Question

Context: An experiment investigated the reaction between crystal violet (a purple colored dye) and hydroxide ion using spectrophotometry. The collected data was then analayzed to determine the concentration dependence of the speed of the reaction which was used to establish the rate law of the reaction which was used to determine the value of its rate constant. Since the reaction of interest involved two charged species, an aqueous solution of NaCl was used alongside an aqueous solution of sodium hydroxide to enforce the same electrical nature of the reaction medium in the determination of reaction order. Since NaCl is not involved in the reaction, it is called an intert electrolyte used to maintain the same electrical nature of the medium when [OH^-] is varied.

Question: What was the role of sodium chloride in the experiment and what kind of problem(s) would be encountered if sodium chloride was left out?

Answer 1

Part-1: Role of NaCl in the given experiment:

Since NaCl is not involved in the reaction, it is called an inert electrolyte. But it is used to maintain the same electrical nature of the medium when the concentration of OH^- (i.e. [OH^-]) is varied.

Part-2:

The rate of the crystal violet/NaOH reaction can be expressed as follows.

Rate = k [OH^-]^m [CV]ⁿ ...................... Equation 1

Where k is the rate constant of the reaction, CV is an abbreviation for crystal violet, C₂₅ H₃₀ N₃⁺, m is the reaction order with respect to OH^-, and n is the reaction order with respect to CV. The values of m and n will be determined experimentally. Possible m values are 1 or 2 (first order or second order). Possible n values are also 1 or 2.

In this experiment, the initial [OH^-] is made much greater than the initial [CV]. Thus, the [OH^-] change, during the time that the CV is consumed, is negligible. For this reason, [OH^-]^m can be treated as a constant and Equation 1 can be rewritten as follows.

Rate = k' [CV]ⁿ ............... Equation 2

Where k' = k [OH^-]^m, which is termed a pseudo rate constant.

Hence, there would be no effect if sodium chloride was left out.