December 13, 2019, 07:39:31 AM
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### Topic: Heat of Neutralization  (Read 264 times)

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#### macmillanasia21

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• Mole Snacks: +0/-0 ##### Heat of Neutralization
« on: November 20, 2019, 12:08:33 AM »
I have a lab tomorrow and I'm looking over the pre-lab, and not fully understanding how to do this calculation. Here's the question and data.

The heat of neutralization is the heat of reaction when one mole of acid reacts with one mole of base to create one mole of salt and one mole of water. Use the following data to determine the heat of neutralization (in kj/mole) of HCl by NaOH

For this problem you may assume that the densities of the solutions of HCl, NaCl and NaOH are 1.000g/mL

Data:
Mass of 1.00M aq. HCl used = 100.0g
Mass of 1.00M aq. NaOH used= 100.g
Temp of solutions before mixing= 18.29 °C
Temp of solutions immediately upon mixing= 25.34 °C
Specific heat of 0.500M aq. NaCl= 3.955 J/gram.°C
Heat capacity of the calorimeter= 43.40 J/°C

Any help would be awesome

#### macmillanasia21

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• Mole Snacks: +0/-0 ##### Re: Heat of Neutralization
« Reply #1 on: November 20, 2019, 12:18:57 AM »
Just want to add that it is mainly the calorimeter factor that is confusing me
So far I've got

qneut+qsol+qcal=0
qneut+(200.0g)(3.955 J/g°C)(7.05°C) + Ccal(7.05°C)

So I'm needing to solve for qneut, but can't figure out where the calorimeter constant comes in (qcal=mcal,ccal,t change cal)

#### Borek ##### Re: Heat of Neutralization
« Reply #2 on: November 20, 2019, 03:08:48 AM »
can't figure out where the calorimeter constant comes in (qcal=mcal,ccal,t change cal)

It is equivalent to mcal×ccal (should be obvious once you look at the units).
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