Question

Will someone help me, dilute solutions are commonly made from more concentrated solutions. How many mL of a 12 M HCl solution would be required to prepare 500 mL of a .25 M HCl solution?

Answer 1

@pooja195 @zepdrix

Answer 2

@Luigi0210

Answer 3

For these questions there are two ways.
You could calculate the dilution factor or use a formula.

Dilution factor
12 M original solution becomes diluted to 0.25 M
Dilution factor = 12/0.25 = 48.
This means the new solution is 48 times more dilute than the original.
We could achieve this my pouring 1 part of the 12 M solution into 47 parts of the 0.25 M solution by volume.

If the final volume is 500 mL, 500mL/48 parts = 10.417 mL = 10.4 mL
So you would add this amount of 12 M solution and fill with water to 500 mL

Answer 4

The formula method.
The number of moles of solute is conserved from your original volume and when you dilute it (since you only add water to what you started with). We write this as

\[n = c_{old}V_{old} = c_{new}V_{new}\] or c1v1 = c2v2
Remember to use mol/L for concentrations and L for the your volume.

Sub in a find the unknown required
\[12 M \times V = 0.25 M \times 0.5 L \rightarrow V = \frac{ 0.25 M \times 0.5L }{ 12} = \frac{ 0.125 }{ 12 } = 10.4 mL\]

Answer 5

Thank you so much @mww