When the reaction equation is:
HgBr2(s) ↔ Hg2+(aq) + 2Br-(aq)
So Ksp expression = [Hg2+] [Br-]^2
assume the solubility S = X = 2.66 x 10^-7 M
and from the reaction equation :
we can see that [Hg2+] = X
and the [Br-] = 2 X
so by substitution in Ksp formula will can get the Ksp value:
∴ Ksp = X * (2X)^2
= 2.66 x 10^-7 * (2*2.66 x 10^-7)^2
= 7.53 x 10^-20
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First, we have to know that we can put [H+] instead of [H3O+]
so, according to the reaction equation:
by using ICE table:
H2O ↔ H+ + OH-
initial 0 0
change +X +X
Equ X X
when Kw = [H+] [OH-]
and when we have Kw = 2.4 x 10^-14
and when [H+] = [OH-] = X
∴ 2.4 x 10^-14 = X^2
∴ X = √(2.4 x 10^-14)
= 1.55 x 10^-7
∴[H+] = 1.55 x 10^-7
