Question

Learning Goal:

To learn how to calculate ion concentrations in an aqueous solution of a strong diprotic acid.

Sulfuric acid, H2SO4, is a strong acid. Its complete dissociation in aqueous solution is represented as

H2SO4→H++HSO4−

A HSO4− anion can dissociate further by

HSO4−⇌H++SO42−

but the extent of dissociation is considerably less than 100%. The equilibrium constant for the second dissociation step is expressed as

Ka2=[H+][SO42−][HSO4−]=0.012

Part A

Calculate the concentration of H+ ions in a 0.010 mol L−1 aqueous solution of sulfuric acid.

Express your answer to three decimal places and include the appropriate units.

[H+] =

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Part B

Calculate the concentration of SO42− ions in a 0.010 mol L−1 aqueous solution of sulfuric acid.

Express your answer to four decimal places and include the appropriate units.

[SO42−] =

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Part C

Calculate the concentration of HSO4− ions in a 0.010 mol L−1 aqueous solution of sulfuric acid.

Express your answer to two significant figures and include the appropriate units.

[HSO4−] =

Answer 1

1.0x10^-2M solution of H2SO4.

Sulfuric acid is a strong acid for it's first dissociation, which means it will dissociate completely to give us H+ and HSO4-. HSO4- will also disassociate, but not completely. We will find out how much:

initial species in solution: 1.0x10^-2M H+, 1.0x10^-2M HSO4-, H2O.

Since the initial concentration of SO4 2- is 0, we know that the equilibrium will progress in the forward direction. Let X equal the change in concentration.

..................................... HSO4-.................<--->. H+....................+ SO4 2-
Initial concentration........1.0x10^-2............ 1.0x10^-2.................. 0
Change in concentration .... -X......................... +X....................... + X
Equilibrium concentration.. 1.0x10^-2-X..........1.0x10^-2+X........... X

The Ka2 value (the Ka value for the second dissociation) is 1.2 x 10^ -2. Which means that 1.2x10^-2 = [H+][SO4 2-] / [HSO4-]

Therefore 1.2x10^-2 = [1.0x10^+2X][X] / [1.0x10^-2-X]

Use the quadratic (or the TI83 solver if your teacher lets you) to solve for X, and you get X= 4.5x10-3 (the only positive value for X).

Therefore the concentration of H+ is (1.0x10^-2) + (4.5x10-3) = 1.45x10^-2M

The concentration of HSO4- is (1.0x10^-2) - (4.5x10-3) = 5.5x10^-3M

The concentration of SO$ 2- is just 4.5x10-3M

The pH of the solution is 1.84