Question

Calculate the equilibrium constant K_n for the given data . Use the average volume of NaOH rather than doing the calculation for each trial.

A student titrated 5.00 mL of 50,00 mL reference solution with 0.698 M NaOH aq) The information of the reference solution is given as Chemical nitia Mols ethyl 0.101 acetato cthanol 0.258 acetic 0.062 acid H20 1.034 HCI 0.066 The titration data for 4 trials is given by A student prepared 4 samples for titration and titrated the samples with NaOH. Here are the results: Volume Trial NaOH 21.48 21.43 22.53 mL 21.39 4 Calculate the equilibrium constant Kn for the given data. Use the average volume of NaOH rather than doing the calculation for each trial

please show work!

Answer 1

For the reaction:

$CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3CH_2OCH_2CH_3 + H_2O$

the equilibrium constant Kn is given by:

Kn

Taking into account the initial moles of each of the species we can find the moles in equilibrium:

	$CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3CH_2OCH_2CH_3 + H_2O$
Initial	0.062 moles 0.258 moles 0.101 moles 1.034moles
Reaction	-x -x +x +x
Equilibrium	0.062-x 0.258-x 0.101 +x 1.034+x

To find each of the concentrations and moles in the equilibrium, we started with the titration with 0.698M NaOH, the average volume spent of the base is:

VNaOH =ー(21.48 22.53 21.39:n L-21.71mL 21.43

So the moles of the base are:

1L y * M = 21.7 * 0.698mol/L = 0.015moles 1mL * rNao 11 = 1000mL

Now, from the volume of base spent in the titration obtain the moles of acetic acid and HCl in the equilibrium:

$\eta_{NaOH}= \eta_{CH_3COOH} + \eta_{HCl}$

$\eta_{CH_3COOH} + \eta_{HCl}= 0.015moles$

HCl acts as a catalyst in the reaction and does not intervene in equilibrium. Starting from the initial moles of HCl and ethyl acetate, we can calculate the moles of NaOH if hydrolysis occurs at 100%:

$\eta_{CH_3CH_2OCH_2CH3} + \eta_{HCl}= \eta_{NaOH (T)}$

NaoH(T)(0.101 0.066)moles 0.167moles

For the difference between the moles of NaOH if the hydrolysis occurs at 100% and the moles of NaOH spent in the titration, we can obtain the moles of ethyl acetate in the equilibrium:

$\eta_{CH_3CH_2OCH_2CH_3} = \eta_{NaOH (T)}- \eta_{NaOH}= (0.167-0.015)moles=$

$\eta_{CH_3CH_2OCH_2CH_3} = 0.152moles$

Therefore the value of x is:

(0.152-0. 101)772 oles 0.051772 0les

So the moles in the balance of each of the species are:

$\eta_{CH_3COOH}= (0.062-0.051)moles= 0.011moles$

IC, H,CH,OH-(0.258-0.051)Inoles = 0.207moles

$\eta_{CH_3CH_2OCH_2CH_3} = 0.152moles$

$\eta_{H_2O}= (1.034-0.051)moles= 1.085moles$

Finally the equilibrium constant Kn for the reaction is:

Kn 7t nCHscooH nCHsCH20H0.011 0.207