## Chemistry 10th Edition

$pH = 3.096$ $Ka = 8.68\times 10^{- 6}M$
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x = 3.5 \times 10^{- 5} - x$ For approximation, we consider: $[Acid] = 3.5 \times 10^{- 5}M$ 2. Write the percent ionization equation, and find 'x': %ionization = $\frac{x}{[Initial Acid]} \times 100$ 1.07= $\frac{x}{ 0.075} \times 100$ 0.0107= $\frac{x}{ 0.075}$ $8.025\times 10^{- 4}= x = [H_3O^+]$ 3. Calculate the pH: $pH = -log[H_3O^+]$ $pH = -log( 8.025 \times 10^{- 4})$ $pH = 3.096$ 4. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{( 8.025\times 10^{- 4})^2}{ 0.075- 8.025\times 10^{- 4}}$ $Ka = \frac{ 6.44\times 10^{- 7}}{ 0.0742}$ $Ka = 8.68\times 10^{- 6}M$