Acids and Bases calculation problems

[kenshi64]

Homework Statement

Data for Question B:
(initial)[CH3COOH] = 0.500 mol dm–3 and) eqm [CH3COOH] = 0.200 mol dm–3;
(initial)[CH3COO–] = 0.300 mol dm–3 and) eqm [CH3COO–] = 0.300 mol dm–3;

The Attempt at a Solution

Question A) " FInd the pH of a mixture of 50.0 cm3 of 0.100 mol dm–3 aqueous ammonia and 50.0 cm3 of 0.0500 mol dm–3 hydrochloric acid solution"

The teacher told us to do this by equating [NH3] to [NH4]. I just don't get the logic! Ammonia is a weak base, so it's initial concentration in undissociated form is unchanged and stoichiometrically its products are formed in 1:1 ratios, but I haven't ever seen THIS relationship anywhere, can someone please explain this to me?

Question B) " Determine the pH of a solution formed from adding 50.0 cm3 of 1.00 mol dm–3 ethanoic acid, CH3COOH(aq), to 50.0 cm3 of 0.600 mol dm–3 sodium hydroxide, NaOH(aq)." Refer data provided.

One would assume if a weak base behaved as seen above a weak acid would too but the answer booklet implies that ethanoic acid doesn't! Which is peculiar!
[CH3C00H] DOESN'T EQUAL [CH3COO-] apparently!? And how on Earth will [CH3COO-] have a initial concentration at all?!

How does that make sense? I can understand the equilibrium concentration being hte difference between acid and base concentration, 0.3 mol dm3 makes sense, but the INITIAL concentration MUST be an error!?

Thank you a lot! :D

 

[Borek]

The teacher told us to do this by equating [NH3] to [NH4]. I just don't get the logic! Ammonia is a weak base, so it's initial concentration in undissociated form is unchanged and stoichiometrically its products are formed in 1:1 ratios, but I haven't ever seen THIS relationship anywhere, can someone please explain this to me?

This is only an approximation, but it is known to work quite well as long as the weak acid/base is not too weak nor too strong. You just assume neutralization went to completion so concentrations of the products are given by the reaction stoichiometry.

Try to apply this logic to both questions, you will see why [NH₃] = [NH₄⁺], and why [CH₃COOH] ≠ [CH₃COO^-].

 

[kenshi64]

This is only an approximation, but it is known to work quite well as long as the weak acid/base is not too weak nor too strong. You just assume neutralization went to completion so concentrations of the products are given by the reaction stoichiometry.

Try to apply this logic to both questions, you will see why [NH₃] = [NH₄⁺], and why [CH₃COOH] ≠ [CH₃COO^-].

Aah Borek! My old friend! ;)

Well I know the logic doesn't apply because the answer book says it doesn't but the first para answered by question. However this is so silly because I cannot tell with any certainty which rule I should follow when!?

 

[Borek]

What is the other rule you can follow?

You can assume if pKa (pKb) is somewhere between 3.5-10.5 (note 10.5=14-3.5) and concentration not below 10^-3M assumption about neutralization being stoichiometric is correct. Note that it covers 99% of buffers used in practice.

 

[DeeNos]

Dear student,

I understand your confusion about these problems. Let's start with Question A. In this case, we are dealing with a mixture of a weak base (ammonia) and a strong acid (hydrochloric acid). When weak bases react with strong acids, the weak base is considered to be completely consumed in the reaction. This means that all of the ammonia will be converted to ammonium ions (NH4+). So, we can equate the initial concentration of ammonia to the final concentration of ammonium ions, which will be equal to the concentration of the hydrochloric acid solution. This is why we can set [NH3] = [NH4+] = 0.0500 mol dm-3. From there, we can use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution.

For Question B, we are dealing with a mixture of a weak acid (ethanoic acid) and a strong base (sodium hydroxide). In this case, the weak acid will not be completely consumed in the reaction, so we cannot equate the initial concentration of ethanoic acid to the final concentration of acetate ions (CH3COO-). The initial concentration of ethanoic acid will remain the same, while the concentration of acetate ions will increase due to the reaction with sodium hydroxide. This is why we cannot simply set [CH3COOH] = [CH3COO-] in this case.

I hope this helps to clarify the logic behind these calculations. It's important to remember the differences between strong and weak acids and bases, and how they behave in reactions with each other. Keep practicing and asking questions, and you will continue to improve your understanding of these concepts. Good luck with your studies!