Question

Determine the volume of distilled water that should be added to 20.0 mL of 12.0 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution.

Answer 1

Answer:

$V_{H_2O}=460 mL$

Explanation:

Hello!

In this case, in a dilution process, water is added to a solute in order to decrease its concentration but increase the volume of the solution. It means that if we have 20.0 mL of a 12.0-M solution of HCl and we want a 0.500-M solution, we need to apply the following formula considering that the moles remain unchanged:

$M_1V_1=M_2V_2$

Thus, solving for the final volume is:

$V_2=\frac{M_1V_1}{M_2}$

So plugging in the values we obtain:

$V_2=\frac{12.0M*20.0mL}{0.500M}\\\\V_2=480 mL$

Now, since the initial volume of acid was 20.0 mL and the final volume is 480 mL, the added volume of distilled water is:

$V_{H_2O}=480-20=460 mL$

Best regards!