I'm doing a titration lab and writing a lab report where I'm sort of stuck on how exactly to find the concentration of HCl. These are the following info I have from the lab,
Equation:
HCl(aq)+NaOH(aq)-> H2O(l)+NaCl(aq)
-Calculated molarity of NaOH solution is .103M
-Volume of NaOH solution used was 15.95mL
-Volume of HCl used was 20.01mL
Now the question is what is the molarity of HCl in this case?
Knowing the molarity of NaOH, I converted from the volume of NaOH solution used to moles of NaOH.
(0.01595L NaOH)(0.103M NaOH)~0.00164mol NaOH
Since the balance equation is a 1 to 1 ratio between NaOH and HCl, does that mean I also have ~0.00164mol of HCl as well??
I assumed that that was the right answer and proceeded to calculate the molarity of HCl by (0.00164mol HCl)/(0.02001L HCl)~0.0821 [HCl]
Am I doing this correctly or am I missing an important step? Thanks in advance!
The titration equation is a lot easier to use (normality of both solutions is same as Molrity)
Macid*Vacid=Mbase*Vbase
Macid=Mbase*Vbase/Vacid
= .103M*15.9ml/20.01ml which is not exactly the same as you got.
It looks like it's the same answer, but how we got to it was different. With your equation, that's what I would take on for this lab, but my chemistry teacher warned us a million times to NOT do it that way, don't know why, but she wanted us to do it the way I described above, more confusing in a way, but was just double checking if I'm doing it correctly or not, thanks
Your way is correct but I would suggest you not use the approximations but actual calculated values and keep the number of significant figures correct.
Bob Pursley is absolutely correct, also. The main reason today's teachers don't use that method is because normality is not taught in the classroom anymore. Mr. Pursley and I agree that's a huge mistake; however, normality and chemical factors (along with a few other concepts) have been abandoned. Pity. But progress? moves along. We "old" guys are about the only ones now that know how to use those abandoned concepts.
Yes, you are on the right track and your calculations are correct! Let's walk through the steps to find the molarity of HCl.
Step 1: Calculate the number of moles of NaOH used.
To find the moles of NaOH, you multiply the volume of NaOH solution used by the molarity of NaOH:
(0.01595 L NaOH) × (0.103 M NaOH) = 0.00164 mol NaOH
Since the balanced equation shows a 1 to 1 ratio between NaOH and HCl, you are correct in assuming that you also have 0.00164 mol of HCl.
Step 2: Calculate the molarity of HCl.
To find the molarity of HCl, divide the moles of HCl by the volume of HCl used:
(0.00164 mol HCl) / (0.02001 L HCl) = 0.0821 M HCl
Based on your calculations, the molarity of HCl in this case is 0.0821 M.
It seems like you have followed the correct steps and got the right answer. Just make sure to double-check your calculations and units to ensure accuracy. Remember to record your final answer with the appropriate number of significant figures.
Keep up the good work with your lab report! If you have any more questions, feel free to ask.