A 0.26 M solution of a diprotic acid, itaconic acid, has an [H+] of 6.0 10-3 M. What is the value of K1?
I got answer like 2.6 10-1 ,I am not realy sure if u help me
C5H6O4--> C5H5O4 + H+

K1 is constant

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  1. I suppose we make the assumption that the second ionization (k2) doesn't contribute to the H^+. To help with typing, lets call tnis diprotic acid H2T.
    H2T ==> H^+ + HT^-

    K1 = (H^+)(HT^-)/(H2T)

    Substitute 6.0 x 10^-3 fror H^+ and for HT^-, and substitute 0.26-0.006 for H2T, solve for k1. My answer is 1.4 x 10^-4 but check my work. Check my arithmetic.

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