Question

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

Answer 1

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$