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Homework Help: Enthalp of a neutralization reaction

  1. Mar 10, 2004 #1
    i'm doing a lab on enthalpy of a neutralization reaction, and am having a little trouble with some of the questions, here is the info....

    The neutralization of a hydrochloric acid with sodium hydroxide solution is represented by the following equation.

    HCL(aq) + NaOH(aq) ----> NaCl(aq) + water (H2O)

    Using a coffee-cup calorimeter you will determine the enthalpy change for this reaction.

    Procedure -

    1)Rinse the graduated cylinder with a small quantity of 1.00 mol/L NaOH(aq) Use the cylinder to add 50.0 ML of a 1.00 mol/L NaOH(aq) to the calorimeter. Record the initial temperature of the NaOH. ( this will also represent the initial temp of the HCl(aq)

    2) Rinse the graduated cylinder with tap water, then rinse it with a small quantity of 1.00 mol/L HCl to the NaOH in the calorimeter.

    3) cover the colorimeter and record the temp every 30s stirring gently and continuously.

    4) when the temp levels off record the final temp


    1) Determine the amount of heat that is absorbed by the solution in the calorimeter,

    2) use the following equation to determine the amount of heat that is released by the reaction :

    -Qreaction = Qsolution

    3) determine the number of moles of HCL and NaOH that were involved in the reaction,

    4)Explain what happens during a neutralization reaction. use equations in your answer. Was heat released or absorbed during the neutralization reaction? explain your answer.

    5)Use your results to determine the enthalpy change of the neutrailization reaction. in KJ/mol of NaOH. write the thermochemical equation for the neutralization reaction.


    1) When an acid gets on your skin why must you flush the area with plenty of water rather than neutralization the acid with a base?

    2) Supposed that you had added solid sodium hydroxide pellets to hydrochloric acid, instead of adding hydrochlodric acid to sodium hydrocide solution

    a) do you think you would have ibtained a different enthalpy change?
    b) would the enthalpy change have been higher or lower?
    c) what change od u need to make to the thermochemical equation if you perform the invertigation using solid sodium


    thx alot for anyone that can hep me out, i have a pretty good idea of what to do first the first couple, i'll tell later i g2g now thx
  2. jcsd
  3. Mar 10, 2004 #2


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    Nobody's going to do the whole homework for you. I'll just help you with some to get you started:

    a strong base is harmful to your skin. Also, think la chatelier's principle, refluxing with water would hinder an acid base reaction since water is the end product.

    Yes, think about the enthalpy process, in this case there is an extra step (with a few substeps), solvation has a few substeps to it (look in your chemistry book).

    You should know from above that it would be lower; work is needed for some of the substeps.

    deduce from above...look in your book for the solvation of substances (NaCl should be included).
  4. Mar 10, 2004 #3
    ya im not asking to do the WHOLE hw, i just need someone to verify my answers and helpme out with a couple., thx i just dident have time to post my answers here i go

    ok, For my observations i got:

    initial temperature was 25 degrees celcius

    Time Temperature
    30s 26 degrees
    60s 27 degrees
    90s 28 degrees

    the final temperature was 28 degrees

    for number one, since theres 2 50 ml solutions mixed, i used the dentsity of water to convert it to grams, which is a 100g, now using that and heat capaicty of water, i got:

    Q = mct
    Q = (100g)(4.185)(28-25)
    Q = 1255.2 Joules

    for number 2, im pretty sure i got it right, so no need to bother u guys about it,

    for number 3, i got 1.37 mol for HCl and 1.25 mol for NaOH, can someone confirm this, im pretty sure its right, i used n = m*mm

    ok now am really having trouble with #4 i got no idea where to start

    i got number 5, it took me a while but i think i got it

    and thats all the help i need thx for anyone that can bye
    Last edited: Mar 10, 2004
  5. Mar 10, 2004 #4


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    Nerp. 50cm3 of 1.0 M solution contains how many moles?
    Think "neutralization" --- neutral pH --- water --- HCl neutralizing NaOH, or NaOH neutralizing HCl.
    Not with the answer you came up with on (3).

    "Harmful?" Yeah, but this isn't the answer being sought --- the heat of reaction is sufficient to cause thermal burns in situations more concentrated than the experiment just completed. Le Chatelier hasn't got a thing to do with it.

    re. solid NaOH pellets ---

    Did you get to watch your instructor prepare the NaOH solution? If not, what you would have seen is addition of pellets to a partially filled flask, and the immersion of that flask in a water bath to keep it cool; there is an enormous exothermic heat of solution in preparation of NaOH solutions.
  6. Mar 10, 2004 #5
    damn, now am completly lost

    Nerp. 50cm3 of 1.0 M solution contains how many moles?

    i dont understand

    Think "neutralization" --- neutral pH --- water --- HCl neutralizing NaOH, or NaOH neutralizing HCl.

    could u expand on that?

    and no i dident get to see my instructor prepare it.
  7. Mar 10, 2004 #6


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    Step at a time. How much NaOH solution? What is the concentration of the solution? Concentration times volume is the amount of NaOH you added to the calorimeter. Same thing for the HCl.

    NaOH is a base. In solution it yields a high concentration of OH-. How do you "neutralize" a high concentration of hydroxide ion?

    Too bad you didn't get to watch --- people do regularly burn their fingers preparing NaOH or KOH solutions --- it's not something that gets emphasized nearly so much as heat of solution of sulfuric acid in water (always add acid to water), and people are forever picking up incredibly hot flasks of base solutions while they are in preparation.
  8. Mar 10, 2004 #7
    hmmm.. i c, ok well u'v beena great help, am gonna go work real hard on em for a couple hours, and see what a come up with the i'll post it back here, thx alot again
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