reach the first equivalence point (all
number relating to acid go on one side, and all
relating to base go on the other)
A Sample Titration Problem

20.00 mL of 0.250 M H2SeO3 is titrated
with 0.250 M NaOH
 Ka1 = 2.7 x 10-3
 Ka2 = 2.5 x 10-7
 Find pH when 0, 10, 17, 20, and 45 mL of
NaOH have been added
Zero mL
H2SeO3 + H20 ↔ HSeO3- + H3O+
I .250M*20mL
0
0
5mmol
C
-x
+x
+x
E 5mmol/20mL
≈.250M
x
x
Zero mL con’t
Ka1 = [HSeO3-]*[H3O+]/[H2SeO3]
2.7 x 10-3 = x*x/.250
x=.026 M
-log(x) = pH = 1.59
Ten mL

At the half-way point of the titration the
pKa = pH
 So…
-log (2.7x10-3) = 2.57
Seventeen mL
H2SeO3 + H20 ↔ HSeO3+ H3O+
I 5mmol
0
0
-(17mL*.250M) + (17mL*.250M)
.750mmol
4.25mmol
C
-x
+x
+x
E ≈.750mmol
≈4.25mmol
x
Seventeen mL con’t
Ka1 = [HSeO3-]*[H3O+]/[H2SeO3]
2.7 x 10-3 = 4.25*x/.750
x=4.76x10-4 M
-log(x) = pH = 3.32
Twenty mL

At each equivalence point before the end is
reached, it must be determined whether the
Ka or the Kb should be used.
 Ka2 = 2.5x10-7
 Kb1 = 1x10-14 / 2.7x10-3 = 3.7x10-12
 So, the acidic properties are stronger, and
the solution continues as an acid (Ka>Kb)
Twenty mL con’t
H2SeO3 + H20 ↔ HSeO3- + H3O+
I
5mmol
0
0
-5mmol
+5mmol
0
5mmol
Because all of the H2SeO3 is used up, and it
acts as an acid, we move on the the next
equation…
HSeO3- + H20 ↔ SeO32- + H3O+
Twenty mL con’t
I
C
HSeO3- + H20 ↔ SeO32- + H3O+
5mmol
0
0
-x
E 5mmol/40mL
.125M
+x
+x
x
x
Twenty mL con’t
Ka2 = [SeO32-]*[H3O+]/[HSeO3-]
2.5 x 10-7 = x*x/.125
x=1.77x10-4 M
-log(x) = pH = 3.75
Forty-five mL

Since all of the acid has been neutralized
(beyond last equivalence point), the
calculation does not involve neutralization.
5mL NaOH (not used to neutralize acid)
*.250M = 1.25mmol
1.25 mmol / 65 mL = .0192 M
-log (.0192) = pOH = 1.72
14 – pOH = pH = 14 – 1.72 = 12.28
Helpful Hints





You can use the concentration of the acid or base in the ice
chart as long as the concentrations of both products are
unknown.
pH= pKa + log([A-]/[HA])
pOH= pKb + log ([HB+]/[B])
Remember that the pH of a solution after the acid or base
has been completely neutralized does not require an ice
chart.
Determining whether the Ka or Kb should be used requires
finding both and finding the larger quantity.
Titrating a Base with an Acid

Virtually no difference except:
– when solving for x, x= [OH-]
– so, when the -log(x) is taken, the result is the
pOH
– therefore, to find the pH, subtract pOH from 14