Q4(a). Absorbance of an unbuffered indicator solution at 430 nm
The equilibrium constant for the given conjugate acid–base pair is 2.5 × 10⁻⁵: HIn + H₂O ⇌ In⁻ + H₃O⁺. Determine the absorbance at 430 nm for the concentration of HIn = 3.0 × 10⁻⁴ M (up to 3 decimal places), using the following data: HIn, absorption maximum 430 nm, molar absorptivity 4.0 × 10³ at 430 nm and 7.25 × 10² at 600 nm; In⁻, absorption maximum 600 nm, molar absorptivity 3.0 × 10² at 430 nm and 9.65 × 10³ at 600 nm. Given: path length of the cuvette = 1 cm; molar absorptivity in M⁻¹ cm⁻¹.
Answer in brief
A₄₃₀ = 0.923 (exact 0.9225). Without a buffer 25 % of the indicator is ionised: [HIn] = 2.25 × 10⁻⁴ M and [In⁻] = 7.5 × 10⁻⁵ M.
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CoversBeer's lawadditivity of absorbanceindicator absorbancechemical deviationmolar absorptivity
All parts of Question 4
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