Here's some helpful hints for the calculations:
Part I, Calorimetric analysis of an unknown HCl solution
We get DT from the first plot above - here, the temperature change is (41.6C-23.5C)=18.1C.
In this experiment, we use q = 57 321 J/mol and c=3.89 J/g oC to solve for the per cent HCl in the unknown. This is done as follows:
% HCl = [(mass HCl + mass NaOH, g)(3.89 J/g oC)(D T)(36.45 g/mol) / (mass HCl, g)(57 321 J/mol)]*100
Part II, Thermometric titration of an unknown HCL solution
Plot 2 above shows the data from Logger Pro. Use the plot to determine the volume of NaOH delivered at the equivalence point (it's 10.0 mL on fig 2 above.) You'll need to know the density and what per cent NaOH the solution is, so be sure to record these in lab. At the equivalence point, mol base = mol acid; to calculate how many mol of NaOH were delivered,
(volume NaOH, mL)(density NaOH, g/mL)(per cent NaOH)(1 mol NaOH / 40.0 g NaOH) = mol NaOH
since NaOH and HCl are 1:1 in the reaction, mol HCl = mol NaOH.
To get the per cent HCl, convert mol HCl to grams and divide by the mass of the HCl sample (you should have recorded this quantity in your notebook!)
Be sure that we can find your results (the calculated % HCl for both methods) and be sure to include your unknown number in your lab report!
This page was last updated on
Questions? Comments? Drop me a note...
This page is maintained by Doug Chapman