What is the concentration of [H+] in a 0.050 M solution of acetic acid? The Ka v

Question

What is the concentration of [H+] in a 0.050 M solution of acetic acid? The Ka value for acetic acid is 1.8 Times 10 -5 . What is the solubility of barium sulfate in a solution containing 0.020 M sodium sulfate? The Ksp value for barium sulfate is 1.1 Times 10 -10 Aluminum metal is refined in a very energy-intensive process from bauxite ore using he Hall process, which requires electrolysis of aluminum oxide, AI2O3, in molten cryolyte it more than 1012 degree C. In an industrial electrolysis cell, 1900 kg of aluminum oxide was educed with a current of 1500 amps. How long did it take to reduce all the reactant to aluminum metal? (Amp = C/t) 1 Faraday = 96500 C/mole Phosphorus-32 is a radioactive isotope used as a tracer in the liver. How much Phosphorus-32 was originally used if there is only 3.50 mg left in a sample after 288 h? The half-life of phosphorus-32 is 14.3 days.)

Explanation / Answer

1) HC2H3O2 = H+ + C2H3O2-

     0.05- x       x         x              at equilibrium

Ka = [H+] [C2H3O2-] / [HC2H3O2] = 1.8 X 10-5

or x * x / 0.05 - x = 1.8 X 10-5

or x2 / 0.05 = 1.8 X 10-5       x is much smaller than 0.05, so can be neglected in denominator

or x2 = 0.09 x 10-5

or x = 0.000949 = 9.49 x 10-4 M = [H+]

2) BaSO4(s) <==> Ba2+(aq) + SO42-(aq)

    Ksp = [Ba2+][SO42-] = 1.1 x 10-10

    = [x][0.020 + x]     0.02 comes from common ion effect

   = [x][0.020]       neglecting x

or x = 5.5 x 10-9 M = [Ba+2] and [SO4-2] = 0.020 M

3) MW of Al2O3 = 101.96 g/mol

So 1 Faraday = 96500 Coulomb reduces 101.96 g of the oxide.

So to reduce 1900 kg = 19 x 105 g of the oxide, charge required = (96500 / 101.96) * 19 x 105

= 17983 x 105 Coulomb

time (sec) = charge (C) / current (amp)

= 17983 x 105 / 1500

= 11.99 x 105 sec

= 333 hours

4) t1/2 = 14.3 days

or 0.693 / k = 14.3

or k = 0.0485 day-1

N = No e-kt

or No = N / e-kt = N * ekt = 3.5 * e (288/24) * 0.0485 = 3.5 * e 0.582 = 3.5 * 1.7896 = 6.26 mg

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