Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.
Explanation :
Step 1 : Find moles of crystal violet and NaOH.
The molarity formula is
![Molarity = \frac{mol}{L}](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7Bmol%7D%7BL%7D)
Molarity of crystal violet = ![3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}](https://tex.z-dn.net/?f=3.00%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20%5Cfrac%7Bmol%20%28CrystalViolet%29%7D%7BL%7D)
The volume of crystal violet solution is 18 mL which is 0.018 L.
Moles of crystal violet = ![3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}](https://tex.z-dn.net/?f=3.00%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%200.018%20%3D%205.4%20%5Ctimes%2010%5E%7B-5%7D)
Moles of crystal violet = 5.4 x 10⁻⁵
Moles of NaOH = ![Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}](https://tex.z-dn.net/?f=Molarity%20%5Ctimes%20L%20%3D%200.250%20%5Ctimes%200.00200%20%3D%205.00%20%5Ctimes%2010%5E%7B-4%7D)
Moles of NaOH = 5.00 x 10⁻⁴
Step 2 : Find total volume of the solution
The total volume of the solution after mixing NaOH and crystal violet is
0.018 L + 0.00200 = 0.020 L
Step 3 : Use molarity formula to find final concentrations
Molarity of crystal violet = ![\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}](https://tex.z-dn.net/?f=%5Cfrac%7Bmol%28CrystalViolet%29%7D%7BTotal%20Volume%28L%29%20%7D%20%3D%20%5Cfrac%7B5.4%20%5Ctimes%2010%5E%7B-5%7D%7D%7B0.020%7D%20%3D%202.7%20%5Ctimes%2010%5E%7B-3%7D)
Final concentration of CV = 0.0027 M
Molarity of NaOH= ![\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}](https://tex.z-dn.net/?f=%5Cfrac%7Bmol%28NaOH%29%7D%7BTotal%20Volume%28L%29%20%7D%20%3D%20%5Cfrac%7B5.00%20%5Ctimes%2010%5E%7B-4%7D%7D%7B0.020%7D%20%3D%200.025%20%5Ctimes%2010%5E%7B-3%7D)
NaOH is a strong base and dissociates completely as follows.
![NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)](https://tex.z-dn.net/?f=NaOH%20%28aq%29%20%5Crightarrow%20Na%5E%7B%2B%7D%20%28aq%29%20%2B%20OH%5E%7B-%7D%20%28aq%29)
The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.
Concentration of OH⁻ = 0.025 M