## Chemistry: The Central Science (13th Edition)

Ka ($HN_3$): $1.9 \times 10^{-5}$ 1. Drawing the ICE table, we get: -$[H^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x$ For approximation, we consider: $[HA] = [HA]_{initial}$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H^+][A^-]}{ [HA]}$ $Ka = 1.9 \times 10^{- 5}= \frac{x * x}{ 0.4}$ $Ka = 1.9 \times 10^{- 5}= \frac{x^2}{ 0.4}$ $7.6 \times 10^{- 6} = x^2$ $x = 2.75 \times 10^{- 3}$ %Ionization : $\frac{ 2.75 \times 10^{- 3}}{0.4} \times 100\% = 0.689\%$ Since ionization < 5%, the acid concentration have a right approximation.