Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calculate the volume of the 1.475 M stock NaOH solution needed to prepare 250.0 mL of 0.1374 M dilute NaOH solution.

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

[tex] C_1*V_1=C_2*V_2[/tex]

Now we can identify the variables:

[tex]C_1=~1.475_M[/tex]

[tex]V_1=~?[/tex]

[tex]C_2=~0.1374_M[/tex]

[tex]V_2=~250.0~mL[/tex]

If we plug all the values into the equation:

[tex]1.475_M*V_1=0.1374_M*250.0~mL[/tex]

And we solve for [tex]V_1[/tex]:

[tex]V_1=\frac{0.1374_M*250.0~mL}{1.475_M}[/tex]

[tex]V_1=23.3~mL[/tex]

I hope it helps!