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

You should write the balanced equation for the reaction between HCl and NaOH.

HCl + NaOH ==> NaCl + H₂O ... balanced equation for neutralization reaction

You can calculate moles NaOH that you used from the volume (6.8333 mls ???) and the molarity of 0.7 moles/liter (0.7 M). NOTE: I'm going to use 6.83 mls for the calculation because I'm sure you didn't really measure the volume to 4 decimal places. Also, 0.7 M has only 1 significant figure, so we can't use more than that for the final answer.

moles NaOH used: 6.83 ml x 1 L / 1000 ml x 0.7 mol/L = 0.004781 mols NaOH

moles HCl present: 0.004781 mols NaOH x 1 mol HCl / mol NaOH = 0.004781 mols HCl (0.005 M)

Since the volume of HCl is 25 ml (0.025 L), we can now find the concentration of the HCl ...

0.004781 mols HCl / 0.025 L = 0.191 M = molarity of HCl (0.2 M)

The pH = -log[H+] and [H+] = 0.191 M since HCl --> H+ + Cl-

pH = -log 0.191 = 0.72 = pH