## General Chemistry: Principles and Modern Applications (10th Edition)

1. Convert the pKa into Ka: $Ka = 10^{-pKa} = 10^{-4.89} = 1.29 \times 10^{-5}$ 2. Drawing the ICE table we get: -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x$ For approximation, we consider: $[HA] = [HA]_{initial}$ 3. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 1.29 \times 10^{- 5}= \frac{x * x}{ 4.5 \times 10^{- 1}}$ $Ka = 1.29 \times 10^{- 5}= \frac{x^2}{ 4.5 \times 10^{- 1}}$ $5.8 \times 10^{- 6} = x^2$ $x = 2.4 \times 10^{- 3}$ 4. Find the degree of ionization: Degree of ionization: $\frac{ 2.4 \times 10^{- 3}}{ 4.5 \times 10^{- 1}} = 0.00533$ 5. Calculate the percent ionization. 0.00533 * 100% = 0.533%