CHEMISTRY 31

Fall, 2017 - Dixon

Homework Set 3.1 Solutions

Ch. 23: 28, 32, 44

28. Which of the following columns will provide:

a) the highest number of plates?

Col 2 (biggest N value)

b) the greatest retention?

Col 3 (largest k value)

c) highest relative retention?

Col 3 (largest a value)

d) best separation?

Col 4 (only one with baseline resolution)

Column 1: N = 1000; k₂ = 1.2; a = 1.16; resolution = 0.6

Column 2: N = 5000; k₂ = 3.9; a = 1.06; resolution = 0.8

Column 3: N = 500; k₂ = 4.7; a = 1.31; resolution = 1.1

Column 4: N = 2000; k₂ = 2.4; a = 1.24; resolution = 1.5

32. Chromatograms of compounds A and B were obtained at the same flow rate with two columns of equal length. [Refer to plot in text]

(a) Which column has more theoretical plates?

Column 1 (peaks eluting at same times are narrower in column 1)

b) Which column has a larger plate height?

Column 2 H = L/N. Since the length is the same and the N for col. 1 is larger, H for col. 2 will be larger

c) Which column gives higher resolution?

Column 1 (there is better separation between peaks)

d) Which column gives a greater relative retention?

There is no difference (since the retention times and the mobile phase elution time are the same for both columns)

e) Which compound has a higher retention factor?

Compound B (elutes later)

f) Which compound has a greater partition coefficient?

Compound B (elutes later)

g) What is the numerical value of the retention factor of peak A?

k = (8 – 1.3)/1.3 = 5.2

h) What is the numerical value of the retention factor of peak B?

k = (10 – 1.3)/1.3 = 6.7

i) What is the numerical value of the relative retention?

a = 6.7/5.2 = 1.30

44. A chromatogram with ideal Gaussian bands has t_r = 9.0 min and w_1/2 = 2.0 min.

a) How many theoretical plates are present?

N = 5.55(t_r/w_1/2)² = 5.54(9/2)² = 110 – an excellent column, if made in the 1950s.

b) Find the plate height if the column is 10 cm long.

H = 10 cm/110 = 0.089 cm = 0.89 mm

Ch. 8: 1, 4, 5a, b, 10, 13

1. Explain why the solubility of an ionic compound increases as the ionic strength of the solution increases (at least up to ~0.5 M).

The solubility increases because the ions dissolving in solution are stabilized by the presence of other ions (due to lower energy of ion-ion interactions as opposed to ion-dipole interactions).  Alternatively, (mathematical approach), as the ionic strength increases, activity coefficients decrease, requiring increases in concentrations so that the equilibrium constant stays the same.

4. Calculate the ionic strength of a) 0.0087 M KOH and b) 0.0002 M La(IO₃)₃ (assuming complete dissociation at this low concentration).

a) μ = 0.5*(0.0087*1 +0.0087*1) = 0.0087 M

b)      0.0002 M La(IO₃)₃  = 0.0002 M La³⁺ + 0.0006 M IO₃^-

μ = 0.5*[0.0002*(3)² + 0.0006*1] = 0.5(0.0018+0.0006) = 0.001 M

5. Find the activity coefficient of each ion at the indicated ionic strength:

Note this problem can be "solved" by looking up values in Table 8-1 or by using the Debye-Hückel equation.  The solutions using the Debye-Hückel equation are shown below.

a)      SO₄^2- m = 0.01 M

logg = -[0.51*(-2)²*(0.01)^0.5]/[1 + 400*(0.01)^0.5/305] = -0.204/1.13 = -0.1805

g = 0.660

b)   Sc³⁺ m = 0.005 M

logg = -[0.51*(+3)²*(0.005)^0.5]/[1 + 900*(0.005)^0.5/305] = -0.325/1.21 = -0.268

g = 0.540

10. Find [Hg₂²⁺] in saturated Hg₂Br₂ in 0.001 00 M KBr.

If we start with the approach we learned in Chapter 6, we first set up an ICE table:

Hg₂Br₂ (s) ↔ Hg₂²⁺ + 2Br^-

Initial 0 0.00100

Change +x +2x

Equilibrium x 0.00100 + 2x ~ 0.00100 if 0.00100 >> 2x

Next we normally would use the equilibrium equation using concentrations:

K_sp (Hg₂Br₂) = 5.6 x 10^-23 = [Hg₂²⁺][Br^-]²

However, this ignores the solution activity which makes the equation:

K_sp (Hg₂Br₂) = 5.6 x 10^-23 = g(Hg₂²⁺)[Hg₂²⁺]g(Br^-)²[Br^-]² = g(Hg₂²⁺)(x)g(Br^-)²(2x)²

The activity coefficients can be solved by initially assuming that the Hg₂²⁺ and Br^- originating from dissolution of Hg₂Br₂ is small.

m = 0.5*(0.001*1+0.001*1) = 0.00100 M

g(Hg₂²⁺) = 0.867; g(Br^-) = 0.964 (Values from Table 8-1)

Now, solving for x, 5.6 x 10^-23 = g(Hg₂²⁺)(x)g(Br^-)²(0.00100 M)² = 0.867(0.964)²(1.00 x 10^-6x)

x = [5.6 x 10^-23/(0.867*.9293*1.00 x 10^-6)] = 7.0 x 10^-17 M

Note: since x and 2x are much less than 0.00100 M, both assumptions are true (that 2x can be neglected and that Hg₂²⁺ and Br^- originating from dissolution of Hg₂Br₂ is negligible for calculating ionic strength).

13. Using activities, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO₃. What would be the pH if you neglected activities?

Following dissociation, we expect [Na⁺] = [OH^-] = 0.010 M and [Li⁺] = [NO₃^-] = 0.0120 M

m = 0.5[(0.010 M)(+1)² + (0.0120)(+1)² + (0.010)(-1)² + (0.0120)(-1)²] = 0.022 M

pH + pOH = pK_w or pH = pK_w – pOH = -log(1.0 x 10^-14) – [-log(g(OH^-)[OH^-])

using the Debye-Hückel Equation, we can calculate g(OH^-):

-log(g(OH^-)) = -0.51(1)(0.022)^0.5/[1 + (350)(0.022)^0.5/305] = -0.07565/1.170 = -0.06464

or g(OH^-) = 0.862

pH = 14 – log[(0.862)(0.010)] = 11.94 (vs. 12.00 if no activity is considered)