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In: Chemistry

Calculate the concentration of all species present (H2SO4, HSO4^-, H^+, SO4^-2) in a 0.285 M H2SO4...

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.

Solutions

Expert Solution

H2SO4 H+ + HSO4-    Ka1 = very large (about 1 x 103)

HSO4- H+ + SO4-2    Ka2 = 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

Ka2 = [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

      


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