Question

In: Chemistry

1) One of the major problems is the large amount of phthalic acid (see your text...

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

B) (12 points) What would the pH of the pond be, assuming there are no other acid/base species present except the phthalic acid?

C) (15 points) What would the fraction of the three different acid species (H₂A, HA^- and A^2-) be in solution at that pH?

2) 15 points) 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.

Expert Solution

You say you have a 3 L tank with phtalic acid with 3M concentration so

Molarity = moles / volume for moles

moles = Molarity * volume = 3 * 3 = 9 moles of Acid

calculate volume of the pond as if it was a cylinder

volume = 3.14*r² * h = 3.14*1*1 = 3.1416 m³ or 3141 Liters, just multiply it by 1000

So the new concentration of the acid is the moles / total volume

Total volume = 3 L (from acid tank) + 3140 = 3144 Liters

Concentration = 9 / 3144 = 0.00286 M

Ka1 = 1.12x10^-3

Ka2 = 3.9x10^-6

Lets calculate dissociations with ICE chart

H₂A ========== H⁺ + HA^-

I.........0.00286....................0............................0

C...........-x.........................x...............................x

E......0.00286-x...................x..............................x

Equilibrium constant is the relation of products and reactants

K = [ products] / [reactants]

[HA] [ A] / [H2A] = Ka or

[X]*[X] / [0.00286-X] = 1.12x10^-3

x² / (0.00286 - x ) = 1.12x10^-3

x² = (0.00286 - x ) * 1.12x10^-3

x² = 3.206x10^-6 - 1.12x10^-3 x

x² + 1.12x10^-3 x - 3.206x10^-6 = 0

If we solve this equaation for x we get a value of 0.001315 for H at first dissociation and HA^-

since the Ka 2 is very small we dont need to make calculation for this one because it wont bring any significant contribution

so remember that x = [H] and x = [HA^-]

[H] = 0.001315 M

[HA^-] = 0.001315 M

[H₂A] = 0.00286 - 0.001315 = 0.001547 M

Total conc mix is 0.001315 + 0.001315 + 0.001547 = 0.0041779 M

calculate fracion

fraction [H] = 0.001315 / 0.0041779 = 0.3148

Fraction [HA^-] = 0.3148

fraction [H₂A] = 0.37

For The PH remember that you have to apply

PH = -log [H] = -log [ 0.001315] = 2.88

queen_honey_blossom answered 2 years ago

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