Answer
$pH = 3.096$
$Ka = 8.68\times 10^{- 6}M$
Work Step by Step
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:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 8.025\times 10^{- 4})^2}{ 0.075- 8.025\times 10^{- 4}}$
$Ka = \frac{ 6.44\times 10^{- 7}}{ 0.0742}$
$Ka = 8.68\times 10^{- 6}M$
