[CHM 111] Calorimetry Lab

[Andrew J. Pounds]

On 6/12/21 10:20 AM,
> Good morning Dr. Pounds, 
> I’m confused about how I use the density of NaCl and water to find the
> mass of the solution. Do I add the densities of the products together
> then multiply by the mass of the reactants or is there a ratio I’m
> supposed to know? And when we find the Enthalpies the equations are
> balanced so does that mean the moles are 1?
> Thank you for your help
>
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So in the following experiments...

NaOH (aq) + HCl (aq) -> H2O (l) + NaCl (aq)

NaOH (s) + HCl (aq) -> H2O (l) + NaCl (aq)

You immediately form H2O and NaCl because sodium hydroxide and
hydrochloric acids are strong electrolytes.  To computer the mass of the
solution for the first reaction above you assume that both volumes (for
HCl and NaOH) are additive and use the density of the NaCl solution
shown on the report form.

For the second reaction you could make the same assumption that the
final solution will be the same NaCl as for the first experiment above
an assume that the volume of the NaOH pellets is negligible.  

So in both cases the mass of the water would simply be the total volume
of each solution multiplied by the density given in the report form.

The heat in each case would be q=mcsΔTq=m c_s \Delta T .  

However the ENTHALPY in each case is based of the heat and the moles of
NaOH consumed in the reaction.

For the reaction...

NaOH (s) -> Na+ + OH-

Just compute the moles of NaOH by dividing the mass of NaOH you used by
the molar mass of NaOH.  Then divide the heat by that number of moles to
get the enthalpy.

For the reaction

NaOH (s) + HCl (aq) -> H2O (l) + NaCl (aq)

You do the exact same thing.

For the reaction

NaOH (aq) + HCl (aq) -> H2O (l) + NaCl (aq)

You have to compute the moles of NaOH you used by multiplying the volume
of NaOH in liters by the molarity of the NaOH.   Then divide the heat by
that number of moles to get the enthalpy.

Remember -- enthalpy is an extensive property so we are having to base
our calculation off the amount we used.  If we were to use and entire
mole of NaOH, then you would use "1" in your calculations.




-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University,  Macon, GA 31207   (478) 301-5627

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