Question

There has been a major accident at the Hairum Scairum Costume Hair Company. Due to an explosion in the hair dye line, several tanks were overturned, spilling their solutions which seep out into the soil and into the pond (with a nice fountain and plastic flamingos in it) that is in front of the headquarters building. The Fire department puts out the fire and now it’s time for the clean up to begin. The company calls in the world-famous consulting chemists, Cornelius and Cordelia, to help them figure out what to do. You are part of their consulting company and are assigned the following problems to solve. Warning! Cornelius and Cordelia are sticklers for good use of sig figs and always expect you to define your assumptions.

1)     One of the major problems is the large amount of phthalic acid (see your text for K_as) managed to make its way into the pond. The Company wants to know the pH of the pond. Cornelius makes some quick measurements and finds the pond is perfectly circular and has a radius of 1.000 meter and a depth of 1.000 meter. (Remember that 1 mL = 1 cm³). The tank that held the phthalic acid was a 3.00 L tank. The concentration in the tank was 3.000M.

A) What is the formal concentration, (H₂A)_0, in the pond if all of the phthalic acid solution is added?

B) What would the pH of the pond be, assuming there are no other acid/base species present except the phthalic acid?

C) What would the fraction of the three different acid species (H₂A, HA^- and A^2-) be in solution at that pH?

2)     Another problem is the large amount of Cobalt in the vats that spilled. Cordelia has the brilliant idea that the cobalt can be removed from the pond by precipitating it out and then dredging the bottom. There are several possibilities (see your text) for this operation. Your task is to choose a precipitation agent. It has been found that the cobalt is in the +2 state. After consulting your handy textbook, you see that the possible precipitants are carbonate, sulfide and hydroxide ions. Remember that the pH of the pond is already set.

A) Based on what you know about equilibria, write the pertinent chemical reactions that would occur for each of the possible precipitants (not the precipitation reaction). Does the pH of the solution matter?

B) You take 25.00mL samples of the contaminated pond water back to the lab. There, using a Cobalt sensitive electrode, you titrate some samples with 0.01000M Na₂S, some samples with 0.01000M Na₂CO₃ and the remaining samples with 0.010000M NaOH. You collect the following data:

	Volume needed to reach end point (mL)
Titration #	0.01000M Na2S	0.01000M Na2CO3	0.01000M NaOH
1	22.12	22.32	44.46
2	22.16	22.31	44.54
3	22.25	22.28	44.39
4	22.22	22.23	44.44

Calculate the molarity of cobalt in the 25.00ml sample that was titrated in each case (make sure you are paying attention to the chemistry….write the reactions out!). Calculate the standard deviation for each set of titrations. Are the cobalt concentrations given in each case statistically different? Compare each pair of precipitants (you should have three comparisons)

3) The C&C Consulting Company determines that the ionic strength is going to play a role in the equilibria that are above. (I couldn't let you get away without an activity problem, could I??) After some measurements, they calculate the ionic strength to be 0.1000M. Recalculate the pH of the solution in Question 1. You may need to go back and look at the assumptions that were made in getting the form of the equation you previously used in order to make the correct use of the activity coefficients.

Phthalic acid:

pKa1 = 2.76; pKa2 = 4.92 at 25 deg C

Answer 1

A)

Volume of pond = π r²d

= 3.1416 * 1² * 1 = 3.1416 m3

= 3141.6 L

Total volume of pond after pthalic acid addition = 3141.6 + 3 L

= 3144.6 L

Moles of pthalic acid added = 3 M * 3 L = 9

Formal concentration of pthalic acid solution, [H₂A]₀ = Moles/ Total Volume

= 9 / 3144.6 = 0.00286 M

B)

H₂A --> HA^- + H⁺

(0.00286 – x) --> (x – y) + (x + y)

HA^- --> A^2- + H⁺

(x – y) --> y + (x + y)

Ka1 = 10^-2.76 = 0.00174

Ka2 = 10^-4.92 = 1.202 x 10^-5

Ka1 = [HA-] [H+] / [H2A]

(x – y) (x + y) / (0.00286 – x) = 0.00174

Ka2 = [A2-] [H+] / [HA-]

y (x + y) / (x – y) = 1.202 x 10^-5

Solving we get,

x = 0.001525 M

y = 1.18 x 10^-5 M

[H⁺] = x + y = 0.001537 M

pH = -log(0.001537) = 2.81

C)

[H2A] = 0.00286 – x

[HA-] = x – y

[A2-] = y

Species	Concentration, M	Fraction
H2A	0.00134	0.467
HA-	0.00151	0.529
A2-	1.18E-05	0.004