Calculate the pH of sulphuricacid whose concentration is 5 x 10-3mol/lit
Answers
Answer:H2SO4is a diprotic acid, dissociating in aqueous solution in two steps.
H2SO4+ H2O ==> H3O+ + HSO4 -
HSO4 - + H2O <==> H3O+ + SO4 - -
The first dissociation is complete because pK1 = - 3. The second dissociation is partial , since pK2 = +2. If the stoichiometric molarity of the solution is C mol/L, [H+] < 2*C.
[H+] = C+x, C is due to the first dissociation and x due to the second.
[HSO4 -] =C - x , [SO4 - -] = x
K2 = [H+][SO4 - -]/[HSO4 -] = (C+x)*x/C-x
Rearrangement gives the quadratic equation
x^2 +(C+K2)*x - K2*C = 0
Solving for x gives
x = [√{(C+K2)^2 +4*K2*C} - (C+K2)]/2
In the present question , C =10^-3 and K2=1*10^-2
(pK2 =+2 leads to K2 =1*10^-2)
Substituting the values of C and K2, we get
x = 8.5*10^-4
[H+] = C+x =10^-3 +8.5*10^-4 = 1.85*10^-3
pH = - log [H+] = - log(1.85*10^-3) = 3 - log1.85
pH = 3 - 0.27 = 2.73
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