Question

How would you prepare $500.0mL$ of $0.2500M$ $NaOH$ solution starting from a concentration of $1.000M$?

Answer 1

Remember, when it comes to diluting of solutions the number of moles of the solute should remain constant at all the time. So, here simply consider, the number of moles of the solute present in the solution to be equal to the number of moles of solute present in the concentrated sample.

Complete step by step solution:
It is important to keep in mind that the number of moles of solute remains constant at all times when solutions are diluted. So, by putting the number of moles of solute present in the dilute solution must be equal to the number of the moles of the solute present in the concentrated sample, we can get the concentration or volume of the solution. And this is the key to the dilution calculations.
We know that, the formula for calculating molarity is
$\implies Molarity=\dfrac{\text{moles of solute}}{\text{volume(litres) of solution}}$
So, here molarity of the $NaOH$ solution is $0.2500M$.
Volume of the $NaOH$ solution is $500.0mL$ which is also represented as $0.5$ litre.
Therefore, the moles of the solute in $NaOH$ solution will be $(0.25\times 0.5)moles=0.125moles$.
Again, the concentration of the $NaOH$ solution is $1.000M$.
Keeping the number of moles of solute constant, the volume of the $NaOH$ solution will be
$Volume(in\text{ litres)}=\dfrac{\text{moles of solute}}{\text{molarity of the solution}}$
So, the volume will be $=\dfrac{0.125}{1}=0.125\text{ litre}=125mL$
This question can also be calculated using a simple formula of ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ where,
${{M}_{1}}$ is the initial concentration of the solution,
${{V}_{1}}$ is the initial volume of the solution in litres,
${{M}_{2}}$ is the final concentration of the solution and
${{V}_{2}}$ is the final volume of the solution.

Hence, the initial volume should be $125mL$ so that we can prepare $500.0mL$ of $0.2500M$ $NaOH$ solution.

Note: It is important to note that, when we dilute a solution, we tend to decrease the concentration of that solution by increasing its volume, which is done by adding more solvent to the solution. So, here in the question the concentration at beginning is more than that at later and the volume is found to be less than at later.