Calculate the equilibrium constant for a reaction, USING individual ionization c

ID: 873980 • Letter: C

Question

Calculate the equilibrium constant for a reaction, USING individual ionization constants.

The reaction is H2PO4^- + HS^- <--> HPO4^2- + H2S

The first question was to identify conjugate acid-base pairs, which was easy.

But the next question asked to calculate the EQUILIBRIUM CONSTANT for the reaction, using individual ionization constants.

Listed were:

H3PO4: Ka1 = 7.5 x 10^-3, Ka2 = 6.2 x 10^-8, Ka3 = 4.2 x 10^-13

H2S: Ka1 = 9.5 x 10^-8, Ka2 = 1.0 x 10^-19

My question is - how can you solve for an equilibrium constant while using acid-dissociaion constants?

My professor said the setup was Koverall = Ka2 (H3PO4) x (Kw/Ka1(H2S)) =

Can anyone explain this to me and simply state how he got to the Koverall equation?

I'd thought it would've been K = ([HPO4^2-][HSS]) / ([H2PO4^-][HS^-]), but now I'm confused.

H2PO4- + HS- <--> HPO42- + H2S

The equilibrium constant is

Keq = ([HPO42- ] [H2S])/([ H2PO4- ] [HS-]) (you had a good start!)

Using the expressions of each Ka, replace:

[HPO42- ]/ [ H2PO4- ] by Ka2,H3PO4 / [H+] and

[H2S]/ [HS-] by [H+]/Ka1,H2S

You will obtain

Keq = (Ka2,H3PO4 / [H+]) ([H+]/Ka1,H2S) =

= Ka2,H3PO4/ Ka1,H2S =

= 6.2 x 10^-8 / 9.5 x 10^-8 = 0.65

Note:

The suggested answer Ka2 (H3PO4) x (Kw/Ka1(H2S)) is not correct.

But you can write also Keq = Ka2,H3PO4

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