## Chemistry and Chemical Reactivity (9th Edition)

(a) $[H_3O^+] = 1.585 \times 10^{- 4}M$ (b) This acid is moderately weak.
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x$ 2. Calculate the hydronium concentration. $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.8}$ $[H_3O^+] = 1.585 \times 10^{- 4}M$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[InitialAcid] - x}$ $Ka = \frac{( 1.585\times 10^{- 4})^2}{ 0.0025- 1.585\times 10^{- 4}}$ $Ka = \frac{ 2.512\times 10^{- 8}}{ 2.342\times 10^{- 3}}$ $Ka = 1.073\times 10^{- 5}$ - Therefore, it is a moderately weak acid.