Problem Set #12, Chem 340, Fall 2013

Problem Set #12, Chem 340, Fall 2013 ? Due Wednesday, Dec 4, 2013 Please show all work for credit

To hand in: Ionic equilibria:

1. Calculate the value of m? in 5.0 10?4 molal solutions of (a) KCl, (b) Ca(NO3) 2, and (c) ZnSO4. Assume complete dissociation.

a) KCl

1

m =

v v v v

vm

m = 1 m 5.0 104 mol kg1

b) Ca(NO3)2

1

m =

v v v v

vm

m

=

4

1 3

m

4

1 3

5.0

104

mol

kg1

7.9

104

mol

kg1

c) ZnSO4

1

m =

v v v v

vm

m

=

1

1 2

m

5.0

104

mol

kg1

2. Calculate ?, and a?for a 0.0325 m solution of K4Fe(CN) 6 at 298 K.

I m 2

v z2 v z2

1 2

m z2 m z2

I 1 4 0.0325 mol kg1 42 0.0325 mol kg1 0.325 mol kg1 2

ln 1.173 z z

I mol kg1

1.173 4

0.325 2.6749

0.069

1

1

m =

v v v v

v m 44

5 0.0325 mol kg1 0.099 mol kg1

a

m m

0.099 0.069

0.0068

3. Chloroacetic acid has a dissociation constant of Ka = 1.38 10?3. (a) Calculate the degree of dissociation for a 0.0825 m solution of this acid using the Debye?H?ckel limiting law. (b) Calculate the degree of dissociation for a 0.0825 m solution of this acid that is also 0.022 m in KCl using the Debye?H?ckel limiting law.

a) Ka = 1.38 10?3 0.0825 m

Ka

m

m

m

m2

2

0.0825

m

1.38 103

when = 1

m2 1.1385 104 1.38 103 m m2 1.38 103 m 1.1385 104 0

1.38 103 1.38 103 2 41 1.1385 104

m

0.0100 mol kg1

2

I m 2 m 0.0100 mol kg1

2

ln 1.173 1 0.0100 0.1173 0.8893

when = 0.8893 m2 1.43956 104 1.7449 103 m

m2 1.7449 103 m 1.43956 104 0

1.7449 103 1.7449 103 2 41 1.43956 104

m

0.0112 mol kg1

2

I m 2 m 0.0112 mol kg1

2

ln 1.173 1 0.0112 0.1239 0.8835

Same value, done!

when = 0.8835 m2 1.4587 104 1.7681103 m

m2 1.7681103 m 1.4587 104 0

1.7681103 m I 0.0112 mol kg1

1.7681103 2 41 1.4587 104 0.0112 mol kg1 2

ln 1.173 1 0.0112 0.1239 The degre of dissociation is 0.0112 100% 13.6%

0.0825

b)

K+ Cl

1

1

z 1 z 1

I 0.022 mol kg1 from the KCl alone.

We first calculate the activity coefficient using this value.

I 0.022 1 0.022 mol kg1

2 ln 1.173 1 0.022 0.1740 0.8403

We next calculate the concentration of the ions produced through the dissociation of

the acid in the solution.

m2 0.84032 1.38 103

0.0825 m

m2

1.9544 103

0.0825 m

m2 1.6124 104 1.9544 103 m

m2 1.9544 103 m 1.6124 104 0

1.9544 103 1.9544 103 2 41 1.6124 104

m

0.01176 mol kg1

2

The ionic strength is the sum of that due to the chloroacetic acid and the KCl or

I = 0.022 + 0.012 = 0.034 mol kg1.

We recalculate using this value for the ionic strength.

ln 1.173 1 0.034 0.2163 0.8055

and recalculate m

m2 0.80552 1.38 103 giving

0.0825 m m 0.01222 mol kg1

Given the fact that the concentration of KCl is only known to 2 significant figures, this value of m is sufficiently close to the previous value that a further iteration is not

0.01222 14.8%. necessary. The degree of dissociation is 0.0825

4. The principal ions of human blood plasma and their molar concentrations are

mNa 0.14 m, mCl 0.10 m, mHCO3 0.025 m. Calculate the ionic strength

of blood plasma.

Note: obviously something missing, pos-neg imbalance

I 1 2

i

mi zi 2

1 2

0.14 m 12

0.10 m 12

0.025 m 12

0.1325 m

5. The oxidation of NADH by molecular oxygen occurs in the cellular respiratory system:

O2 (g) + 2NADH(aq) + 2H+ (aq) 2H2O(l) + 2NAD+ (aq) Using the information in Table 9.7, calculate the standard Gibbs energy change that results from the oxidation of NADH by molecular oxygen.

The standard reduction potential for nicotine adenine dinucleotide is

NAD aq H aq 2e NADH aq E 0.320V

Using the half cell reactions: Oxidation: 2 NADH 2 NAD 2 H 4 e

E0 0.320 V

Reduction: O2 4H 4 e 2 H2O

E0 0.815 V

Ecell = Ered + Eox = 1.135 V

G reaction

n F 4 96485 C mol-1

1.135 V 438.0 kJ mol-1

6. Determine Ksp for AgBr at 298.15 K using the electrochemical cell described by

Ag s AgBr s Br aq, aBr Ag aq, aAg Ag s . The half cell and overall

reactions are

AgBr + e? Ag + Br?

E? = +0.07133 V

Ag Ag+ + e?

E? = ?0.7996 V

AgBr Ag+ + Br?

E? = ?0.72827 V

log10

Ksp

nE 0.05916

0.729 V 0.05916 V

12.310

Ksp 4.90 1013

7.

For a given overall cell reaction,

S

R

17.5

J

mol1

K 1

and

H

R

225.0

kJ

mol1.

Calculate E? and

E T P .

E

GR nF

HR T SR nF

225000 J mol1 298.15 K 17.5 J K1mol1

1 96485 C mol1

2.38 V

E

T

P

SR nF

17.5 J K 1 96485

1mol 1 C mol1

= 1.81104V K1

8. The following data have been obtained for the potential of the cell

 Pt s H2 g, f 1 atm HCl aq, m AgCl s Ag s as a function of m at 25?C.

m (molkg?1) E (V)

m (mol kg?1) E (V) m (mol kg?1) E (V)

0.00100

0.597915 0.0200

0.43024 0.500

0.27231

0.00200

0.54425 0.0500

0.38588 1.000

0.23328

0.00500

0.49846 0.100

0.35241 1.500

0.20719

0.0100

0.46417 0.200

0.31874 2.000

0.18631

Calculate E? and ? for HCl at m = 0.00100, 0.0100, 0.100, and 1.000.

Cell reaction: 2AgCl(s) + H2(g) 2Ag(s) + 2H+(aq) + 2Cl?(aq)

RT

2

RT

E

E

ln 2F

a a H Cl

E F ln a a H Cl

a a H Cl

a2

2

m2

EE

2RT F

ln

m m

2 RT F

ln

In the low concentration limit we can use the Debye-H?ckel result

ln 0.50926 log10

m 1.172614 ln m

m m

Therefore, for dilute solutions

E

2RT F

ln

m m

E

2RT 1.172614 F

m m

Using this result, a plot of

E

2RT F

ln

m m

(y axis) vs.

m m (x axis) will have an

intercept of E . We use the data up to m = 0.100, as the Debye-H?ckel model is not

valid for more concentrated solutions.

m

E

m

m

m

E

2RT F

ln

m m

0.001 0.002 0.005 0.010 0.020 0.050 0.100

0.031623 0.044721 0.070711 0.1 0.141421 0.223607 0.316228

0.57915 0.54425 0.49846 0.46417 0.43024 0.38588 0.35241

0.224212 0.224909 0.226203 0.227531 0.229218 0.231943 0.234090

The data in the table is graphed below. The best fit line gives a value for E of

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