The indicator methyl red has a \(\mathrm{pK}_{\mathrm{HIn}}=4.95 . \mathrm{It}\) changes from red to yellow over the pH range from 4.4 to 6.2 (a) If the indicator is placed in a buffer solution of \(\mathrm{pH}=4.55,\) what percent of the indicator will be pre- sent in the acid form, HIn, and what percent will be present in the base or anion form, In \(^{-2}\) (b) Which form of the indicator has the "stronger" (that is, more visible) color- -the acid (red) form or base (yellow) form? Explain.

Short Answer

Expert verified
a) 39.81% of the indicator will be in the anion form (In-) and 60.19% will be in the acid form (HIn). b) The acid form (red color) will be more visible.

Step by step solution

01

Calculate the ratio of the concentrations of base and acid form

Using the Henderson-Hasselbalch equation: \(pH = pKa + log(\frac{[In^-]}{[HIn]})\), the ratio of the concentrations of the base and acid forms can be determined. It comes out to be \(\frac{[In^-]}{[HIn]} = 10^{pH - pKa} = 10^{4.55 - 4.95} = 10^{-0.4}\).
02

Calculate the percentage of each form

The fraction of the form is given by the ratio of its concentration to the total concentration. Here, the percentage of In^- is given by \(\frac{[In^-]}{[HIn] + [In^-]} * 100\%\). Substitute the value from step 1 into this expression to yield the percentage: \(\frac{10^{-0.4}}{1 + 10^{-0.4}} * 100\% = 39.81\%\). The remaining (100%- 39.81% = 60.19%) will be in the acid form HIn.
03

Decide which form of the indicator has stronger color

Looking at the pH range where methyl red changes color (4.4 to 6.2), the indicator will exhibit an intense red color (in the acidic form) in solutions of pH less than 4.4 and an intense yellow color (in the base form) in solutions with pH greater than 6.2. At pH 4.55, the solution is closer to 4.4. This suggests that the acid form with red color will be more visible than the base form with yellow color at this pH.

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