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vekshin1
3 years ago
14

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)

Chemistry
1 answer:
rewona [7]3 years ago
5 0

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}

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