Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL
Explanation:
In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.
Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.
Molarity (<em>M)</em> is calculated from the following equation:
<em>M</em> = <em>n</em> ÷ <em>V</em>
where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.
Accordingly, the number of moles of the solute is given by
<em>n</em> = <em>M</em> x <em>V</em>
Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:
<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)
<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)
As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,
<em>n</em> (1) = <em>n</em> (2)
and
<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)
For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.
The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:
<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)
<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M
<em>V </em>(1) = 26.16 mL
The volume of the 1.224 M NaOH solution needed is 26.16 mL.
