Chem 222
#9 Ch 7Sep 21, 2004
Announcement
• The answer for quiz 2 with a grading guide line is at the web site
• The due date for report of Exp 5 is this W/R
Strengths of Acids and BasesStrong Acids and BasesA strong acid or base is completely
dissociated in aqueous solution.HCl(aq) H+ + Cl- (~ 100%)KOH(ac) K+ + OH-(~ 100%)
Remember all the acids and bases in Table 6-2.
Q. Calculate the concentration of H+ and the pH in the following solutions at 25 °C: a. 0.056 M HCl, b. 0.25 M KOH, c. 5.60 M HCl. Kw = 1.01 × 10-14
a. [H+] = 0.056 M pH = -log[H+]= -log(0.056)= 1.252
b. [OH-] = 0.25 M [H+] = Kw/[H+] a. pH = -log[H+] = -log(5.60) = - 0.7482
←→
←→
Weak acids and basesHA H+ + A- a weak acid
Ka = [H+][A-]/[HA] = x2/(0.056 –x)
B + H2O BH+ + OH- a weak base
Kb = [BH+][OH-]/[B]
←→
0.056 M - x x x
←→
7-1 Titration
The color changes indicates “end point”, which is close to the “equivalent point”
The titrant functions as an “indicator”
At “equivalent point”
Indicator•BB (Blue Yellow)•MR (Yellow Red)
• pH meter (pH is detected byvoltage; see P 323)
Standardization of KMnO4• Suppose that 0.3562g of Na2C2O4 is dissolved in a
250.0mL volumetric flask. If 10.00 mL of this solution require 48.36 mL of KMnO4 solution for titration, what is the molarity of KMnO4 solution ?
Oxalate concentration COX is
COX = 1.0633 mM
Lmolgg
VolumeMWWeight
2500.0)/00.134/(3562.0
solution of /OCNa of 422 =
moles 2moles 5
KMnO4 of Oxalate of
44
==MnOMnO
OXOX
VCVC
MolesMoles
mmol8794.0000.5000.2
36.4810001063.0
52
7
3
44
=
××
==mL
mLMVVCC
KMnO
OXOXKMnO
Analysis of unknown
• CKMNO4 = 0.8794 mM
Calcium in a 5.00 mL urine sample was precipitated, was redissolved, and then required 16.17mL of standard MnO4- solution.
Find the concentration of Ca2+ in the urine.
Ca2+: C2O42-: MnO4
- = 5 : 5: 2
moles 2moles 5
KMnO4 of Ca of
44
==MnOMnO
CaCa
VCVC
MolesMoles
VCa = (5/2) × (CMnO4VMnO4)/VCa= (5/2) × (0.8794 mM × 16.17 mL)/5mL
Ex. (p132) (Related to Exp 8)
y g
(1.372 – y) g
Na2CO3 moles × 2 + NaHCO3 moles = HCl moles
{y/105.99 + (1.372 – y)/84.01} = CHCLVHCl = 29.11mL × 0.734 M= 0.02138 moles
(1/104.99 – 1/84.01)y + 1.372/84.01 = 0.02138
Do not forget to weigh your sample !Without 1.372 g you can never obtain the answer.
Spectrophotometric Titration (Lab. Exp 19)
• Absorbance: A = εbC
ConcentrationPathlength
(p411)
Fe3+ Binding Site of Transferrin
Apo-transferrin + 2Fe3+ (Fe3+)2transferrin
Show absorbance at 465 nm !
Titration of 2.00 mL of apotransferrinWith 1.79 × 10-3 M ferric nitrilotriacetate
Titration of 2.00 mL of apotransferrinWith 1.79 × 10-3 M ferric nitrilotriacetate
moles 1moles 2
protein of Fe of 3
==+
proteinprotein
FeFe
VCVC
MolesMoles
Apo-transferrin + 2Fe3+ (Fe3+)2transferrin
↑ 203 µL
mLLmM
VVCCprotein
FeFeprotein 00.22
20379.12 ×
×==
µ
To construct the above graph the dilution effect must be corrected.
initial
total
VVAbsObservedAbsCorrected =
Ex.
The absorbance measured after adding 125 uL of ferric nitrilotriacetate to 2.000 mL of apotransferrin was 0.260. Calculate the corrected absorbance that should be plotted in the Figure 7-5.
mLmL
VVAbsObservedAbsCorrectedini
total
000.2125.2260.0 ==
7-4 The precipitation Titration Curve
• The titration curve is a graph showing how the concentration of one of the reactions varies as a titrant is added
• Because concentration of the reagent varies over many orders, it is most useful to plotpfunction: pX = -log10[X]
Q. Consider the titration of 25.00 mL of 0.1000 M I- with 0.05000 M Ag+
I- + Ag+ AgI(s) (7-9)
Suppose we are monitoring Ag+ with an electrode. (7-9) is the reverse of the dissolution of
AgI(s) I- +Ag+ Ksp=[I-][Ag+]=8.3×10-17
←→
Q1. What volume of Ag+ titrant is needed to reachthe equivalence point?
Almost 100 % forms AgI(s)
CIVI = CAgVAge
VAge = CIVI/CAg = (0.1000 M)25.00 mL/0.05000 M
= 50.00 mL
titrant
How does [Ag+] change before the end point?
At the equivalence point at VAge = 50.00 mL.
A. Before the equivalence point, we have an excess of I-. [Ag+] is obtained from
[Ag+] = Ksp/[I-] = 8.3×10-17/[I-]
[I-] =
AgI
AgAgII
VVVCVC
+
−=
+ volumeTotal
)Ag of (Moles - I-) Original of (Moles
If VAg = 49.00 mL
Moles of I- = (25.00 mL) (0.1000 M) - (49.00 mL)(0.05000 M)= 2.500 mmol - 2.450 mmol =0.050 mmol
Vtotal = (25.00 mL + 49.00 mL) = 74.00 mL[I-] = [Q1]/74.00 mL = 6.76 × 10-4 M See P135
[Ag+] = [Q2]/6.76 ×10-4 M = 1.23 × 10-13
pAg+ = -log[Ag+] = 12.91
Q. What is [Ag+] when VAg = 49.00 mL ?
At the equivalence point, what is [Ag+]?
• At equivalence point, VAg
eCAg = VICI
Little l- and Ag+ are left, and[Ag+] = [l-]. We define [Ag+] = x.
Then,Ksp = 8.3×10-17 = [Ag+][I-] = x2
x = 9.1×10-8 MpAg+ = -log(9.1 ×10-8 ) = 7.041
Q. What is [Ag+] when VAg = 51 mL
After the end point,Q. What is [Ag+] when
VAg = 51 mL
Now, all the I- ions are consumed, we have an excess of Ag+.
• Moles of Ag+ = (CAgVAg – CIVI) =
(51.00 mL)(0.05000 M) - (25.00 mL) (0.1000 M) =1.00 mL × 0.05000 M
= 0.05 mmol[Ag+] = 0.05 mmol/(total volume)
= 0.05 mmol/(51 mL + 25 mL)= 6.6 × 10 -4 M
pAg+ = 4 – log6.6 = 3.18
Titration curve
↑For 1:1 stoichiometry of reactants, the equivalence pointcan be indicated by
Excess I- Excess Ag+
Ion dependence(pAg+) e = -Log(Ksp
1/2)
12-1 Strong acid-strong base Titration
Excess OH- Excess H+
pH
Va (mL)
♦Excess OH- Calculate [OH-] [H+] =Kw/[OH-]
♦Equivalence Point Kw = x2 pH = 7.0
♦Excess H+
Calculate [H+] = (CHCLVHCl – CNaOHVNaOH)/(VHCl +VNaOH)
NaOH + HCl H2O + NaCl
P141 Calculation of titration from spread sheet
Known [M+] [X-] = Ksp/[M+] Then [M+] & [X-] VM
Home work
• Read Ch 13 1-3
• Ch7: 7-A, 7-B, 7-4, 7-7, 7-8, 7-9, 7-10, 7-14, 7-21, 7-23