Ethanoic acid can be reacted with alcohols to form esters, an equilibrium mixture being formed.
CH3CO2H + ROH <--> CH3CO2R + H2O
The reaction is usually carried out in the presence of an acid catalyst.
In an experiment to determine Kc a student placed together in a conical flask 0.10 mol of ethanoic acid, 0.10 of an alcohol ROH, and 0.005 mol of hydrogen chloride catalyst.
The flask was sealed and kept at 25°C for seven days.
After this time, the student titrated all of the contents of the flask with 2.00 mol/dm3 NaOH using phenolphthalein indicator.
At the end point, 22.5 cm3 of NaOH had been used.
a) i) Calculate the amount, in moles of NaOH used in the titration.
ii) What amount, in moles, of this NaOH reacted with the hydrogen chloride?
iii) Write a balanced equation for the reaction between ethanoic acid and NaOH.
iv) Hence calculate the amount, in moles, of NaOH that reacted with the ethanoic acid.
b) i) Use your results from a) to calculate the amount, in moles, of ethanoic acid present at equilibrium.
Sorry for the huge chunk of boring text, but here's what I got:
i) moles = concentration x volume, which is 2 x 22.5/1000 = 0.045 mol
ii) Ehhh, not too sure about this one, some help would be appreciated :)
iii) CH3COOH + NaOH --> CH3CO2Na + H2O
iv) Not too sure 'bout this one either
b) i) Can't work this one out
Thanks in advance for your time and help ^_^
Hey there! I'm here to help with your chemistry questions, and add a little humor along the way. Let's tackle this step by step, shall we?
a) i) Great job! You've correctly calculated the moles of NaOH used in the titration as 0.045 mol.
a) ii) Now, to find the moles of NaOH that reacted with the hydrogen chloride, remember that the reaction between NaOH and HCl is a 1:1 ratio. So, the moles of NaOH that reacted with HCl is also 0.045 mol.
a) iii) You nailed it! The balanced equation for the reaction between ethanoic acid and NaOH is CH3COOH + NaOH --> CH3COONa + H2O. Well done!
a) iv) To find the moles of NaOH that reacted with the ethanoic acid, simply subtract the moles that reacted with HCl from the total moles used in the titration. So, it would be 0.045 mol - 0.045 mol = 0 moles. The ethanoic acid didn't react with NaOH in this case.
b) i) Alrighty, let's calculate the amount of ethanoic acid at equilibrium. Since the moles of NaOH reacted with ethanoic acid is 0, all the moles of NaOH used in the titration must've reacted with the alcohol to form esters. So, the amount of ethanoic acid remaining at equilibrium is equal to the initial amount. Therefore, the amount is 0.10 mol.
I hope that clears things up! If you have any more questions, feel free to ask, and I'll be here with my clown jokes to lighten things up.
Let's break down the questions step-by-step:
a) i) To calculate the amount in moles of NaOH used in the titration, you can use the formula:
moles = concentration x volume
Given that the concentration of NaOH is 2.00 mol/dm3 and the volume used is 22.5 cm3 (convert to dm3 by dividing by 1000), you can calculate:
moles = 2.00 mol/dm3 x 22.5 cm3 / 1000 cm3/dm3 = 0.045 mol
ii) To determine the amount in moles of NaOH that reacted with the hydrogen chloride (HCl) catalyst, we need to recognize that the HCl catalyst does not participate in the chemical reaction. Therefore, all the NaOH used in the titration reacted with the ethanoic acid, not the catalyst. Thus, the moles of NaOH used in the titration is also the moles of NaOH that reacted with the ethanoic acid.
iii) The balanced equation for the reaction between ethanoic acid (CH3COOH) and NaOH is:
CH3COOH + NaOH -> CH3COONa + H2O
iv) Since the balanced equation for the reaction between ethanoic acid and NaOH is 1:1, and the moles of NaOH used in the titration is 0.045 mol (as determined in part i), then the amount in moles of NaOH that reacted with the ethanoic acid is also 0.045 mol.
b) i) To calculate the amount in moles of ethanoic acid present at equilibrium, we need to consider the stoichiometry of the balanced equation and the fact that it is an equilibrium reaction.
From the balanced equation, we can see that there is a 1:1 stoichiometric ratio between ethanoic acid and NaOH:
CH3COOH + NaOH -> CH3COONa + H2O
Since 0.045 mol of NaOH reacted with the ethanoic acid (as determined in part iv), then the amount in moles of ethanoic acid present at equilibrium is also 0.045 mol.
Therefore, the amount in moles of ethanoic acid present at equilibrium is 0.045 mol.
I hope this helps! Let me know if you have any further questions.
No problem! Let's break it down step by step:
a) i) To calculate the amount of NaOH used in the titration, you correctly used the equation moles = concentration x volume. So the moles of NaOH used would be 2 mol/dm3 x 22.5 cm3 /1000 cm3/dm3 = 0.045 mol.
a) ii) To determine the moles of NaOH that reacted with the hydrogen chloride, we need to remember that the reaction between NaOH and HCl is a 1:1 stoichiometric ratio. This means that for every mole of HCl, one mole of NaOH reacts. Since the amount of NaOH used is 0.045 mol, the moles of NaOH that reacted with HCl would also be 0.045 mol.
a) iii) The balanced equation for the reaction between ethanoic acid (CH3COOH) and NaOH is:
CH3COOH + NaOH -> CH3COONa + H2O
a) iv) The reaction between ethanoic acid (CH3COOH) and NaOH follows a 1:1 stoichiometric ratio as well. So the amount of NaOH that reacted with the ethanoic acid would also be 0.045 mol.
b) i) Now, to calculate the amount of ethanoic acid present at equilibrium, we need to consider the balance of the reaction. The initial amount of ethanoic acid was 0.10 mol, and we know that 0.045 mol of NaOH reacted with it. Since the reaction follows a 1:1 stoichiometric ratio, 0.045 mol of ethanoic acid also reacted.
Therefore, the amount of ethanoic acid remaining at equilibrium would be the initial amount minus the amount reacted: 0.10 mol - 0.045 mol = 0.055 mol.
So, at equilibrium, 0.055 mol of ethanoic acid is present.
I hope this clears things up! Let me know if you have any further questions.
i is correct at 0.045 mols NaOH.
ii. The NaOH neutralized all of the HCl (remember the catalyst doesn't change in concentration) which is 0.005 mols. This means that 0.045-0.005 = 0.040 mols of the NaOH reacted with whatever CH3COOH was left after equilibrium was attained.
iii. Your equation is correct.
iv.See my response to ii.
b.
........CH3COOH + EtOH ==> CH3COOEt + H2O
I.............0.1..............0.1...............0....................0
C.......- 0.040.........-0.040..........+0.040........+0.040
E............?.................? ? ?
Plug the E line into the Kc expression and solve for Kc.