Answer :
Answer: the concentration of the diluted solution is 0.125 mol/L.
Explanation:
The question requires us to calculate the final concentration of a diluted solution prepared by adding 25 mL of water to 125 mL of a 0.15 mol/L NaOH solution.
In a dilution process, the number of moles of a substance does not change, only the volume in which this amount of moles is. Thus, we can write:
[tex]n_{initial}=n_{final}[/tex]Considering the definition of molarity (also called molar concentration), we can write the following equation to determine the number of moles of a substance:
[tex]\begin{gathered} molarity=\frac{number\text{ of moles of solute \lparen mol\rparen}}{volume\text{ of solution \lparen L\rparen}}\rightarrow C=\frac{n}{V} \\ \\ n=C\times V \end{gathered}[/tex]Therefore, we can write the first expression as:
[tex]n_{\imaginaryI n\imaginaryI t\imaginaryI al}=n_{f\imaginaryI nal}\rightarrow C_{initial}\times V_{initial}=C_{final}\times V_{final}[/tex]Rearranging this expression, we can calculate the final concentration (Cfinal) that is required by the question.
Remember that the question provided us with the concentration of the initial solution (Cinitial = 0.15 mol/L), the volume taken from the initial solution (Vinitial = 125 mL) and we can calculate the volume of the final solution as a sum of the Vinitial and the volume of water used (Vfinal = 25 + 125 = 150 mL).
Therefore, applying these values to the equation above, we'll have:
[tex]\begin{gathered} \begin{equation*} C_{initial}\times V_{initial}=C_{final}\times V_{final} \end{equation*} \\ \\ 0.15mol/L\times125mL=C_{final}\times150mL \\ \\ C_{final}=\frac{(0.15mol/L)\times(125mL)}{(150mL)}=0.125mol/L \end{gathered}[/tex]Therefore, the concentration of the diluted solution is 0.125 mol/L.