chapter 8
DESCRIPTION
Chapter 8. Activity. Homework Chapter 8 - Activity. 8.2, 8.3, 8.6, 8.9, 8.10, 8.12. Question 8.2. Q: Which statements are true? In the ionic strength, m , range of 0-0.1 M, activity coefficients decrease with: a) increasing ionic strength b) increasing ionic charge - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 8
ActivityActivity
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HomeworkHomeworkChapter 8 - Chapter 8 - ActivityActivity 8.2, 8.3, 8.6, 8.9, 8.10, 8.12
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Question 8.2Q: Which statements are true?
In the ionic strength, , range of 0-0.1 M, activity coefficients decrease with:
a) increasing ionic strengthb) increasing ionic chargec) decreasing hydrated radius
All are true!!
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Question 8.3 Calculate the ionic strength of
a) 0.0087 M KOHb) 0.0002 M La(IO3)3 (assuming complete
dissociation at low concentration)
Remember for +1/-1 systems: Ionic strength, Remember for +1/-1 systems: Ionic strength, = Molarity, M = Molarity, M
Ionic strength () = ½ (c1z12+ c2z2
2 + …) = ½ [0.0087M(+1)2+ 0.0087M(-1)2] = 0.0087 M
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Question 8.3 (cont’d) Calculate the ionic strength of
a) 0.0087 M KOHb) 0.0002 M La(IO3)3 (assuming complete
dissociation at low concentration)Ionic strength () = ½ (c1z1
2+ c2z22 + …)
= ½ [0.0002M(+3)2+ 0.006M(-1)2] = 0.0012 M
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Question 8.6Calculate the activity coefficient of Zn2+
when M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1.
xx
xz
3.31
51.0log
2
305083.0)600(1
083.0)2(51.0log
2
x =-0.375 =0.42=0.4222
a)
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Question 8.6 (cont’d)Calculate the activity coefficient of Zn2+
when M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1.
245.0)245.018.0(05.01.0
05.0083.0
xknown
yunknown
interval x intervaly
= 0.432
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Question 8-9Calculate the concentration of Hg2
2+ in saturated solutions of Hg2Br2 in 0.00100 M 0.00100 M KNOKNO3.3.Hg2Br2(s) Hg2
2+ + 2Br- Ksp=5.6x10-23
232 106.5 BrHgsp AAKIICCEE
somesome -- - --x-x +x+x +2x+2xsome-xsome-x +x+x +2x+2x
2322 106.5964.0]2[867.0][ xxK spMHgx 82
2 106.2][][
232222 106.5][][ BrBrHgsp BrHgK
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8-10. Find the concentration of Ba2+ in a
0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+
Ba(IO3)2 Ba2+ + 2IO3 Ksp = 7.11 x 10-11IICCEE
somesome -- 0.100 0.100-x-x +x+x +2x+2xsome-xsome-x +x+x +2x+2x
223
223
23
2 ][][ IOBaIOBasp IOBaAAK
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8-10. Find the concentration of Ba2+ in a
0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+
Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9IICCEE
somesome -- 0.1 0.1-x-x +x+x +2x+2xsome-xsome-x +x+x 0.1+2x0.1+2x
223
223
23
2 ][][ IOBaIOBasp IOBaAAK 22
322 775.0][38.0][
32
IOBaAAK IOBasp222 775.0]1.02[38.0][
32 xxAAK IOBasp
22 775.0]1.0[38.0][xK sp
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8-10. Find the concentration of Ba2+ in a
0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+
Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9IICCEE
somesome -- 0.1 0.1-x-x +x+x +2x+2xsome-xsome-x +x+x 0.1+2x0.1+2x
22 775.0]1.0[38.0][xK sp X = 6.57 x 10-7
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Question 8-12 Using activities correctly, correctly, calculate the
pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities?
() = ½ (c1z12+ c2z2
2 + …) = ½ [0.010M(+1)2+ 0.010M(-1)2+ 0.0120M(+1)2+ 0.0120M(-1)2]
= 0.0220 M
OH = 0.873 pH = AH = [H+]H
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Question 8-12 (cont’d) Using activities correctly, correctly, calculate
the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities?14100.1 xAAK OHHw
OHOH
wH OH
xAKA
][100.1 14
)873.0](010.0[100.1 14
x
AKAOH
wH
121015.1 HA pH = 11.94
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Question 8-12 (cont’d) Using activities correctly, correctly, calculate the
pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities?
pH ~ -log[H+] = 00.12][
log
OH
Kw
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Finally Calculate the pH of a solution that
contains 0.1 M Acid and 0.01 M conjugate base
Calculate the pH of a solution containing 0.1 M Acid and 0.05 M conjugate base.
Calculate the pH of a solution containing 0.1 M Acid and 0.1 M conjugate base.
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Acid/Base TitrationsAcid/Base Titrations
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Titrations Titration Curve – always calculate
equivalent point first Strong Acid/Strong Base
Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point
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Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
First -find Volume at equivalence M1V1 = M2V2 (0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL
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Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30
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Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) +
KBr(aq) BeforeAfter
0.001000 mol 0.0006000 mol0.000400 mol 0 mol
Limiting Reactant
0.0006000 mol0.0006000 mol
pH = 11.8
(~6 ml)
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Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)
BeforeAfter
0.001000 mol 0.0010000 mol0 mol 0 mol 0.0010000 mol0.0010000 mol
pH = 7.0
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Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr
Finally – find pH after equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq)
BeforeAfter
0.001000 mol 0.001200 mol0 mol 0.0002000 mol 0.0010000 mol
pH = 2.5
12 ml
Limiting Reactant
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Typical pH titration
0246
8101214
0 5 10 15 20
mL of HBr
pH
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Titration of WEAK acid with a strong base
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Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
First, calculate the volume at the equivalence-point M1V1 = M2V2 (0.0250 L) 0.1000 M = 0.1000 M (V2) V2 = 0.0250 L or 25.0 mL
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Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Second, Calculate the initial pH of the acetic acid solution
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Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Third, Calculate the pH at some intermediate volume
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Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Fourth, Calculate the pH at equivalence
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Titration of a weak acid solution with a strong base.
25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH
Finally calculate the pH after the addition 26.0 mL of NaOH
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