Answer
$K_a (HA) = 5.56 \times 10^{-3}$
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.
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x = 0.45M$
2. Calculate the initial acid concentration:
$c(mol/L) = \frac{n(moles)}{v(L)} = \frac{1 mol}{2L} = 0.5 mol/L$
3. Now, we can find "x":
$[HA] = [HA]_{initial} - x$
$0.45M = 0.5M - x$
$0.45 - 0.5 = - x$
$-0.05 = -x$
$x = 0.05M$
So, $[H_3O^+] = [A^-] = x = 0.05M$
3. Write the Ka expression, and find its value:
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
$K_a = \frac{x * x}{0.45} = \frac{x^2}{0.45}$
$K_a = \frac{(0.05)^2}{0.45} = 5.56 \times 10^{-3}$
