2. An Indicator Dye Crystal violet is a base (pK, 13.05) that is sometimes used
ID: 1032273 • Letter: 2
Question
2. An Indicator Dye Crystal violet is a base (pK, 13.05) that is sometimes used as an indicator dye. In strongly basic solutions, the indicator is blue, wheréas in strongly acidic solutions, it is yellow. (a) Which form of the indicator dye, protonated or unprotonated, is blue? Explain. (b) What is the nominal pH of the yellow-to-blue transition? (c) Since yellow is a much paler color than blue, suppose that you can detect a color change when 2.0% (or more) of the dye is in its blue form. As you add base to an acid solution, at what pH will you just begin to see a color change?Explanation / Answer
a) Since the dye itself is a base and given that the acidic conditions make it yellow, it can be deduced that the dye in is prisitne unprotonated form is blue in colour. So adding a base does not cause any change in its state and it remains blue. However when an acid is added, it gets protonated and that protonated form is yellow.
b) From the pKb of the dye as 13.05, which is the equilibrium constant between the yellow and blue forms of the dye, we can say that exactly at a pH = pKb, there will be a 50% population of the yellow and blue forms in the dye. Thus a transition from yellow to blue can be expected to occur when the pH goes greater than 13.
c) For 2% of the dye to be in its blue form it required 98% of it to be protonated implying roughly a very acidic pH close to and slightly greater than the pKb of the dye.
A 98% protonation means the ratio of concentrations of protonated dye to unprotonated dye is 0.98.
Since crystal violet is a monobasic base, the equilibrium expression can be written using B for base as Kb = [OH-][HB+]/[B] where HB+ is the protonated form of the dye. pH is related to pOH as pH = 14 - pOH.
Now that we determined the percentage dissociation of the dye, we can say that [HB+] and hence [OH-] (1:1 molar ratio) is 0.98x when [B] is x.
This gives pKb = 10-13.05 = 8.9125x10-14 and Kb = 13.05 = (0.98x)2/x giving 13.05 = 0.9604x.
This gives x = 13.5881 resulting in [OH-] = 13.3163 which is the pH at which one can begin to see the colour change.
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