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Ostwald Dilution Law Mathematical Interpretation

Nicotinic aci...

Question

Nicotinic acid (Ka = 1.4 x $10^{- 5}$ ) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is:

4.2%

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12.77%

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1.66%

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3.33%

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Solution

The correct option is C
1.66%

Let the initial concentration of Nicotinic acid be C molL^-1
Let us write down the equation:

HNic $⇋$ ${N i c}^{-}$ + $H^{+}$

(1- $α$ )C $α C$ $α C$

$K a = \frac{[C α] [C α]}{C (1 - α)}$
$1 - α = 1$
$K a = C α^{2}$
$= 1.4 \times 10^{- 5} = 0.05 \times α^{2}$
$α^{2} = \frac{1.4 \times 10^{- 5}}{0.05}$
$= 0.28 \times 10^{- 3}$
$α = 0.01673$

Percentage dissociation=1.67 %

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Nicotinic acid (Ka = 1.4 x $10^{- 5}$ ) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is:

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Nicotinic acid (Ka = 1.4 x 10−5) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is:

The correct option is C 1.66% Let the initial concentration of Nicotinic acid be C molL-1 Let us write down the equation: HNic ⇋ Nic− + H+ (1- α)C αC αC Ka=[Cα][Cα]C(1−α) 1−α=1 Ka=Cα2 =1.4×10−5=0.05×α2 α2=1.4×10−50.05 =0.28×10−3 α=0.01673 Percentage dissociation=1.67 %

Nicotinic acid (Ka = 1.4 x $10^{- 5}$ ) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is: