So I did a titration lab at school with NaOH and HCl. i have to find:
- volume used, NaOH
- Moles of NaOH
- Moles of HCl
-Volume used unknown HCl
- Molarity of HCl solution
My concentration of NaOH stock solution was 0.100 mol/lL
Final buret reading, NaOH: 14.20 ml
Initial buret reading, NaOH: 50 ml
What I did :
Volume used NaOH: 35.8 mL (I subtracted 50 mL from 14.20 mL.. Am I doing this right??)
Moles of NaOH: 3.58 mol
Because n= C xV
n= (0.100 mol/L)(35.8 mL)
HCl (aq) + NaOH (aq) = H2O (l) + NaCl (aq)
That's what I think the chemical equation would be. So it's already balanced. I would do:
Moles HCL over moles of NaOH= 1 mole over 1 mole...
So wouldn't HCl have the same number of moles??
How could I find the volume used for unknown HCl if I don't know the concentration or volume???
I really appreciate the help, it means a lot
Volume used NaOH: 35.8 mL (I subtracted 50 mL from 14.20 mL.. Am I doing this right??)
You did it right but said it wrong. Volume is 50-14.20 = 35.8 mL NaOH BUT you subtracted 14.20 from 50 not the other way around.
Moles of NaOH: 3.58 mol
Because n= C xV
n= (0.100 mol/L)(35.8 mL)
Exactly right.
HCl (aq) + NaOH (aq) = H2O (l) + NaCl (aq)
That's what I think the chemical equation would be. So it's already balanced. I would do:
Moles HCL over moles of NaOH= 1 mole over 1 mole...
So wouldn't HCl have the same number of moles??
Yes, mols HCl = mols NaOH = 3.58
How could I find the volume used for unknown HCl if I don't know the concentration or volume???
You should have the volume of the HCl if this were a titration in the lab. You probably started with a known quantity of HCl, measured either with a buret or some other device.
Then since M HCl = mols HC/L HCl. You know mols HCl from above and L HCl you started with, solve for M HCl. Look in your lab notes or think back. You HAD to have some HCl in the titration flask when you started titrating with the NaOH. The only way it can get there is for you to have added it OR for your instructor to have handed you a flask containing the HCl. If you added it you know how much you added. If the instructor gave it to you s/he must tell you how much is there.
To determine the volume used for NaOH, you subtract the initial buret reading from the final buret reading. So, in your case, it would be 50 mL - 14.20 mL = 35.80 mL (which you already calculated correctly).
To find the moles of NaOH, you can use the formula n = C x V, where n is the number of moles, C is the concentration in mol/L, and V is the volume in L. So, you correctly calculated the moles of NaOH as (0.100 mol/L) x (35.8 mL/1000 mL/L) = 0.00358 mol.
Since the balanced chemical equation shows a 1:1 ratio between HCl and NaOH, the moles of HCl would also be 0.00358 mol.
To find the volume used for the unknown HCl, you need to use the stoichiometry of the reaction. Since you know the ratio between the moles of NaOH and HCl is 1:1, you can assume that the volume of HCl used is the same as the volume of NaOH used. So, the volume used for the unknown HCl is 35.8 mL.
Finally, to calculate the molarity of the HCl solution, you need to use the formula Molarity (M) = moles/volume. In this case, since you have the moles (0.00358 mol) and the volume (35.8 mL), you just need to convert the volume to liters by dividing by 1000 mL/L. So, the molarity of the HCl solution is (0.00358 mol)/(35.8 mL/1000 mL/L) = 0.100 mol/L.
It seems like you have followed the correct steps and calculations, and your answers are indeed correct. Well done!