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What is the pH of an aqueous solution of 0.10 M NaHC(3)H(2)O(4)?

Answer: 10.9

What I did:

NaHC(3)H(2)O(4) + H(2)O--><--
Na + H(2)C(3)H(3)O(4) + OH

H(2)C(3)H(3)O(4) is Malonic acid with a Ka value of 1.5*10^(-3)

initial-change-equilibrium (i.c.e) table results: (x^2)/(0.10-x)

x= 8.16*10^(-7)

[OH] = x
pOH = -log(x)
pH = 14-pOH
pH = 7.91

not sure what I did wrong.

  • chemistry -

    I don't believe 10.9 is the correct answer.
    If this is malonic acid, then NaHmal is the half neutralized salt of H2Mal and as such behaves as an amphiprotic salt. For simplified solutions the pH is given by 1/2(pK1 + pK2). For solutions that may not meet the criteria for that equation, one may use
    (H+) = sqrt[(k2*C + Kw)/(1+ term)] where
    term = C/k1
    I worked the problem both ways using 2.83 for pk1 and 5.69 for pK2 and obtained pH = 4.26 for both equations. For what it's worth, your math is right but this is not the simple hydrolysis of a base HC3H2O4^- which is the basis of your solution.

  • chemistry -

    After thinking about this for 2-4 hours, are you sure this is malonic acid? Malonic acid is H2C3H2O4 and the salts would be NaHC3H2O4 and Na2C3H2O4.

  • chemistry -

    I guess not; I thought it was, but I could be wrong. If it's not malonic what do you think it is?

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