R RHC1987 10+ Year Member 5+ Year Member May 11, 2008 21 0 Status Pre-Medical May 12, 2008 #1 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...

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...

T TypeSH07 Junior Member 7+ Year Member 15+ Year Member Mar 20, 2004 76 0 Visit site Status May 12, 2008 #2 no other info is given?

scottyT Real Member 10+ Year Member Mar 19, 2006 412 0 38 Mountain View, CA Status Medical Student May 13, 2008 #3 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.

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.

Zerconia2921 Bring your A-game! 10+ Year Member Mar 24, 2008 207 0 California Status Pre-Pharmacy May 14, 2008 #4 RHC1987 said: 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... Click to expand... 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

RHC1987 said: 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... Click to expand... 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

OP R RHC1987 10+ Year Member 5+ Year Member May 11, 2008 21 0 Status Pre-Medical May 14, 2008 Thread Starter #5 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...

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...

Zerconia2921 Bring your A-game! 10+ Year Member Mar 24, 2008 207 0 California Status Pre-Pharmacy May 14, 2008 #6 RHC1987 said: 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... Click to expand... 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.

RHC1987 said: 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... Click to expand... 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.

S Scarletblack 10+ Year Member Jun 3, 2008 130 0 Status Pre-Medical Jun 8, 2008 #7 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 Last edited: Jun 8, 2008

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

M minhaj Awesomeness Incarnate 10+ Year Member 7+ Year Member Dec 3, 2007 200 2 Birmingham,AL Status Medical Student Jun 12, 2008 #8 Zerconia2921 said: Neutralization reaction requires a strong base and strong acid resulting product water plus salt. Click to expand... I am not sure if that is true. Neutralization reaction occurs regardless of the strength of the acid or the base.

Zerconia2921 said: Neutralization reaction requires a strong base and strong acid resulting product water plus salt. Click to expand... I am not sure if that is true. Neutralization reaction occurs regardless of the strength of the acid or the base.

ahs4n MS0 10+ Year Member 5+ Year Member Mar 7, 2009 74 1 Status Pre-Medical May 17, 2009 #9 Scarletblack said: 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 Click to expand... 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

Scarletblack said: 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 Click to expand... 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

F Fort 10+ Year Member 7+ Year Member Nov 25, 2008 93 0 Status Pre-Medical May 20, 2009 #10 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.

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.

ahs4n MS0 10+ Year Member 5+ Year Member Mar 7, 2009 74 1 Status Pre-Medical May 22, 2009 #11 That helps tremendously! Thank you for doing that.

G gunj122 5+ Year Member Aug 10, 2010 154 1 Status Pre-Medical Jan 11, 2011 #12 i'm sorry if i'm missing something here but how is 10^-12 * 10^14 = 10^12? shouldn't it equal 10^2? Fort said: 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. Click to expand...

i'm sorry if i'm missing something here but how is 10^-12 * 10^14 = 10^12? shouldn't it equal 10^2? Fort said: 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. Click to expand...