A flask containing 5.00mL of 3 M HCl solution requires 14.45mL of 1.00 M NaOH for titration. How many moles of HCl are present in the solution? The density of 3 M HCl is 1.05g/mL
2 answers
I rechecked this question, and I wrote it down correctly. I couldn't understand the process of finding the correct answer, so i would appreciate the help.
Unless I have really missed something, and I don't think I have, there is no correct answer. There is no unknown to determine. 5.00 mL of 3 M HCl CAN'T require 14.45 mL of 1.00 M NaOH. moles = M x L. For HCl that is 0.005 x 3.00 M = 0.015 moles.
NaOH titrated it.
0.01445 x 1.00 M NaOH = 0.01445 moles. So if you believe the HCl is labeled correctly the answer is 0.015 moles. If you believe the titration to be correct, there is 0.01445 mole. WAIT!
Could this be a problem with significant figures? 3 M x 5.00 = 0.015 mole but to only one significant figure (from the 3M) this would be rounded to 0.02 mole (or 0.01 depending upon the rule you follow).
14.45 mL x 1.00 M = 0.01445 and you are allowed three s.f. which would round to 0.0144 mole NaOH = 0.0144 mole HCl. The NaOH is more accurately measured; therefore, the NaOH titration would rule. Significant figures COULD be the problem in which case the answer is 0.0144 mole HCl.
NaOH titrated it.
0.01445 x 1.00 M NaOH = 0.01445 moles. So if you believe the HCl is labeled correctly the answer is 0.015 moles. If you believe the titration to be correct, there is 0.01445 mole. WAIT!
Could this be a problem with significant figures? 3 M x 5.00 = 0.015 mole but to only one significant figure (from the 3M) this would be rounded to 0.02 mole (or 0.01 depending upon the rule you follow).
14.45 mL x 1.00 M = 0.01445 and you are allowed three s.f. which would round to 0.0144 mole NaOH = 0.0144 mole HCl. The NaOH is more accurately measured; therefore, the NaOH titration would rule. Significant figures COULD be the problem in which case the answer is 0.0144 mole HCl.
1 answer