## Chemistry 9th Edition

Published by Cengage Learning

# Chapter 15 - Acid-Base Equilibria - Exercises - Page 753: 71

Phenolphthalein

#### Work Step by Step

$\dfrac{0.5g}{204.22g/mol} = 2.45 \cdot 10^{-3}$ mol $/dfrac{2.45 \cdot 10^{-3} mol}{0.100L} = 2.45 \cdot 10^{-2}$ M $P^{2-}$ $K_b = \dfrac{1.0 \cdot 10^{-14}}{10^{-5.51}} = 3.24 \cdot 10^{-9}$ $K_b = \dfrac{[HP^-][OH^-]}{[P^{2-}]}$ $3.24 \cdot 10^{-9} = \dfrac {(x){x}}{2.45 \cdot 10^{-2}}$ $7.92 \cdot 10^{-11} = x^2$ $x = [OH^-] = 8.90 \cdot 10^{-6}$ M $OH^-$ $pOH = -log(8.90 \cdot 10^{-6}) = 5.051$ pH = 14 - pOH pH = 8.949 pKa of Phenolphthalein = 9.3 pH at the equivalence point = 8.9 Phenolphthalein is the best choice.

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