what is the molarity of unknown HCL from the known NaOH concentration.

Question

we did this titration lab and I got the following info. We used 6.8333 mL average of 0.7 molarity NaOH to neutralize 25.00 mL of HCL. what are the moles, molarity, ph the of the HCL in this lab? Anything that you may find will really help. Thank you.

Answer 1

From V1S1= V2S2

V1= volume of NaOH

S1= Strength of NaOH

V2= volume of HCl

Strength of HCl=S2=V1S1/V2

=6.8333×0.7/25

=0.191 N= 0.191 M

No of moles of NaOH= No. Of moles of HCl = 0.191×25/1000 = 0.00478

pH= -log[H+]

= - log(.191)

=O.72