What is the Keq of the neutralization reaction between oxalic acid (H2C2O4; KA1 = 5.9 × 10^-2) and sodium hydroxide? ANSWER: 5.9 x 10^12 Can't seem to get this one...

I'm just taking a guess here but I think this is going on: H2C2O4 + NaOH --> H2O + HC2O4 Ka=5.9*10^-2 Kw=1*10^-14 We know Ka*Kb=Kw. If you have achieved neutralization then you can reword the question as follows: What is the Kb of the base dissociation reaction? HC2O4 + H2O --> H2C2O4 + NaOH Please correct me if I'm wrong but I think this should help.

Neutralization reaction requires a strong base and strong acid resulting product water plus salt. Ka = 5.9x10^-2 Kb= x Kw= 1.0x10-^14 Solve for Kb 1.0x10-^14/5.9x10^-2= 5.9x10-^12

the answer to this question is 5.9x10^+12 not 5.9 x10^-12... which is what threw me off.. i originally did it your way too...

Raised to the power of positive 12? I am thinking about it and if the Kb was positive power than means its a VERY VERY VERY strong base so it will completely dissocate so I am not sure if that is an typo error or what. Sorry i wasnt helpful.

Uh well here's my attempt H2C2O4 + NaOH --> H2O + HC2O4 so [HC2O4-][H20] ................. (Call this eq 1) ----------------- = Keq [-OH][H2C2O4] ok but [HC2O4-][H+] ---------------- = Ka = 5.9*10^-2 [H2C2O4] You can do algebra to get: 5.9*10^-2.....[HC2O4-] (2) ------------ = ----------- [H+]............ [H2C2O4] Substitute (2) into (1), you get: 5.9*10^-2 [H2O] ------------------- = Keq [-OH] [H+] But at equilibrium [H+][OH-]/[H2O] = 10^-14 so 5.9*10^-2 ------------ = 5.9 * 10^12 10^-14

I am not sure if that is true. Neutralization reaction occurs regardless of the strength of the acid or the base.

bringing back an old thread... does this look right.. I'm trying to figure this problem out from the Kaplan FL 9 on the PS problem 12

Here's another way. It's a little less mathematical, but you still arrive at the final answer. We know the dissociation constant for oxalic acid to be 5.9e-12. So we can write H2C2O4 --> HC2O4- + H+ We want to get the final equation to look like H2C2O4 + OH- --> HC2O4- + H2O Well, we can actually add another equation to the first one: H2C2O4 --> HC2O4- + H+ K1 = 5.9e-12 OH- + H+ --> H2O K2 = 1.0e14 (it's e14 because we've reversed the water dissociation reaction) If we add these two equations, we get the desired reaction. To find Keq, we simply multiply K1 and K2. K1*K2 = (5.9e-12)(1.0e14) = 5.9e12 Hope this helps.