"A student started a titration by placing 20.0 mL of 0.24 M NaOH into a flask with 3 drops of phenolphthalein. The student then titrated to the endpoint using 31.0 mL of HCl. Calculate the concentration (molarity) of the HCl that the student used. Show all work for credit."

The answer to your question is 0.15 M

Explanation:

Data

volume of NaOH = 20 ml

[NaOH] = 0.24 M

Volume of HCl = 31 ml

[HCl] = ?

Process

1.- Write the balanced chemical reaction

                    NaOH + HCl  ⇒   NaCl + H₂O

2.- Calculate the moles of NaOH

Molarity = moles / volume (L)

- Solve for volume

moles = Molarity x volume

           = 0.24 x 0.02

           = 0.0048

3.- Use proportions to find the moles of HCl

        1 mol of NaOH 1 mol of HCl

      0.0048 moles of NaOH   x

         x = (0.0048 x 1)/1

         x = 0.0048 moles of HCl

4.- Calculate the concentration of HCl

Molarity = 0.0048 / 0.031

Molarity = 0.15        

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.