Assume that the reaction of cv with oh- ions proceeds to completion, that is, the solution turns colorless. What percentage of oh- ions will remain at the end of the reaction if the initial crystal violet to sodium hydroxide mole ratio is 1:1 (reaction 1)? What is the initial ratio is 1:1000 (reaction 2)?
The Correct answer and Explanation is:
To analyze the reactions of crystal violet (CV) with hydroxide ions (OH⁻) and the resulting concentration of OH⁻ ions at the end of each reaction, we need to consider both the stoichiometry of the reactions and how the concentrations change as the reactions proceed to completion.
Reaction Overview
Reaction 1 (CV⁻ = 1:1): In this scenario, one mole of crystal violet reacts with one mole of hydroxide ions. The balanced equation for the reaction can be represented as:CV+OH−→colorless products\text{CV} + \text{OH}^- \rightarrow \text{colorless products}CV+OH−→colorless products
Given that the initial molar ratio of CV to OH⁻ is 1:1, if we start with xxx moles of CV, we also start with xxx moles of OH⁻. As the reaction proceeds to completion, all xxx moles of CV will react with xxx moles of OH⁻, resulting in:
- Remaining OH⁻ = Initial OH⁻ – Consumed OH⁻
- Remaining OH⁻ = x−x=0x – x = 0x−x=0
Thus, 0% of OH⁻ will remain at the end of Reaction 1.
Reaction 2 (CV⁻ = 1:1000): In this case, one mole of crystal violet reacts with 1000 moles of hydroxide ions. The reaction proceeds similarly:CV+1000 OH−→colorless products\text{CV} + 1000 \, \text{OH}^- \rightarrow \text{colorless products}CV+1000OH−→colorless products
Assuming we have 111 mole of CV, we start with 100010001000 moles of OH⁻. As the reaction progresses to completion, all 111 mole of CV will react with 111 mole of OH⁻, resulting in:
- Consumed OH⁻ = 111 mole
- Remaining OH⁻ = 1000−1=9991000 – 1 = 9991000−1=999 moles
To find the percentage of OH⁻ that remains, we use the formula:Percentage remaining=(Remaining OH−Initial OH−)×100\text{Percentage remaining} = \left(\frac{\text{Remaining OH}^-}{\text{Initial OH}^-}\right) \times 100Percentage remaining=(Initial OH−Remaining OH−)×100
So,Percentage remaining=(9991000)×100=99.9%\text{Percentage remaining} = \left(\frac{999}{1000}\right) \times 100 = 99.9\%Percentage remaining=(1000999)×100=99.9%
Conclusion
- For Reaction 1, 0% of OH⁻ remains.
- For Reaction 2, 99.9% of OH⁻ remains.
These outcomes illustrate how varying the molar ratios between reactants influences the concentrations of unreacted species after the reaction has reached completion. In Reaction 1, the stoichiometric balance leads to complete consumption of both reactants, while in Reaction 2, the excess OH⁻ results in a significant remaining concentration.