Citric acid H 3 C 6 H 5 O 7 isa weak polyprotic acid. It is described by the fol

Question

Citric acid H3C6H5O7 isa weak polyprotic acid. It is described by the followingseries of deprotonation reactions:
H3C6H5O7 <-> H+ +H2C6H5O7-Ka1=7.4*10^-4
H2C6H5O7 <->H+ +HC6H5O72-Ka2=1.7*10^-5
HC6H5O7 <-> H+ +C6H5O73- Ka3=4.0*10^-7

You are given an initial concentration of citric acid of 0.45M.Determine the equilibrium concentrations ofH3C6H5O7,H2C6H5O7,HC6H5O7,C6H5O73-, H+ and the pHof the solution.

I started out with an Ice table for the first equation and got theconcentration of 0.4318, and thats right but all the otherconcentrations after I got wrong, I was wondering how I solve thisand if theres an easier way than an Ice table.

Explanation / Answer

Would you mind writing your answers? I would like to compare. I amaware of a more accurate way to obtain the answer (mass and chargebalance, followed by nonlinear equation solving), but that takes along time. I would like to try something else and compare with youranswers. Ka1 = 7.4 e-4 = (x)(x)/(0.45 - x) x2 + 4.7e-4 x - 0.000333 = 0 x = 0.0574716307 = [H+] =[H2C6H5O7-] [H3C6H5O7 ] = 0.45 -0.004747 = 0.447 M Ka2 = [H+][HC6H5O72-]/[H2C6H5O7-]~ (0.0574)*[HC6H5O72-]/(0.0574) =1.7e-5 [HC6H5O72-] = 1.7e-5M Ka3 =[H+][C6H5O73-]/[HC6H5O72-]= 4e-7 4e-7 ~(0.0574)*[C6H5O73-]/(1.7e-5) [C6H5O73-] = 1.18 e-10 pH = -log[H+] = -log(0.0574) = 1.24 I made the assumption of pseudo steady. But I am not sure, so theseare not the final answers. Please provide your answers so that Ican make a comparison and then we can go from there.

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