Question

In a titration of 12.5 ML of 0.20m NaOH is used to neutralize 10.0 ML of HCL. Calculate the acid concentration

Answer 1

=> 0.25 M

The question is asking us to find the molarity/concentration of the acid (HCl).

We have been provided with;

12.50 mL of 0.20 M NaOH

10.0 mL of HCl

We know that molarity (M) of a solution is contained in 1 L or 1000 mL or 1000 cm³.

This means that;

0.20 M is contained in 1000 mL.

X mol is contained in 12.50 mL

`x \: mol \: = (12.50 * 0.20)/(1000) \\ = 0.0025 \: moles`

= 0.0025 moles

These 0.0025 moles is contained in 10 mL of HCl.

To find the molarity of the acid;

0.0025 moles is contained in 10 mL

x mol is contained in 1000 mL

`x \: mol \: = (0.0025 * 1000)/(10) \\ = 0.25 \: mol`

= 0.25 M

Therefore the Concentration of the acid (HCl) is 0.25 M