Question

In a titration of 25.0 ML of 0.40m NaOH is used to neutralize 10.0 ML of HCL. Calculate the molarity of the acid.

Answer 1

=> 1.0 M

Step-by-step explanation:

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

We have been provided with;

• 25.0 mL of 0.40 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;

0.40 M is contained in 1000 mL.

X mol is contained in 25.0 mL

`= (25 \: * \: 0.40)/(1000) \\ = 0.01 \: moles`

These 0.01 moles is contained in 10 mL of HCl.

To find the molarity of the acid;

0.01 moles is contained in 10 mL

x mol is contained in 1000 mL

`x \: mol \: = (0.01 * 1000)/(10) \\ = 1.0 \: mol`

= 1.0 M

Therefore the molarity of the acid (HCl) is 1.0 M