Preparing Acidic solutions.

$\text { Molarity}= \frac{n \text{ mols of solute}}{V \text{ mL of solution}}$

Using the above equation and the given information, in order to find the number of mL needed of 12.0 HCl, we must first find the number of mols, $n$ .

$\text { V mL}= \frac{n \text{ mols of solute}}{12.0 \text{ M}}$

To find the # mols of HCl:

It is given that, for the prepared solution…

$0.23 \text{ M} = \frac{n \text{ mols}}{250 \text{ mL}}$

So, solving for $n$ , we get…

$n=(0.23 \text{ M})(250 \text{ mL}) \rightarrow n = 57.5 \text{ mols}$

So now we have…

$\text { ? mL}= \frac{57.5 \text{ mols}}{12.0 \text{ M}} \rightarrow V=4.79 \text{ mL}$

We need $0.25\text{ L} \times \frac{0.23\text{ mol}}{1\text{ L}} = 0.0575\text{ mol}$ of HCL. Thus we must use $0.0575\text{ mol}\times \frac{1\text{ L}}{12\text{ mol}} \approx 0.00479\text{ L} = \boxed{4.79\text{ mL}}$ of our original solution.

C1V1 = C2V2

C1 = 12.0 M

C2 = 0.23 M

V2 = 250.0 mL = 0.250 L

V1 = (0.23 M)(0.250) /12.0 = 4.8 mL