What is the molarity of an HCl solution if 125 mL is neutralized in a titration by 76.0 mL of 1.22 M KOH?

0.74176 M HCl

Explanation:

1) Write out a balanced equation.

  KOH + HCl --> KCl + H2O

2) Calculate #mol KOH in 76.0mL needed for the titration.

 1.22 mol KOH in 1000mL

 x mol KOH in 76.0mL

 x = 0.09272 mol KOH

3) Use the balanced equation ratio (from step 1), 1 mol KOH : 1 mol HCl.

0.09272 mol KOH = 0.09272 mol HCl

4) 125mL of the HCl solution was used for the titration, so find the molarity.

0.09272 mol HCl is in 125mL

x mol HCl in 1000mL

x = 0.74176 mol HCL / 1 L

SO... 0.74176 M HCl is the answer.

*hope this helped*

- A Kind Stranger

the answer to this question is 3.

2b2t

explanation:

2b2t