Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: mhameer84 on April 17, 2004, 04:35:33 PM
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Need help with this question:
The dissociation constant of an acid base indicator is 6.31x10^-6. Its 'acid color' is blue, and 'alkaline color' is yellow. Calculate pKa value of this indicator, and state what color it will be WHEN intorduced into aqueous sodium carbonate of pH 5?
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pKa = -lg[ 6.31X10-6 ] = 5.20
pH 5 is in the acidic range, the colour of the indicator will be blue.
However, aq sodium carbonate is alkali in nature. I'm in doubt that its pH will ever be 5.
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Hey thanx for the reply.
I believe this indicator is Methyl Red, which is a weak base. Therefore, shouldn't the dissociation constant be that of the weakbase. And so,
pKb = -log[6.32x10^(-6)]=5.20
pKa + pkb = 14
pKa = 14 - 5.2 = 8.8
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methyl orange indicator dont reflect blue and yellow. methyl orange is red (acid) or yellow (alkali) depending on the solution pH.
the indicator which your previous post describes suits the profile of bromothymol blue.