Question

A.) Write the equation which describes the reaction between crystal violet and hydroxide ion.

B.) A solution is prepared by mixing 25.0 mLs of 4.84*10^-6 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.

C.) After the reaction has gone to completion, calculate the concentration of all species in the reaction.

D.) How does the concentration of sodium hydroxide change over the course of the reaction?

E.) Explain your answer in question D in terms of the intial concentrations of reactants.

Answer 1

ans)

from above data that

a)

Crystal Violet + NaOH ----> Colorless product + Na+

According to the stoichiometry of the reaction 1 mole of crystal violet reacts with 1 mole of NaOH

b)

Given,

The solution is prepared by mixing 25 ml of 4.84 x 10^-6 M crystal violet solution with 20 ml of a 0.10 M sodium hydroxide solution

Moles of Crystal Violet initially = molarity x volume (in L)

=> Moles = 4.84 x 10^-6 x 0.025 = 1.21 x 10^-7 moles

Moles of NaOH initially =0.10 x 0.020 = 0.002 moles

Since 1 mole of crystal violet reacts with 1 mole of NaOH

1.21 x 10^-7 moles of crystal violet will react with 1.21 x 10^-7 moles of NaOH. Rest NaOH will be left over.

We see that crystal violet reacts completely and NaOH is in excess.

Hence the limiting reagent is Crystal Violet

Moles of colorless product formed =1.21 x 10^-7 moles

Moles of NaOH left = 0.002 - 1.21 x 10^-7 = 0.002 moles(approx.)

Final Volume of solution = 100 mL = 0.1 L

=> [Colorless product] = 1.21 x 10^-7 / 0.1 = 1.21 x 10^-6 M

[NaOH] = 0.02 / 0.1 = 0.02 M

The conc. of NaOH at time (t) is C₀ exp (-kt)

where

C₀ = Initial conc. of NaOH

k = rate constant for the reaction