Thermometric titration

15

I've got this assessment later on this week, and I'm not entirely sure about the calcuations. I'm titrating nitric acid with 1M sodium hydroxide, then sulphuric acid with 1M sodium hydroxide, taking the temperature for every 5cm3 of acid added into NaOH. From a graph I will work out the maximum temperature rise, and the volume at which the maximum temperature occurs.

From there I can work out the concentration of acids, but I don't really understand how to work out the enthalpy change of neutralisation. Also, I have to be able to work out the enthalpy change per mole of water as well as per mole of acid produced - how do I go about doing that?

I would really appreciate any help!

Reply 1

Eau

15

Well you can work out the energy evolved from those reactions using

E = mc∆T, where m = mass of acids + base, c = spec heat cap for water, T = temperature

Then ∆H(neut) = E/(amount of water evolved from reaction)

Feel free to ask further questions.

Reply 2

Excalibur

OP

15

Thanks for the quick reply; when you say amount of water evolved, do you mean the moles of water? That's when you work out the enthalpy change per mole of water, right? What would you do when it's per mole of acid?

Perhaps it'll be easier if I used actual numbers? Let's say I used HCl + NaOH --> NaCl + H2O. I used 50cm3 of 1M NaOH, then 23cm3 of unknown HCl was needed for neutralisation. The maximum temperature rise was 9.1ºC. So to work out the enthalpy change per mole of water, assuming that density of the solution is 1gcm-3, it'll be:

[(50+23)÷1000] x 4.18 x 9.1 = 2.78 KJmol-1
Moles of NaOH used = 1 x 0.05 = 0.05 mol = moles of water
Enthalpy change per mole of water = 2.78 ÷ 0.05 = 55.6 KJmol-1

Is that correct? What would I do if it was per mole of acid? Also what would be different if sulphuric acid was used instead of hydrochloric acid (something about it being dibasic and monobasic)?

Thanks!

Reply 3

Eau

15

Excalibur

Thanks for the quick reply; when you say amount of water evolved, do you mean the moles of water? That's when you work out the enthalpy change per mole of water, right? What would you do when it's per mole of acid?

Perhaps it'll be easier if I used actual numbers? Let's say I used HCl + NaOH --> NaCl + H2O. I used 50cm3 of 1M NaOH, then 23cm3 of unknown HCl was needed for neutralisation. The maximum temperature rise was 9.1ºC. So to work out the enthalpy change per mole of water, assuming that density of the solution is 1gcm-3, it'll be:

[(50+23)÷1000] x 4.18 x 9.1 = 2.78 KJmol-1
Moles of NaOH used = 1 x 0.05 = 0.05 mol = moles of water
Enthalpy change per mole of water = 2.78 ÷ 0.05 = 55.6 KJmol-1

Is that correct? What would I do if it was per mole of acid? Also what would be different if sulphuric acid was used instead of hydrochloric acid (something about it being dibasic and monobasic)?

Thanks!

Yep that's correct.

Remember that when working out the enthalpy of neutralisation of a reaction, it is always in terms of the number of moles of water produced.

In the case of H2SO4,

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

In that case, to work out the enthalpy of neut., I recommend working in terms of base used, because moles of base = moles of water.

Reply 4

Excalibur

OP

15

Thanks celeritas. But I just want to get this crystal clear - could you please go through what calculations I will have to do for,

1) when I use sulphuric acid instead of HCl
2) when I have to work out the enthalpy of neut per mole of acid used

I realise that workings in terms of base would make more sense, but I think we will be required to do it per moles of acid. Thank you!!

Reply 5

Eau

15

1. You work out the number of moles of base used, n. And then divide the E = mc∆T value by n. I'm sure.

2. As I said a million times, you don't have to work out the E value in terms of acid. You work it out in terms of the amount of water produced. Write out the equation of the neutralisation always.

Reply 6

goldenshades

19

Original post

by Excalibur

Thanks for the quick reply; when you say amount of water evolved, do you mean the moles of water? That's when you work out the enthalpy change per mole of water, right? What would you do when it's per mole of acid?

Perhaps it'll be easier if I used actual numbers? Let's say I used HCl + NaOH --> NaCl + H2O. I used 50cm3 of 1M NaOH, then 23cm3 of unknown HCl was needed for neutralisation. The maximum temperature rise was 9.1ºC. So to work out the enthalpy change per mole of water, assuming that density of the solution is 1gcm-3, it'll be:

[(50+23)÷1000] x 4.18 x 9.1 = 2.78 KJmol-1
Moles of NaOH used = 1 x 0.05 = 0.05 mol = moles of water
Enthalpy change per mole of water = 2.78 ÷ 0.05 = 55.6 KJmol-1

Is that correct? What would I do if it was per mole of acid? Also what would be different if sulphuric acid was used instead of hydrochloric acid (something about it being dibasic and monobasic)?

Thanks!

You just helped me out a great deal. Thank you soo much! :']

Thermometric titration

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