[Nav]

Hello all,

I have this problem that has got me stumped. Here it is:

The question is asking for the molarities of Ch3COOH and CH3COO- in Buffer A. Here is how Buffer A is made:

1)  20.00ml f 0.100M acetic acid and 20.00ml of 0.100M NaOH are combined
2) Combine 20.00ml of 0.100M acetic acid with 20.00ml of distilled water.
3) Then the solutions made in steps one and two are combined.

I know how to calculate a new molarity with a stock solution diluted with water but I'm not too sure how to deal with step 1. Thanks in advance.

[Yggdrasil]

You need to consider the reaction between acetic acid and sodium hydroxide.  Here, a balanced chemical equation will be useful.

[Nav]

My apologies, here it is:

CH₃COOH + NaOH --> NaCH₃COO + H₂O


CH₃COO^-(aq) + H₂O(l) --> CH₃COOH (aq) + OH^- (aq)

and the other equation given is:

CH₃COOH (aq) + H₂0 (l) --> H₃0⁺  (aq) + CH₃COO^- (aq)

I'm gonna try and take another stab at it. Thanks again.

[Arkcon]

For 1). you'll need to show the reaction, and the conc. of the products.  The last two reactions are kinda MacGuffin's -- technically important, but not really in this case.

For 2). you need to show the final concentration.

For 3). after mixing them you'll use the pKa to determine the pH after mixing

[marvelteen]

Hello

I am also currently having difficulties with this problem.
Could someone please elaborate a bit on the steps required. I'm really lost.
Thanks in advance!

[edit] ok so i set up an ICE table and i currently have the following values,

        CH3COOH  +  NaOH  --> CH3COO  +  H2O
I          0.067M       0.1M             0            0
C            -x             -x               +x         +x
E            0.067-x     0.1-x            x            x

am i on the right track?

[Nav]

That's the same ICE table I got as well..

[Nav]

Ok, I tried this different approach; I abandoned the ICE table as well. Here is what I did:

For Step 1) Calculated the # of moles of acetic acid. I got 0.002mol. Then I divided that by 0.04L (which is 40ml converted to litres) to get 0.05mol/L

For Step 2) I did the same thing by using M1V1=M2V2 to get 0.05mol/L

For Step 3)Then I added 0.05mol/L with 0.05mol/L to get 0.1mol/L. Then I assumed 1L and multiplied that by 0.1mol/L (to cancel out the Litres).

Finally, divided 0.1mol/ by 0.08L (which is 80ml converted to litres) to get a final answer of 1.25mol/L

I'm fairly certain this is incorrect but I gave it yet another shot. Any feedback would be greatly appreciated. Thanks in advance.

[LQ43]

Connect these ideas:

You need to consider the reaction between acetic acid and sodium hydroxide.  Here, a balanced chemical equation will be useful.

CH₃COOH + NaOH --> NaCH₃COO + H₂O

The question is asking for the molarities of Ch3COOH and CH3COO- in Buffer A. Here is how Buffer A is made:

1)  20.00ml f 0.100M acetic acid and 20.00ml of 0.100M NaOH are combined
2) Combine 20.00ml of 0.100M acetic acid with 20.00ml of distilled water.
3) Then the solutions made in steps one and two are combined.

Consider the components of the buffer as above

What compound is being prepared in 1)? in 2)?

keep thinking about moles and total volume

0.08L (which is 80ml converted to litres)