Question

9. Sulfuric acid (H2S04), a diprotic acid, is a strong acid in its first deprotonation (deprotonation- loss of H'). The first deprotonation is considered complete (complete ionization). It is, however considered a weak acid for its second deprotonation with a Ka value of 1.2 X 102, For a 0.0456 M H2S04(aq) solution determine: (16 pts total) a. The chemical reaction equations for the successive loss of protons. (4 pts b. The concentration of HSO4-1 and H (H30') after the first deprotonation.(4pts) c. The concentration of So4 2andH'(H0') after the second deprotonation. (4 pts) d. Determine pH of the solution after the first deprotonation. (4 pts)

Answer 1

9 . sol:-

(a)

First deprotonation of H₂SO₄ is :

H₂SO₄(aq) ------------------> H⁺(aq) + HSO₄^-(aq)

second deprotonation of H₂SO₄ will be partial because it is more difficult to remove one hydrogen from an anion i.e HSO₄^-(aq) as :

HSO₄^-(aq) <-----------------> H⁺(aq) + SO₄^2-(aq)       , k_a = 1.2 X 10^-2 (given )

(b)

after the first deprotonation the concentration of HSO₄^- and H⁺ are equal to concentration of H₂SO₄ i.e 0.0456 M , because H₂SO₄ is the strong dibasic acid and therefore its first deprotonation or dissociation will be complete i.e H₂SO₄ has very high value of dissociation constant ( k_a).

H₂SO₄(aq) ------------------>                 H⁺(aq)           +              HSO₄^-(aq)

0.0456 M                         0.0456 M                                0.0456 M

Hence, [H⁺] = 0.0456 M

and [HSO₄^-] = 0.0456 M

(c)

concentration of SO₄^2- and H⁺ after the second deprotonation can be calculated by using ICE table .

because ,value of k_a for HSO₄^- is small , therefore its dissociation will be partial as :

       HSO₄^-(aq) <----------------->                H⁺(aq)       +        SO₄^2-(aq)       , k_a = 1.2 X 10^-2 (given )

I       0.0456 M                                        0.0456 M                  0

C          -x                                       +x             +x

E      (0.0456 - x) M                             (0.0456 + x ) M       x M

Now , the expression of k_a for HSO₄^- will be :

k_a =[ H⁺] [ SO₄^2-] / [ HSO₄^-]

1.2 X 10^-2 = (0.0456 +x ) x /(0.0456 - x)

on cross multiplication , we have

1.2 X 10^-2 (0.0456 - x) = (0.0456 +x)x

5.47 X 10^-4 - 0.012x = 0.0456x + x²

as x is very very small , therefore neglect x²

0.0456x + 0.012x = 5.47 X 10^-4

0.0576x = 5.47 X 10^-4

x = 5.47 X 10^-4 / 0.0576

x = 0.0095

hence , [ SO₄^2-] = x = 0.0095 M

and [ H⁺] = 0.0456 + x = 0.0456 + 0.0095 = 0.055 M

(d)

From first dissociation , we have [ H⁺ ] = 0.0456 M

because pH = - log [ H⁺]

therefore , pH = - log 0.0456

pH = - ( - 1.34 )

pH = 1.34