Question

What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?

Answer 1

`V_1=0.0714L`

Step-by-step explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

`n_1=n_2`

And in terms of volume and concentration:

`C_1V_1=C_2V_2`

Thus, we can solve for the volume of the concentrated HCl as shown below:

`V_1=(C_2V_2)/(C_1)`

Therefore, we plug in the data to get:

`V_1=(0.250M*0.500L)/(1.75 M)\\\\V_1=0.0714L`

Best regards!