Question

Part A Calculate the pH of a 0.60 M H2SO3, solution that has the stepwise dissociation constants Ka1 = 1.5 x 10-2 and K 2 6.3 x 108 1.02 1.06 2.04 1.82 Submit Previous Answers Correct

I solved it myself and i got 1.06 but for some reason thats not correct. 1.02 is and i dont know how to get that

Answer 1

PH of 0.60M H2SO3

H2SO3 ========== H+ + HSO3- Ka1= 1.5*10^-2

HSO3- ========== H+ + SO3-- Ka2 = 6.3*10^-8

Because the Ka1 and Ka2 are so different need do only the first dissociation. The H+ formed from the second equilibrium ek ne depressed by that formed in the first.

[H+] from second dissociation will approximately equal to Ka2 (6.6*10^-8) , so it is negligible.

H2SO3 ========== H+ + HSO3-

Initially 0.60M 0 0

equilibrium (0.60-x) x x

Ka1=[H+] [HSO3-] / [H2SO3]

1.5*10^-2 = X * X / (0.60-x)

x​​​​​​2 = { (1.5 * 10^-2) * 0.60 }

x= (90 * 10^-4 ) ^½

X= [H+] = 9.49*10^-2 M

pH= -log [H+]

pH= - ( log 9.49 + log 10^-2)

pH= - ( 0.98 + (-2 log 10)

pH= - ( 0.98 -2)

pH= 1.02

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