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marin [14]
2 years ago
5

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

Chemistry
1 answer:
gogolik [260]2 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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Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

\tt \dfrac{18.24}{12}=1.52

mol H(MW=1 g/mol) :

\tt \dfrac{0.51}{1}=0.51

mol F(MW=19 g/mol)

\tt \dfrac{16.91}{19}=0.89

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

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