Transcript EXAM_1_answers.ppt

EXAM #1
EXAM #1a: Find moles of OH- in 8.00mL of 0.10M NaOH.
NaOH  OH- + Na+
M = mol
#L
0.10 = mol
0.008
Mol (OH-) = 8 * 10-4
EXAM #1b: Find molarity of OH- AFTER of 8.00mL of 0.10M NaOH is
titrated in. NOTE total volume = Vacid(50mL) + Vbase(8 mL) = 58 mL
M = mol
#L
M = 8*10-4mol
.058L
M = 0.0138
EXAM #1c: Find the pH after 8.00 mL of 0.10 M OH- is added to the 50.0
mL acid sample.
FIRST FIND MOLES OF ACID ORIGINALLY PRESENT.
M = mol
#L
0.10 = mol
0.050L
Mol HA= 0.005
THEN EVALUATE NEUTRALIZATION STOICHIOMETRY; REMEMBER THE
OH- IS LIMITING UP TO EQ POINT. Working in moles alleviates the need to
keep track of dilutions!
HA
OH-

0.005 mol 8.0*10-4mol
-X
-X
0.005-X
0
OH- IS LIMITING
REAGENT AND IS
CONSUMED TOTALLY
A-
H2O
0
X
8*10-4
NOW USE THE FINAL MOLES
IN HENDERSON-HASSLEBACH
EXAM#1-c CONTINUED;
HA 
0.005
-X
0.005-X
A8.0*10-4
X
8.0*10-4 +X
H+
0
X
X
pH = pKa + LOG [A-]
[HA]
pH = 9.21 + LOG [8.0*10-4 ]
[0.0042]
pH = 8.49
Ok to work
in moles for
this
equation.
EXAM #1d and e: AT THE pKa (HALFWAY) POINT:
Moles OH- added =:
1. mole HA neutralized
2. moles HA remaining
3. A- produced
*********AND mol OH- added = ½ HA INITIAL ********
HA
0.005 mol
-X
0.005-X=0.0025
OH-

0.0025
-X
0
A-
0
X
0.0025
Neutralization stoichiometry: base
added limits, OH- added = X and
is completely consumed
#1e) Moles OH- at halfway point = ½
HA original ½ of 0.005 = 0.0025
H2O
#1-f) AT THE pKa (HALFWAY POINT) point pH = pKa
pH = pKa + LOG [A-]
[HA]
pH = 9.21 + LOG [0.0025]
[0.0025]
pH = pKa = 9.21
#1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0
MOLE, AND A- IS THE ONLY ION REMINING.
OH- added total) = HA original = A- produced
#1-h) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0
MOLE, AND A- IS THE ONLY ION REMINING.
OH- added (total) = HA original = A- produced = 0.005 mol
HA
OH-
0.005 mol
-X
0.0050
-X
0
0
M = mol
#L

A-
H2O
0
X
0.005
0.10 = 0.005 mol vol OH- = .050 L = 50.0mL
XL
#1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0
MOLE, AND A- IS THE ONLY ION REMINING.
OH- added total) = HA original = A- produced, The Ph at equivalence
is calculated by the hydrolysis equilibrium of A-.
OH-
A-
0.005 mol
0.0050
0
-X
-X
X
0
0
0.005
HA

THE EQUIVALENCE
POINT
STOICHIOMETRY
GIVES THE AMOLARITY FOR
THE EQUIVALENCE
(HYDROLYSIS OF A) EQUILIBRIUM.
H2O
HYDROLYSIS EQUILIBRIUM OF AAT EQUIVALENCE PT.
A-
HOH 
HA
OH-
0
0
-X
X
X
0.005-X
X
X
0.005 mol
IGNORE SOLVENT
HYDROLYSIS EQUILIBRIUM OF A- AT
EQUIVALENCE POINT
AHOH 
HA
OH0.005 mol
IGNORE SOLVENT
0
0
-X
X
X
0.005-X
X
X
Kb = Kw/Ka = [HA][OH-]
[A-]
Kb = 1.6 x 10-5 = (X)(X)
; X = 8.86 x 10-4 = [OH-]
(0.005/0.1L)
MUST CONVERT MOLES TO
MOLARITY, TOTAL VOLUME IS
0.10 L
(0.05L ACID AND 0.05L BASE)
pOH- = -LOG(8.86 x 10-4) = 3.05
pH = 10.95 at equivalence