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Topic: Enthalpy of Neutralization  (Read 4016 times)

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Offline Boxxxed

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Enthalpy of Neutralization
« on: October 28, 2010, 05:20:48 PM »
In a lab I've mixed 50 ml of each 1.00M/L of NaOH and 50 ml of 1.10 M/L of HCl.


To find qreaction do I use the specific heat of water?

Offline Borek

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Re: Enthalpy of Neutralization
« Reply #1 on: October 28, 2010, 05:27:42 PM »
Yes.

It would be better to use specific heat of NaCl solution of correct concentration, but difference is not that large.
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Offline Boxxxed

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Re: Enthalpy of Neutralization
« Reply #2 on: October 28, 2010, 05:40:28 PM »
Also..

I've already calculated the calorimeter constant using hot and cold water and the excess energy that the calorimeter absorbed.

They gave us the formula Ccal = qcal / DT

Why is there no mass?

To calculate qcal for the acid-base reaction would I use qcal = Ccal (from previous experiment) / DT?

I'm lost as to where the mass component went.


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Re: Enthalpy of Neutralization
« Reply #3 on: October 29, 2010, 03:26:09 AM »
Mass is necessary when dealing with specific heats, heat capacity already takes mass into account.
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Re: Enthalpy of Neutralization
« Reply #4 on: October 29, 2010, 01:22:24 PM »

Last related question

When they are asking for qwater am I calculating mass from moles of reactants and subtracting it from the final mass of the mixture to find mass of water then using q=mcDT? I find it odd that they aren't asking for q of the whole solution instead.

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Re: Enthalpy of Neutralization
« Reply #5 on: October 29, 2010, 02:49:03 PM »
We are asked to calculate qwater, qcal and qrxn for this acid-base reaction.

qrxn is heat produced by 0.0501 mol NaOH reacting with equal number of moles of HCl.

qcal is heat absorbed by calorimeter


What exactly is qwater?

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