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marin [14]
3 years ago
5

When 2.0 x 10-2 mole of nicotinic acid (amonoprotic

Chemistry
1 answer:
gogolik [260]3 years ago
5 0

Answer:

Ka of nicotinic acid = 1.41 \times 10^{-5}

Explanation:

pH = 3.05

pH = -log [H^+]

H^+ = (10)^{-3.05}=0.00089125 M

No. of mol of nicotinic acid = 2.0 \times 10^{-2}

Volume of water = 350 mL = 0.0350 L

Molarity = \frac{Moles}{Volume\ in\ L}

Molarity = \frac{2.0 \times 10^{-2}}{0.350} = 0.05714\ M

Nicotinic acid dissoctates as:

HA \rightarrow H^+ + A^-

[H+] = 0.00089125 M

[A-] = 0.00089125 M

[HA} at equilibium = 0.05714 - 0.00089125 = 0.05624875 M

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

Ka = \frac{(0.00089125)^2}{0.05624875} = 1.41 \times 10^{-5}

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