Question

C2H5OH (l) + CH3COOH (l) ↔ CH3COOC2H5 (l) +H2O(l) is the equation given, I am trying to find Kc?

C2H5OH (l) + CH3CO2H (l) ↔ CH3CO2C2H5 (l) +H2O(l)
3.0g of ethanoic acid and 2.3g of ethanol were equilibrated at 100 degrees C for an hour and then quickly cooled in an ice bath. 50 cm3 of 1.0 mol dm-3 aqueous sodium hydroxide were added. When the mixture was titrated with 1.0 mol dm-3 hydrochloric acid, 33.3 cm3 of acid were required.

Why was the mixture cooled in an ice bath and how do you find Kc with this information?

Answer 1

Warning! Long Answer. `K_text(c) = 4.0`

Explanation:

The equation for the equilibrium is

`"CH"_3"COOH" + color(white)(m)"C"_2"H"_5"OH"color(white)(ll)⇌ "CH"_3"COOC"_2"H"_5 + "H"_2"O"` or
`color(white)(ml)"HOAc" color(white)(m)+ color(white)(mm)"EtOH"color(white)(ml) ⇌ color(white)(mm)"EtOAc" color(white)(mll)+ "H"_2"O"`

Step 1. Calculate the moles of starting materials

`"Moles of EtOH" = 2.3 color(red)(cancel(color(black)("g EtOH"))) × "1 mol EtOH"/(46.07 color(red)(cancel(color(black)("g EtOH")))) = "0.0499 mol EtOH"`

`"Moles of HOAc" = 3.0 color(red)(cancel(color(black)("g HOAc"))) × "1 mol HOAc"/(60.05 color(red)(cancel(color(black)("g HOAc")))) = "0.0500 mol HOAc"`

Step 2. Calculate the moles of `"HOAc"` at equilibrium

Once the mixture has reached equilibrium, you cool it in an ice bath to prevent further reaction with the water you are about to add.

(a) Calculate the moles of `"NaOH"` added.

`"Moles of NaOH" = 0.050 color(red)(cancel(color(black)("dm"^3color(white)(l) "NaOH"))) × "1.0 mol NaOH"/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "NaOH")))) = "0.0500 mol NaOH"`

Some of the `"NaOH"` reacted with the `"HOAc"`, and the rest of it was in excess.

You titrated this excess with `"HCl"`

(b) Calculate the excess moles of `"NaOH"`.

The neutralization reaction was

`"NaOH + HCl → NaCl + H"_2"O"`

`"Excess moles NaOH" = 0.0333 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HCl"))) × (1 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "HCl")))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.0333 mol NaOH"`

(c) Calculate the moles of `"NaOH"` that reacted.

`"Moles of NaOH reacted" = "(0.0500 - 0.0333) mol NaOH" = "0.0167 mol NaOH"`

(d) Calculate the equilibrium moles of `"HOAc"`

The `"NaOH"` reacted with the `"HOAc"` that was present at equilibrium.

∴ `"Equilibrium moles of HOAc = 0.0167 mol"`

Step 3. Calculate the equilibrium constant

`color(white)(mmmmml)"HOAc"+ "EtOH" ⇌ "EtOAc" + color(white)(ll)"H"_2"O"`
`"I/mol": color(white)(mll)0.0500 color(white)(mll)0.0499color(white)(mmml)0color(white)(mmmml)0`
`"C/mol": color(white)(m)"-0.0333"color(white)(ml)"-0.0333"color(white)(ml)"+0.0333"color(white)(ll)"+0.0333"`
`"E/mol": color(white)(ml)0.0167color(white)(mll)0.0166color(white)(mml)0.0333color(white)(mll)0.0333`

`K_text(c) = (["EtOAc"]["H"_2"O"])/(["HOAc"]["EtOH"]) = (0.0333 × 0.0333)/(0.0167 × 0.0166) = 4.0`