Chapter 8

ActivityActivity

HomeworkHomeworkChapter 8 - Chapter 8 - ActivityActivity 8.2, 8.3, 8.6, 8.9, 8.10, 8.12

Question 8.2

Q: 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!!

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

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 (c1z12+ c2z2

2 + …) = ½ [0.0002M(+3)2+ 0.006M(-1)2] = 0.0012 M

Question 8.6

Calculate the activity coefficient of Zn2+

when M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1.

x

x

xz

3.31

51.0log

2

305083.0)600(

1

083.0)2(51.0log

2

x =-0.375 =0.42=0.4222

a)

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

x

known

y

unknown

interval x intervaly

= 0.432

Question 8-9

Calculate the concentration of Hg22+ 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

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

22

32

32 ][][ IOBaIOBasp IOBaAAK

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

22

32

32 ][][ IOBaIOBasp IOBaAAK

223

22 775.0][38.0][3

2 IOBaAAK IOBasp

222 775.0]1.02[38.0][3

2 xxAAK IOBasp22 775.0]1.0[38.0][xK sp

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

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

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

x

A

KA

][

100.1 14

)873.0](010.0[

100.1 14

x

A

KA

OH

wH

121015.1 HA pH = 11.94

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

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.

Acid/Base TitrationsAcid/Base Titrations

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

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

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

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)

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

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

Typical pH titration

0

2

4

6

8

10

12

14

0 5 10 15 20

mL of HBr

pH

Titration of WEAK acid with a strong base

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

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

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

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

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