Question

b. A solution is prepared by mixing 25.0 mls of 4.84 x 10% M crystal violet solution with 20.0 mLs of 0.10 M NaOH solution and diluted to a final volume of 100.0 mLs. Determine the limiting reagent. mol CV, 4.84 X10- M 5x 0.025 1.21 x 15 mol mol NaOH = 0.10 X 0.020 = 2.08103 mol Crystal violet is the limiting reagent c. After the reaction has gone to completion, calculate the concentration of all species in the reaction. (2.0 X10-3 - 1.21 X107) 0.1002 11.99x10-4M d. How does the concentration of sodium hydroxide change over the course of the tion! The concentration of NaOH decreases over the course of the reaction e. Explain your answer in question 2d in terms of the initial concentrations of reactants.

I'm not sure if my answer to part d is correct. If it's not then I need part d and part e answered. Thank you!

Answer 1

d.

NaOH reacts with crystal violet hence concentration of NaOH slightly decreases.

Again the solution is diluted so concentration of NaOH decreases sharply.

e.

Initial moles of NaOH = molarity × volume(L)

= 0.10× 20.0×10^-3 = 0.002.

Moles of crystal violet = 4.84×10^-6 × 25×10^-3 = 1.21×10^-7

Final moles of NaOH = initial moles of NaOH - moles of crystal violet

= 0.002 - 1.21×10^-7

= 0.00199.

Final volume of solution = 100 mL = 0.1 L

Now final concenteation of NaOH

= (moles/volume)

= (0.00199/0.1)

= 0.0199 M