Jump to content
Sign in to follow this  

Help with Heat of Neutralization lab!

Recommended Posts

So we have a lab on the heat of neutralization of NaOH and HCl given:


mass of NaOH = 48.01 g

Volume of NaOH = 1000.00 mL

Amount of HCl = 83.3 mL of 12.0 M of HCl in 1000.00 mL solution


And we found the initial temperature, final temperature

Temp initial = 21.5 degrees C

Temp final = 25.0 degrees C


And we have to find the heat of neutralization using this equation:


ΔHθneutralization = (-specific heat capacity of liquid water x mass (acid solution + base solution) x change in temp) / mole of limiting base or acid


And I should be getting an enthalpy of around -58 -ish but I'm wayyy off. Help please!!  :help:



P.S. I'm not that great at writing chem labs... :/

Share this post

Link to post
Share on other sites

Well first I figured out what the limiting reactant was; base or acid.


Then I graphed the data we had collected from the lab for time v. temperature and using the initial and final temperatures, found the change in temperature.

(You find the final temperature by drawing a line from the highest data point back the y-axis)


Then I substituted all of my data into the equation:


∆H(neutralization) = [-specific heat capacity (H2O) x mass (acid solution + base solution) x âˆ†T (H2O)] / n(limiting acid or base)


and then convert from J to kJ


The only part that brought me down in this lab were uncertainties so include those and remember percent error.

Share this post

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Sign in to follow this  

  • Create New...