Question

What is the pH of a 0.305 M solution of H₂SO₄? K_a2 =1.20 x 10^-2

Answer 1

H2SO4 undergoes dissociation in two stages H2SO4----> H+ +HSO4- and this reaction is complete and hence [H+]=[HSO4-] =0.305M

For the second stage HSO4-- ---> H+ +SO4-2

intial concentrations : HSO4- =0.305 [H+] =0.305 [SO4-2]=0

let x= concentration that of HSO4- that has been reduced to reach equilibrium

Equilibrium   HSO4- : 0.305-x           [H+]= 0.305+x and [ SO4-2] =x

K= (0.305+x)*x/ (0.305-x) = 1.2*10^-2

This can be solvend by trial and error method by assuming a value of x <0.305 and matching LHS and RHS

let x=0.2, LHS= (0.305+0.2)*0.2/0.105= 0.505*0.2/0.105=0.96>1.2*10-2

Assmue x=0.1 then LHS= 0.405*0.1/0.205=0.19>1.2*10-2

Assumed x= 0.09   LHS= (0.305+0.09)*0.09/(0.305-0.09)= 0.1

assumed x=0.008 LHS =(0.305+0.008)*0.008/(0.305-0.008) =0.0084

assumed x= 0.011 LHS= (0.305+0.011)*0.011/(0.305+0.011)= 0.011

let x= 0.012 LHS= 0.012

Hence x=0.012M

Tota H+= 0.305+0.012= 0.317M

pH= -log (H+)= -log (0.317)= 0.5