Question

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 M aqueous solution of sulfuric acid.

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

[H+] =

SubmitHintsMy AnswersGive UpReview Part

Part B

Calculate the concentration of SO42? ions in a 0.010 M aqueous solution of sulfuric acid.

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

[SO42?] =

SubmitHintsMy AnswersGive UpReview Part

Part C

Calculate the concentration of HSO4? ions in a 0.010 M aqueous solution of sulfuric acid.

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

[HSO4?] =

Answer 1

H2SO4 is a strong acd and hence dissociates fully

So, 0.010 moles of H2SO₄ -------------> 0.010 moles of H⁺ ions + 0.010 moles of HSO₄^-

[part C answer]

But HSO₄^- is a weak acid and hence we need Ka2 for H2SO4 [which is acid dissociation constant of 2nd proton of H2SO4] which is 1.2 * 10^-2

HSO₄^-  ----------> H⁺ + SO₄^2-

0.010 0.010 - I

-x +x +x C

(0.010 - x) (0.010+x) +x E

Ka₂ = (0.010 + x) * x

(0.10 - x)

Ignoring x in the denominator since x represents moles dissociated from weak acid HSO₄^-  which is too small

1.2 * 10^-2  = (0.010 x + x²)

0.010

0.00012 = 0.010x + x²

x² + 0.010 x - 0.00012 = 0

Solving quadratic equation we get,

x = 0.00475 M = concentration of H⁺ and SO₄^2-

so part B answer = 0.00475 M of SO₄^2-

Total concentration of H+ ions coming from H2SO4 and HSO₄^-  = (0.010 moles + 0.00475)

= 0.01475 [part A answer]