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Can someone please check this please - urgent!

Can someone check this please:

A dibasic acid (H2A), of concentration 1moldm(-3), dissociates in water and it is found that both protons have the same tendency to dissociate from the initial molecule. Associated with this dibasic dissociation is an equilibrium constant of 4.78 * 10-3 mol dm-3.

1 - Calculate [H+](aq) and [A-] (aq) when equilibrium is attained.

my effort

Since both have same tendency - I have assumed ka1 and ka2 both to be 4.78*10-3.

Thus (1st dissociation)

[H+][A-]/[H2A] = 4.78*10-3

4.78*10-3 = [x]^2)/1

Hence [x] = root (4.78*10-3.)

[x] therefore = 0.069

2nd dissociation)

[HA]= [A-] = 0.069

therefore

4.78*10-3 = [x]^2)0.069
[x]^2 therefore = 3.24*10-4
[x] = root Ans
[x] = 0.018

therefore [H+] total = 0.069 + 0.018
Therefore [H+] = 0.087 at equilibrium

2) - calculate pH

= -Log(10) 0.087
pH = 1.06

Is this right?? since 0.069 is over 5% of 1M - is this valid?

Thank you :smile:
I would have said H2A <==>2H+ + A-
Ka = [H+]2[A-]/[HA]

4.78*10-3 x 1 = [H+]2[A-]

but [A-] = [H+]/2

therefore
4.78*10-3 x 2 = [H+]^3
[H+] = cube root[9.56*10-3]
[H+] = 0.212 M
Reply 2
Thank you, I'll probably put both :biggrin:

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