A diprotic acid, H,A,has acid dissociation constants of Ka1= 4.44x 10-4 and Ka2 = 4.18•10-11 Calculate the pH and molar concentrations of H A,HA-,and A2- at equilibrium for each of the solutions below. (a) a 0.193 M solution of H2AA diprotic acid, H,A,has acid dissociation constants of Ka1= 4.44x 10-4 and Ka2 = 4.18•10-11 Calculate the pH and molar concentrations of H A,HA-,and A2- at equilibrium for each of the solutions below. (a) a 0.193 M solution of H2A
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H2 A <===> H+ + HA-
m ...... Ka1 ....m x ....m x
As dissociation is too small we can say
ka = m*x^2, as 1 - x = 1 approximately..
[ H+] = m * x.. x = Dissociation constant.
HA- <====> H+ + A^-2
m*x ... Ka2. ...m*x*y....m x y
Ka2 = m*x*y^2. Ka2 is very small compared to Ka1. So y is very small.
From this equation, now [ H+ ] = m*x*y.
total [ H+] = m*x*(1+y) = m x as y <<1.
[ H+ ] = 0.193* sqrt(4.44×10^-4) M
= 0.004066 M
pH = - Log [ H+] = 2.39 approximately.
m ...... Ka1 ....m x ....m x
As dissociation is too small we can say
ka = m*x^2, as 1 - x = 1 approximately..
[ H+] = m * x.. x = Dissociation constant.
HA- <====> H+ + A^-2
m*x ... Ka2. ...m*x*y....m x y
Ka2 = m*x*y^2. Ka2 is very small compared to Ka1. So y is very small.
From this equation, now [ H+ ] = m*x*y.
total [ H+] = m*x*(1+y) = m x as y <<1.
[ H+ ] = 0.193* sqrt(4.44×10^-4) M
= 0.004066 M
pH = - Log [ H+] = 2.39 approximately.
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