Acid Base Neutralization w/ Weak Acid

[linkin06]

here's my question:

http://g.imagehost.org/0021/naoh.jpg

i was wondering why kaplan didn't take into account the Ka. my approach was to figure on the mols of OH, equate that to the moles of H+. and then since Ka = [conj base][h+]/[molar concentration]

i said the concentration of acid = [OH]^2 / (Ka)

---

Members don't see this ad.

 

[millee3]

Because its at the equivalence point, you don't need to use the Ka...since its 1:1, you would just divide the moles of acetic acid by mL used, since its asking for the initial concentration...then you'd divide by 50ml giving you your M

not sure if thats very clear, im not the best at explaining things

 

[paki20]

here's my question:

http://g.imagehost.org/0021/naoh.jpg

i was wondering why kaplan didn't take into account the Ka. my approach was to figure on the mols of OH, equate that to the moles of H+. and then since Ka = [conj base][h+]/[molar concentration]

i said the concentration of acid = [OH]^2 / (Ka)

because the Ka expression for acetic acid is Ka = [CH3COO^-][H+]/[CH3COOH]..and we dont know the molarity of [CH3COO^-]

in a neutralization reaction [OH-] = [H+] so just figure out the moles of OH- using the info given (20 mg*1g/1000mg * 1 mol/40 g = 20/(40*1000mg)) and divide it by the volume of acetic acid (50 ml *1 L/(1000mL))..the 1000s cancel each other out so you end up with (20/40)/(50)

confusing i know but hope it helps..

 

[linkin06]

but just because i found the H+/the total volume for a molar concentration...doesn't that molar concentration of acid not completely dissociate? hence i'd actually have less H+ than that?

 

[Cjc555777]

Not true. The trick when dealing with a weak acid and a strong base or a strong acid and a weak base is to do the acid base reason first. You're getting buffer's mixed up with this. When you mix these two together the first thing that happens is an acid base reaction that goes to completion. IF there was any acetic acid left over then you would use the Ka. Just remember do the acid base reaction out to completion then see if there's anything left over after the reaction is done. In this case they tell you that the reaction goes to completion, well not directly but they say basically say how much is needed to go to completion. Hope that's helpful?

 

[linkin06]

ok so i'm thinking about this some more...and wanted to check my thinking.

Ka is not a factor because as soon as a dilute amount of H+ dissociates, it gets neutralized by OH, leading to more dissociation of the acid (le chatalier), which keeps on happening until you are completely neutralized.

i hope my logic's right. i'm still kinda curious, then, how Ka could be involved in an acid/base neutralization question.

 

[Cjc555777]

PERFECT! You're thinking is now correct!!!!! Here's an example of how the Ka would be involved.

Q: You titrate 5 L of 1 M NaOH into 10 L of 1 M Acetic Acid. What is the pH of this solution. (I realize this is a LARGE amount of solution but it simplifies our numbers and is very acurate)


A:
Step one: Do the acid/base reaction. From this you start off with 5 mols OH- and 10 mols of Acetic acid. From the acid base reaction all the hydroxide is consumed and you are left with 5 mols of Acetic acid and 5 moles of it's conjugate base (CH3CCOO-)

Step 2: The problem has now turned into a semi sorta buffer problem. Set up the equalibrium equation of acetic acid. Acetic acid <--> CH2COO- + H+ that's your equilibrium equation. Then set up the good old ICE table. You have initial concentrations of Acetic acid and CH3COO- and no H+.

Step 3.Once you do the ICE table you can set up your equation Ka=[CH3COO-][H+]/[CH3COOH] and you can solve for the hydrogen concentration then figure out the pH.

That's how you do it. At first it may seem a little long but with practice it'll come quicker to you. Does that help?

 

[linkin06]

how bout henderson hasselbach?

pH would be pKa + log (dissociated already) / log (yet still in acid)

 

[Cjc555777]

Well that's the same thing. If you take Ka= [CH2COO-][H+]/[CH3COOH] and then solve it for the H+ the equation you get is [H+]=Ka[CH3COOH]/[CH2COO-]. Then you take the -log (which is what the p in all those terms means) of both sides and you get pH=pKa + log[CH2COO-]/[CH3COOH] which is the HH equation. It's just a rearrangment of the equilibrium equation.
Hopefully useful explaination it makes sense?

 

[linkin06]

yea, i think so. thanks for the help.

to be sure, pH = pKa?

 

[Cjc555777]

pH=pKa ONLY when the amount of (to stick with out acetic acid problem) CH3COO- is equal to the amount of CH3COOH. You can confirm this by using the HH equation because the ratio of those two numbers becomes 1, and the log of 1 is zero so the equation simplifies to pH=pKa. Also, I don't recommend memorizing the HH equation, you can just derive it, it's a pretty useless equation that is only used in very very specific instances. Spend your time memorizing other important or harder equations.