Posted by
**Jason** on
.

How do I construct an ICE table for this question.

CH3COOH + NaOH --> CH3COONa + H2O

Fill in the equilibrium line of the table. [Think about the initial pH and think about the pH at the end point of the titration.

Initial pH 2.72

When the solution turned pink pH was 8.49

I forgot to add that the

concentration of acetic acid is 0.2M and volume is 25ml

the concentration of NaOH is 0.2M and the volume is 50mL

Is that all of the data in the problem? If so, the pH = -log(H^+). Use that and ionization of CH3COOH to calculate the concn of CH3COOH at the beginning.

CH3COOH ==> CH3COO^- + H^+

pH at the equivalence point is 8.49. Similarly, determine (H^+) and from that determine (OH^-).

Knowing that the pH at the equivalence point is due to the hydrolysis of the acetate ion,

CH3COO^- + HOH ==> CH3COOH + OH^-

Kb = Kw/Ka = (OH^-)^2/(CH3COO^-). Solve for CH3COO^- (the salt).

Knowing beginning (CH3COOH) and the CH3COO^- at the equivalence point will allow you to determine (NaOH).

I think you need the above data before you can construct and ICE table. But perhaps there is more to the problem than that. Post your work if you get stuck.

I guess all my typing was futile.

Determine #mols CH3COOH

Detemine #mols NaOH.

Then

CH3COOH + NaOH ==>CH3COONa + HOH

initial --you can fill in.

change-- the smaller number of mols will be the mols CH3COONa and mols HOH formed. That same number will be subtracted from the beginning for the reactants.

equilibrium--add the intial to change to find equilibrium.

Post your work if you get stuck.

# of mols for acetic acid = 5mols

# of mols for salt =10mols

CH3COOH + NaOH ==>CH3COONa + HOH

I 0.2M 0.2M 0 0

C -x -x ? ?

E 0.2-x 0.2-x ? ?

do I placed 5mols (since it is the smaller numebr of mols) where CH3COONa and HOH are?

CH3COOH + NaOH ==>CH3COONa + HOH

OK. First, you didn't change volume to liters. You can make a chart on paper but I can't type it onto the screen. It is difficult to leave spaces.

Initially:

CH3COOH = 0.025 x 0.2 M = 0.005 mol.

NaOH = 0.05 L x 0.2 M = 0.01 mol,

CH3COONa = 0

HOH = 0

change:

CH3COOH is the limiting reagent because that is the smaller one. Therefore,

mols CH3COONa formed = 0.005

mols HOH formed = 0.005

mols taken from CH3COOH = 0.005

mols taken from NaOH = 0.005

equilibrium: add the columns.

CH3COONa = 0 + 0.005 mol = 0.005

HOH = 0 + 0.005 = 0.005

CH3COOH = 0.005 - 0.005 = 0 mol

NaOH = 0.01 - 0.005 = 0.005 mol

Therefore, the solution consists of 0.005 mol CH3COONa (sodium acetate) + 0.005 mol NaOH (the amount in excess) after the reaction is complete.

If you want concentrations, divide each of the mol values by 0.075 liters (25 mL + 50 mL = 75 mL). I'm not sure this is exactly what you wanted. If not, clarify and explain what you need.

yes thank you very much, I understand it