Answer to Question #186793 in Organic Chemistry for Eudoxia

Question

The results of a titration experiment were as follows:

 Volume of aqueous NaOH pipetted into the conical flask = 10.00 ml

 Average volume of dilute HCI added from the burette to reach the

endpoint = 12.54 mL.

 Molarity of the dilute HCI = 1.196 M

1- Calculate the number of moles of HCI added during the titration.

-What is the number of moles of NaOH that has reacted during the titration?

-What is the molarity of the aqueous NaOH?

Answer 1

$HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)$

Volume of HCl = 12.54-10.00=2.54ml

$Moles = 1.196×\frac{2.54}{1000}=3.04×10^{-3}Moles$

Moles of NaOH$= 3.04×10^{-3}Moles$

Molarity of NaoH$\frac{3.04×10^{-3}}{10.00×10^{-3}}=0.304M$