Chapter 6: Chemical Activity and Chemical Equilibrium

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Chapter 6: Chemical Activity and Chemical Equilibrium

1.(A) To keep ionic strength constant and maintain a more consistent matrix in a solution where ions could be influencing each other, another component is often added to the solution This is an ionic compound, not involved directly with the reaction, called an ionic strength adjuster (ISA) ingredient. This is an important consideration in investigations using selective electrodes. What is the ionic strength (I) of a solution that contains0.0280 M (NH4)2SO4 as the ISA ingredient? (NH4)2SO4 (s)  2 NH4+ (aq) + SO42-(aq)

(B) One particular use of a selective electrode is in the determination of the level of Cl- in drinking water. What would be the activity coefficient () for the chloride ion Cl- in the solution from part “A”?(Note,Table 6.3 will be necessary for this calculation.)

(C)The chemical activity (a) of an ion in solution is related to the actual concentration of the ion. Which one, if any, of the following statements is actually true? (select one)

(i) The chemical activity of an ion is closest to the actual concentration when the ion concentration is very high.

(ii) Chemical activity and activity coefficient are two names that have the same meaning.

(iii) A substance in its standard state will have a chemical activity (a) = zero

(iv) In a solution with high ionic strength in the background, Cl-ions will have less repulsion for each other.

(v) Actually, none of the above is true.

2. (A) Calcium chloride (CaCl2; very soluble) has been sold as a more effective “de-icer” than NaCl to spread on icy sidewalks. What would be the ionic strength of a solution that was 0.031 M CaCl2?

CaCl2(s) Ca2+(aq) + 2 Cl-(aq)

(B)Nowuse that value and information from Table 6.3 to determine the activity coefficient () of H+ in that solution.

3.Amino acids are the key components of proteins. When fully protonated, amino acids have similar chemical reactions as diprotic acids (such as found in H2CO3). For example, alanine undergoes the following reactions:

H2C3H5NO2+(aq) H+(aq) + HC3H5NO2(aq) K1 = 4.5 x 10-3

HC3H5NO2 (aq) H+(aq) + C3H5NO2-(aq) K2 = 1.3 x 10-10

H2O (l) H+ (aq) + OH-(aq) Kw = 1.0 x 10-14

(A) What would be the [H+] in a 0.010 M solution of H2C3H5NO2+? (** Three important assumptions to use: the second K value is small enough to be ignored for this calculation. The H+ from water is also small enough to be ignored in this calculation. The calculation for H+ should be done with a graphing calculator or successive approximations.)

(B) Looking back at the three reactions, what would be the charge balance expression?

______= ______

(cation) (anions)

(C) Using only the value of the first equilibrium constant value (K1) from part “A” would a solution that had the following concentrations (H2C3H5NO2+ = 0.000050M; H+ = 0.000040M; HC3H5NO2 = 0.000030 M) be at equilibrium?What is the value of Qfor this system? If it is not at equilibrium, in which direction would it proceed to reach equilibrium?

4.(A)The explanation of the effects of ionic strength on activities of ions relies on the concept of the “hydration radius” that surrounds ions in solution. Would the addition of KNO3 to a solution containing CaF2 increase or decrease the solubility of this slightly soluble salt? Briefly explain.

(CaF2(s) Ca2+(aq) + 2F-(aq))

(B)In a certain solution the value for Ca2+ was found to be 0.747, while the  value for F- was found to be 0.926. Using those values, and the equilibrium value of 3.3 x 10-4, what is the molarity (M) of dissolved F- in the solution? (CaF2(s) Ca2+(aq) + 2 F-(aq))

(C) Slightly soluble CaF2salt may have a connection to music. (A recent study of the wood used in the famous Stradivari violins suggested that the unique sound quality may be due to an antifungal wood treatment containing CaF2 and borax.) Using the following reactions, write the proper mass balanceequation for Ca2+.

CaF2(s) Ca2+(aq) + 2 F-(aq)

Ca2+(aq) + H2O Ca(OH)+(aq) + H+(aq)

Ca2+(aq) + F- (aq) CaF-(aq)

F- (aq) + H+(aq) HF (aq)

( MASS BALANCE) CF- = ______

5. Nickel (II) hydroxide is found in rechargeable batteries (NiCd batteries). The following examples show three pertinent reactions of Ni(OH)2 in water:

Ni(OH)2 (s) [Ni2+] (aq) + 2 [OH-] (aq

Ni2+ (aq) + OH- (aq)  NiOH+ (aq)

H2O (l) H+ (aq) + OH- (aq)

What is the mass balance equation for OH-in this system?

COH- =

6.(A) The antibioticPenicillin G is obtained from a mold and is treated before being ready for human consumption. Penicillin G is a weak acid (Ka = 1.6 x 10-3.) Using HP to represent this weak acid, in water the following reaction represents an equilibrium that could develop.

HP (aq) H+(aq)+ P-(aq)

Use the quadratic equation or a graphing calculator to solve for the equilibrium concentrations (that result from a 1.0 x 10-4 M solution) of each component.

(B)What is the value of Go for this reaction? (Assume 25.0oC.)

7.Barium sulfate (BaSO4) is a nearly insoluble salt used in some X-ray gastrointestinal testing. The solubility equation for the compound is: BaSO4 (s) Ba2+(aq) + SO42-(aq)

(A) Write out the equilibrium expression for this reaction in both the concentration dependent K and the thermodynamic equilibrium expression Ko.

(B) Next show the relationship between these two expressions by converting K to Ko.

(C) Using Table 6.3, determine the activity coefficients for Ba2+and SO42- ina solution that has an ionic strength of 0.02. Use these values to calculate the value of K for the reaction if Ko = 1.08 x10-10.

(D) If the system was in a NaCl solution and the ionic strength was determined to be 0.10 M, would the value of K be higher or lower than that from part C?

8. Solutions with equal concentrations of Ca(NO3)2 or Mg(NO3)2 would have equal values for . However, the value for Ca2+ is smaller than the  value for Mg2+. What causes the difference?

9. On some occasions it is useful to determine the activity coefficient () for a neutral compound. Acetic acid (HC2H3O2) is a weak acid found in vinegar. If the salting coefficient (k) for acetic acid is 0.066, what would be the activity coefficient for neutral acetic acid in a solution that was known to have an ionic strength (I) of 2.0?

10.Lactic acid tends to build up during times of physical exertion. Lactic acid only dissociates slightly in water. The Go valueat 25.0oC for this reaction is 22.1 kJ. What is the value of Ko for the dissociation reaction?

SOLUTIONS TO TEST BANK FOR CHAPTER SIX:

1. (A) I = ½ (cizi2); I = ½[ 2(0.0280)(1)2 +(0.0280)(2)2] = 0.0840

(B) 0.084 is between 0.05 and 0.10 ionic strength. The value may be estimated as:

log () = = = -.114; g = 0.768 = 0.77

The approximatevalue for Cl- may also be interpolated from the table:; X = = 0.77

2 (A) I = ½ (cizi2); I = ½[ (0.031) (2)2 + (0.031)(2) (1)1] = 0.93

(B)log  = = = -.0591;  = 0.873 = 0.87

3. (A) The quadratic equation solution yields: X = [H+] = ; = = 4.86x10-3;

(B) [H+] +[HC3H5NO+] = [HC3H5NO-] + [HC3H5NO-]+[OH-]

4. (A) Adding KNO3 increases the ionic strength of the solution. This would cause the activity coefficients of Ca2+ and F- to decrease. This, in turn, would increase the solubility of CaF2.

(B) K = = [(.747)(X)][(.926)(2X)]2 = 3.3 x 10-4;X = 0.0505 M; F- = 2X; F- = 0.101 M

Ca2+(aq) + F- (aq) CaF-(aq)

F- (aq) + H+(aq) HF (aq)

Ca2+(aq) + H2O Ca(OH)+(aq) + H+(aq)

CCa2+ = [Ca2+] + [CaF-] + [Ca(OH)+]

5. COH-= [OH-] + [NiOH+]

6. (A) Using the quadratic equation will produce “X” then this value can be used as follows:

[H+] = X; [P-] = X; [HP] = 1.0 x 10-4 – X

X =; X = = 6.97x10-5

[H+] = 6.97 x 10-5 M; [P-] = 6.97 x 10-5 M; [HP] = (1.0 x 10-4) – 6.97x10-5 = 3.0 x 10-3 M

(B) Go = -RTlnK; = - 8.3124 x 298 x ln1.6 x10-3 = -15947 J = -16 000 Joules

7. (A) K = = ; Ko =

(B) Ko = Kx

(D) At a higher ionic strength, the activity coefficient values will be smaller, causing K to become larger.

8. The  value for Ca2+ and Mg2+ (and all ions) depends on the hydration layer. The charge (z value) is the same for both Ca2+ and Mg2, however, a calcium ion has a larger ionic radius than does a magnesium ion. Since the same charge is concentrated in a smaller ion, as is the case for Mg2+, it is able to attract more solvent toward itself. This creates a larger hydrated radius and a larger value for .

9. log () = k x I; log () = 0.066 x 2.0 = 0.132; 10132 =  = 1.36

10.Change kJ to J; change oC to K

Go = -RT ln Ko; 22 100 J = - (8.314J/Kmol) (298K) lnKo; 22100/(-8.3124x298) = lnKo;

-8.920 = lnKo; Ko = 1.35 x10-4

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