5.111 Lecture Summary #24 Wednesday, November 5, 2014
Reading for Today: Sections 12.4-12.6 in 5th ed. (4th ed: 11.4-11.6) and 13.1-13.2
Reading Lecture #25: Sections 13.3-13.12 in 5th ( 4th ed: 12.3-12.12)
Topics: Acid-Base Titrations
I. Titration of Weak Acid with Strong Base (Continued)
II. Introduction to Oxidation-Reduction (Redox) Reactions
III. Balancing Redox Reactions
I. TITRATION OF WEAK ACID WITH STRONG BASE CONTINUED
25.0 mL of 0.10 M HCOOH with 0.15 M NaOH (Ka = 1.77 x 10-4 for HCOOH)
3. V = Veq (Point S)
At the equivalence point, the amount of NaOH added is
equal to the amount of HCOOH. The pH is not 7 as it is
for a strong acid and a strong base.
The pH is 7 when a weak acid is titrated with a
strong base.
The pH depends on the properties of the
formed during the neutralization process.
HCOOH and NaOH form NaHCO2 and H2O.
Na+ has on pH and
HCO2- is a .
Thus at the equivalence point, the pH is >7. Figure by MIT OpenCourseWare.
Calculate the pH at the equivalence point
Calculate total volume at equivalence point
moles of HCOOH = 2.5 x 10-3 moles = moles of HCO2- formed = moles of OH- added
2.5 x 10-3 moles of OH- x 1L = 1.67 x 10-2 L of NaOH added
0.15 mol
Total volume = 0.0250 L + 0.0167 L = 0.0417 L
Molarity of HCO2-
2.5 x 10-3 moles of HCO2-/ (0.0417 L) = 0.0600 M HCO2-
1
, HCO2- (aq) + H2O (l) HCOOH (aq) + OH- (aq)
HCO2- (aq) HCOOH (aq) + OH- (aq)
initial molarity 0.0600 0 0
change in molarity -x +x +x
equilibrium molarity 0.0600 -x +x +x
7KLVLVDLQZDWHUSUREOHP
You can take it from here. Simplify if x is small compared to 0.0600 M.
Calculate x, which is equal to [OH- ] = 1.83 x 10-6 M. Then calculate pOH = 5.74.
From pOH, calculate pH. pH = 8.26 (which is >7)
4. V > Veq (Point E)
Beyond the equivalence point, NaOH is added to the
solution of the conj. base HCO2-.
E
Since HCO2- does not give rise to much OH- in
solution (1.83 x 10-6 M), the pOH and pH are
determined by the amRXQWRI1D2+
This problem is similar to a strong base problem.
At 5.00 mL past the equivalence point
0.00500 L x 0.15 M = 7.5 x 10-4 moles excess OH-
7.5 x 10-4 moles OH- /( /
= 0.016 M OH-
pOH = -log [0.016] = 1.79
pH = 12.21 Figure by MIT OpenCourseWare.
IN THEIR OWN WORDS: THE IMPORTANCE OF pKa
Samuel Thompson discusses his research on designing
tools to track the movement of proteins in cells.
Understanding the relationship between pKa and pH
was critical to design a sensor that he hopes will be
used to image proteins in diseased cells.
Samuel’s s video can be found at: http://chemvideos.mit.edu/all-videos/.
Image from "Behind the Scenes at MIT”. The Drennan Education Laboratory. Licensed under a Creative Commons Attribution-
NonCommercial-ShareAlike License. 2
Reading for Today: Sections 12.4-12.6 in 5th ed. (4th ed: 11.4-11.6) and 13.1-13.2
Reading Lecture #25: Sections 13.3-13.12 in 5th ( 4th ed: 12.3-12.12)
Topics: Acid-Base Titrations
I. Titration of Weak Acid with Strong Base (Continued)
II. Introduction to Oxidation-Reduction (Redox) Reactions
III. Balancing Redox Reactions
I. TITRATION OF WEAK ACID WITH STRONG BASE CONTINUED
25.0 mL of 0.10 M HCOOH with 0.15 M NaOH (Ka = 1.77 x 10-4 for HCOOH)
3. V = Veq (Point S)
At the equivalence point, the amount of NaOH added is
equal to the amount of HCOOH. The pH is not 7 as it is
for a strong acid and a strong base.
The pH is 7 when a weak acid is titrated with a
strong base.
The pH depends on the properties of the
formed during the neutralization process.
HCOOH and NaOH form NaHCO2 and H2O.
Na+ has on pH and
HCO2- is a .
Thus at the equivalence point, the pH is >7. Figure by MIT OpenCourseWare.
Calculate the pH at the equivalence point
Calculate total volume at equivalence point
moles of HCOOH = 2.5 x 10-3 moles = moles of HCO2- formed = moles of OH- added
2.5 x 10-3 moles of OH- x 1L = 1.67 x 10-2 L of NaOH added
0.15 mol
Total volume = 0.0250 L + 0.0167 L = 0.0417 L
Molarity of HCO2-
2.5 x 10-3 moles of HCO2-/ (0.0417 L) = 0.0600 M HCO2-
1
, HCO2- (aq) + H2O (l) HCOOH (aq) + OH- (aq)
HCO2- (aq) HCOOH (aq) + OH- (aq)
initial molarity 0.0600 0 0
change in molarity -x +x +x
equilibrium molarity 0.0600 -x +x +x
7KLVLVDLQZDWHUSUREOHP
You can take it from here. Simplify if x is small compared to 0.0600 M.
Calculate x, which is equal to [OH- ] = 1.83 x 10-6 M. Then calculate pOH = 5.74.
From pOH, calculate pH. pH = 8.26 (which is >7)
4. V > Veq (Point E)
Beyond the equivalence point, NaOH is added to the
solution of the conj. base HCO2-.
E
Since HCO2- does not give rise to much OH- in
solution (1.83 x 10-6 M), the pOH and pH are
determined by the amRXQWRI1D2+
This problem is similar to a strong base problem.
At 5.00 mL past the equivalence point
0.00500 L x 0.15 M = 7.5 x 10-4 moles excess OH-
7.5 x 10-4 moles OH- /( /
= 0.016 M OH-
pOH = -log [0.016] = 1.79
pH = 12.21 Figure by MIT OpenCourseWare.
IN THEIR OWN WORDS: THE IMPORTANCE OF pKa
Samuel Thompson discusses his research on designing
tools to track the movement of proteins in cells.
Understanding the relationship between pKa and pH
was critical to design a sensor that he hopes will be
used to image proteins in diseased cells.
Samuel’s s video can be found at: http://chemvideos.mit.edu/all-videos/.
Image from "Behind the Scenes at MIT”. The Drennan Education Laboratory. Licensed under a Creative Commons Attribution-
NonCommercial-ShareAlike License. 2
