Question

a 0.0995 m naoh solution was used to titrate 15.0 ml of a hcl solution. if 33.46 ml of naoh solution were required, was is the molarity of the hcl solution?

Answer 1

The 33.46 ml of  0.0995 molar NaOH required to titrate 15 ml  of HCl solution to the equivalence point. Thus molarity of HCl is 0.22 molar.

What is equivalence point?

Equivalence point in a titration is the point at which the titrant completely reacted with the analyte in the perfect stoichiometric ratio. Equivalence point can be identified using an indicator color change.

If we have M1 molar solution in V1 volume titrated with M2 molar solution of V2 volume, then we can write it as:

M1V1 = M2V2.

Here, the volume and molarity of NaOH is 33.46 ml and 0.0995 molar. The volume of HCl is 15 ml  where the molarity of HCl in this titration is :

molarity of  of HCL  = (volume of NaOH × molarity) / volume of HCl

                            = (0.0995 × 33.46 ml) /15

                            =0.22 M

Therefore, the molarity of HCl required in this titration is  0.22 M

To find more on titration, refer here:

http://brainly.com/question/2728613

#SPJ1