SATHEE: Equilibrium - Result Question 49

Equilibrium - Result Question 49

####51. The rapid change of $p H$ near the stoichiometric point of an acid-base titration is the basis of indicator detection. $p H$ of the solution is related to ratio of the concentrations of the conjugate acid ( $H I n)$ and base $(I n^{-})$ forms of the indicator by the expression

[2004]

(a) $\log \frac{[I n^{-}]}{[H I n]} = p K_{I n} - p H$

(b) $\log \frac{[H I n]}{[I n^{-}]} = p K_{I n} - p H$

(d) $\log \frac{[I n^{-}]}{[H I n]} = p H - p K_{In}$

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Solution:

(d) For an acid-base indicator

$H I n \to H^{+} + I n^{-}$

$∴ K_{I n} = \frac{[H^{+}] [I n^{-}]}{[H I n]}$ or $[H^{+}] = K_{I n} \times \frac{[H I n]}{[I n^{-}]}$

or $\log H^{+} = \log K_{In} + \log \frac{[H I n]}{[I n^{-}]}$

Taking negative on both sides

$\begin{bmatrix} ended with \end{matrix}$

or we can write $p H = p K_{In} + \log \frac{[I n^{-}]}{[H I n]}$

$or \log \frac{[I n^{-}]}{[H I n]} = p H - p K_{In}$

Equilibrium - Result Question 49

Solution:

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Equilibrium - Result Question 49

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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