Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Mds99876 on March 22, 2012, 09:07:40 PM

Title: How do you find Ka1 and Ka2 for maleic acid (lab)? (i attempted it already)
Post by: Mds99876 on March 22, 2012, 09:07:40 PM

If you need more info let me know



The original equation for this is in two steps:

H2C4H2O4 (aq) + NaOH(aq) -- NaHC4H2O4(aq) + H2O(l)

NaHC4H2O4(aq) + NaOH(aq) --- Na2C4H2O4(aq) + H2O(l)

adding them gives you

H2C4H2O4(aq) + 2NaOH(aq) --- Na2C4H2O(aq) + 2H2O(l)

___




On the graph i made, i found that pka1 occured at a pH of 1.9 when 14.5mL of NaOH was added to the solution. The solution was originally made by adding 3.02g of maleic acid to 25ml of distilled water.

i know that Ka = [H][A-]/[HA]

i found [H] by doing 10^-ph, or 10^-1.9 = (0.0126)
i found [A-] because [A-] = [H], so [A-] = 0.0126

i found [HA] by saying that the concentration is 0.302g/(14.5ml+25ml) = 0.00765

So...

Ka = (0.0126)(0.0126)/(0.00765) =0.02075

Is this correct? do i follow the same process for Ka2? Also, where can i find the actual values on the internet? i need them for percent error to analyze my results.


Thanks a ton!

Title: Re: How do you find Ka1 and Ka2 for maleic acid (lab)? (i attempted it already)
Post by: Borek on March 23, 2012, 05:42:09 AM

After you add base to the solution, assumption that [A^-] = [H⁺] no longer holds.

However, you can assume neutralization went to completion - amount of A^- (actually HA^- and later A^2-) is in a simple stoichiometric relation to the amount of base added.