M.Sc.III Sem Physical Chemistry Practical 1

Determination of pKa of an indicator (e.g., methyl red)

(a) in aqueous medium

(b) in micellar medium

Aim

To determine the acid dissociation constant (pKa) of an acid–base indicator (methyl red) using UV–Visible spectroscopy in aqueous solution and in the presence of micelles.

Determination of pKa of Methyl Red

(With Numerical Example & Calculation)

Given (Example Data)

Indicator: Methyl Red
Selected wavelength (λmax): 520 nm

Absorbance values:

Absorbance of acidic form (A_acid) = 0.80
Absorbance of basic form (A_base) = 0.20

Observation Table (Example)

pH	Absorbance (A)
4.2	0.65
4.6	0.55
5.0	0.50
5.4	0.40
5.8	0.30

Formula Used

$\boxed{\text{pKa} = \text{pH} – \log\left(\frac{A – A_{\text{acid}}}{A_{\text{base}} – A}\right)}$

Sample Calculation (at pH = 5.0)

Given:

pH = 5.0
A = 0.50
A_acid = 0.80
A_base = 0.20

Step 1: Substitute values

$\text{pKa} = 5.0 – \log\left(\frac{0.50 – 0.80}{0.20 – 0.50}\right)$

Step 2: Solve numerator & denominator

$\frac{-0.30}{-0.30} = 1$

Step 3: Log value

$\log(1) = 0$

Step 4: Final pKa

$\boxed{\text{pKa} = 5.0}$

Graph Method (Optional but Exam-Friendly)

Plot pH vs $\log\left(\frac{A – A_{\text{acid}}}{A_{\text{base}} – A}\right)$
The point where $\log(\dots) = 0$ $\boxed{\text{pH} = \text{pKa}}$

Micellar Medium (Short Example)

Suppose same experiment in CTAB micellar medium gives:

pKa = 5.6

Reason:

Micelles stabilize one ionic form
Change micro-polarity
Hence pKa shifts

Final Result (Example)

pKa in aqueous medium = 5.0
pKa in micellar medium = 5.6

Conclusion (Exam Ready)

The pKa of methyl red was successfully determined using UV–Visible spectroscopy. The presence of micelles causes a shift in pKa due to microenvironmental effects.