Question

Calculate the concentration of all species present (H2SO4, HSO4^-, H^+, SO4^-2) in a 0.285 M H2SO4 solution. The diprotic acid equilibrium is :

H2SO4 <--> H^+ + HSO4^- Ka = very large

HSO4^- <---> H^+ + SO4^-2 Ka = 1.2 x 10^-2

Please explain why and when we use +x and -x. I am very confused on this.

Answer 1

H₂SO₄ H⁺ + HSO₄^-    K_a1 = very large (about 1 x 10³)

HSO₄^- H⁺ + SO4^-2    K_a2 = 1.2 x 10^-2

Given, concentration of H₂SO₄ = 0.285 M

For the first equibrium,

H₂SO₄ H⁺ + HSO₄^-

let x parts of H₂SO₄ get dissociated

So, from Initial change and equilibrium table or ICE Table

Particulars	H2SO4	H+	HSO4-
Initial	0.285	0	0
change	-x	+x	+x
equilibrium	0.285-x	+x	+x

-x implies that x part of the reactant gets dissociated or lost from it. Here H₂SO₄ loses x part to give x parts of H⁺   and HSO₄^- each

+x implies that x part of the products formed . Here x parts of both H⁺   and HSO₄^- are formed from x part of H₂SO4. let's say 1 unit gets dissociated from H₂SO4. So, 1 unit each of both H⁺   and HSO₄^- are formed.

+x is used when reactant gets dissociated and -x is used products are formed

now, since H₂SO₄ is a strong acid, hence its dissociation is complete,

i.e. 0.285-x = 0 or x = 0.285 M

For second reaction,

HSO₄^- H⁺ + SO4^-2

let y parts of HSO₄^- gets dissociated now

From ICE table,

Particulars	HSO4-	H+	SO4-2
Initial	x	x	0
Change	-y	+y	+y
Equilibrium	x-y	x+y	+y

K_a2 = [H⁺][SO4^-2 ]/[HSO₄^-]

1.2 x 10^-2 = (x+y)(y)/(x-y)

       = (0.285+y) y / (0.285-y)

      or, y² + 0.297 y-0.00342 = 0

y= [-0.297 {0.297² - (4*1*(-0.00342))}^1/2]/2

= (-0.297 0.319) / 2

or, y = -0.616 or 0.022

since, concentration cannot be negative, y= 0.022 M

Hence, at equilibrium,

[H₂SO₄] = 0 M,

[H⁺] = x+y = 0.285 + 0.022 = 0.307 M

[HSO₄^- ] = x-y = 0.285 -0.022 = 0.263 M

[SO4^-2] = y = 0.022 M