IbTrojan Posted February 17, 2015 Report Share Posted February 17, 2015 So we have a lab on the heat of neutralization of NaOH and HCl given: mass of NaOH = 48.01 gVolume of NaOH = 1000.00 mLAmount of HCl = 83.3 mL of 12.0 M of HCl in 1000.00 mL solution And we found the initial temperature, final temperatureTemp initial = 21.5 degrees CTemp 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!! P.S. I'm not that great at writing chem labs... :/ Reply Link to post Share on other sites More sharing options...
IbTrojan Posted February 17, 2015 Author Report Share Posted February 17, 2015 Never mind, I've figured it out Reply Link to post Share on other sites More sharing options...
Mr. Purple Monkey Posted April 4, 2015 Report Share Posted April 4, 2015 Hey I am having the same problem as you (also with a neutralization reaction). How did you figure it out? Reply Link to post Share on other sites More sharing options...
IbTrojan Posted April 4, 2015 Author Report Share Posted April 4, 2015 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 ∆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. Reply Link to post Share on other sites More sharing options...
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