JEE MainChemistryIonic Equilibrium
A 0.1 M solution of a weak acid HA ( K_a = 10⁻⁵ ) is titrated with a 0.1 M solution of NaOH . Four indicators A, B, C, and D are available, having ionization constants ( K_ In ) of 10⁻⁴ , 10⁻⁷ , 10⁻⁹ , and 10⁻¹¹ respectively. Given 5 = 0.7 , which indicator is the most suitable for detecting the end point of this titration?
Options
- AIndicator C
- BIndicator A
- CIndicator B
- DIndicator D
Correct answer
A. Indicator C
Step-by-step solution
At the equivalence point, equal volumes of 0.1 M HA and 0.1 M NaOH react to form the salt NaA . Since equal volumes are mixed, the total volume doubles, making the concentration of the salt C = 0.1 2 = 0.05 M . NaA is a salt of a weak acid and a strong base. The pH at the equivalence point is given by: pH = 7 + 1 2 (pK_a + C) We are given K_a = 10⁻⁵ , so pK_a = - (10⁻⁵) = 5 . The term C = (0.05) = (5 10⁻²) = 5 - 2 = 0.7 - 2 = -1.3 . Substituting these values into the formula: pH = 7 + 1 2 (5 - 1.3) = 7 + 1 2 (3.7)